Search Results

Search found 40248 results on 1610 pages for 'php mysql'.

Page 218/1610 | < Previous Page | 214 215 216 217 218 219 220 221 222 223 224 225  | Next Page >

  • PHP error log does not display script names nor does it display the errors' line numbers [migrated]

    - by gnxtech3
    I think the title is self-explanatory, and my Google-fu isn't bringing up anything useful. I'm working on a new host, and my php error log only displays the error itself, not which script is the offender, nor which line number the error is occurring on. Makes it a tad difficult to debug, especially since there's only 1 error in the script. More info: I'm not using a custom error handler that I'm aware of. This is a standard Wordpress install. The error was [27-Aug-2012 19:22:36 UTC] PHP NOTICE: Trying to get property of non-object. Just no script name or line number in the error I found that Wordpress' error logging contained the information to debug the problem, but that doesn't explain why the log didn't contain line number or script.

    Read the article

  • showing null rows using join

    - by Pradyut Bhattacharya
    Hi, In mysql i m selecting from a table shouts having a foreign key to another table named "roleuser" with the matching column as user_id Now the user_id column in the shouts table for some rows is null (not actually null but with no inserts in mysql) How to show all the rows of the shouts table either with user_id null or not I m executing the sql statement SELECT s.*, r.firstname, r.lastname FROM shouts s left join roleuser r where r.user_id = s.user_id limit 50; which does not executes and shows You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'where r.user_id = s.user_id limit 50' at line 2 but using inner join the sql executes which shows rows which only have user_id values in the shouts table. the nulls are not shown. SELECT s.*, r.firstname, r.lastname FROM shouts s inner join roleuser r where r.user_id = s.user_id limit 50; How can i show all the rows from the shouts table and null values in the firstname and lastname columns where the user_id is null in the shouts table. If not at all possible with sql may be using stored procedures... Thanks Pradyut

    Read the article

  • Getting percentage of "Count(*)" to the number of all items in "GROUP BY"

    - by celalo
    Let's say I need to have the ratio of "number of items available from certain category" to "the the number of all items". Please consider a MySQL table like this: /* mysql> select * from Item; +----+------------+----------+ | ID | Department | Category | +----+------------+----------+ | 1 | Popular | Rock | | 2 | Classical | Opera | | 3 | Popular | Jazz | | 4 | Classical | Dance | | 5 | Classical | General | | 6 | Classical | Vocal | | 7 | Popular | Blues | | 8 | Popular | Jazz | | 9 | Popular | Country | | 10 | Popular | New Age | | 11 | Popular | New Age | | 12 | Classical | General | | 13 | Classical | Dance | | 14 | Classical | Opera | | 15 | Popular | Blues | | 16 | Popular | Blues | +----+------------+----------+ 16 rows in set (0.03 sec) mysql> SELECT Category, COUNT(*) AS Total -> FROM Item -> WHERE Department='Popular' -> GROUP BY Category; +----------+-------+ | Category | Total | +----------+-------+ | Blues | 3 | | Country | 1 | | Jazz | 2 | | New Age | 2 | | Rock | 1 | +----------+-------+ 5 rows in set (0.02 sec) */ What I need is basically a result set resembles this one: /* +----------+-------+-----------------------------+ | Category | Total | percentage to the all items | (Note that number of all available items is "9") +----------+-------+-----------------------------+ | Blues | 3 | 33 | (3/9)*100 | Country | 1 | 11 | (1/9)*100 | Jazz | 2 | 22 | (2/9)*100 | New Age | 2 | 22 | (2/9)*100 | Rock | 1 | 11 | (1/9)*100 +----------+-------+-----------------------------+ 5 rows in set (0.02 sec) */ How can I achieve such a result set in a single query? Thanks in advance.

    Read the article

  • What is optimal hardware configuration for heavy load LAMP application

    - by Piotr Kochanski
    I need to run Linux-Apache-PHP-MySQL application (Moodle e-learning platform) for a large number of concurrent users - I am aiming 5000 users. By concurrent I mean that 5000 people should be able to work with the application at the same time. "Work" means not only do database reads but writes as well. The application is not very typical, since it is doing a lot of inserts/updates on the database, so caching techniques are not helping to much. We are using InnoDB storage engine. In addition application is not written with performance in mind. For instance one Apache thread usually occupies about 30-50 MB of RAM. I would be greatful for information what hardware is needed to build scalable configuration that is able to handle this kind of load. We are using right now two HP DLG 380 with two 4 core processors which are able to handle much lower load (typically 300-500 concurrent users). Is it reasonable to invest in this kind of boxes and build cluster using them or is it better to go with some more high-end hardware? I am particularly curious how many and how powerful servers are needed (number of processors/cores, size of RAM) what network equipment should be used (what kind of switches, network cards) any other hardware, like particular disc storage solutions, etc, that are needed Another thing is how to put together everything, that is what is the most optimal architecture. Clustering with MySQL is rather hard (people are complaining about MySQL Cluster, even here on Stackoverflow).

    Read the article

  • PDO Database Connections Problem

    - by Metropolis
    Hey Everyone, Over a year ago I created my own database classes which use PDO, and handle all preparing, executing, and closing connections. These classes have been working great up until now. There are two different database severs I am grabbing from, MySQL, and MS SQL Express. I am retrieving an employee id from the MySQL server and using it to get that employees information from the MS SQL server. There are about 11k records coming from the MySQL server and my program is only making it through 1200 before crashing with an error like the following. Connection failed (odbc:Driver=FreeTDS;Servername=MSSQLExpress;Database=SMDINC) Class (PDOException) SQLSTATE[08001] SQLDriverConnect: 0 [unixODBC][FreeTDS][SQL Server]Unable to connect to data source It seems like the program is not able to connect to the data source, but it is running the exact same query about 30 times before this and having no problem. Also, I have thoroughly checked all of the data coming into the query and it all looks fine. I believe the issue may be that there are to many connections being created, but I have tried to close all connections in many different places, and nothing seems to be fixing the problem. Any debugging help, or suggestions would be appreciated! Craig Metrolis

    Read the article

  • How do I return a message if $_GET is null or does not match any database entries?

    - by CT
    I am working on designing an IT Asset database. Here I am working on a page used to view details on a specific asset determined by an asset id. Here I grab $id from $_GET["id"]; When $id is null, the page does not load. When $id does not match any entry within the database, the page loads but no asset table is printed. In both these cases, I would like to display a message like "There is no database entry for that Asset ID" How would this be handled? Thank you. <?php /* * View Asset * */ # include functions script include "functions.php"; $id = $_GET["id"]; ConnectDB(); $type = GetAssetType($id); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.1//EN" "http://www.w3.org/TR/xhtml11/DTD/xhtml11.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <link rel="stylesheet" type="text/css" href="style.css" /> <title>Wagman IT Asset</title> </head> <body> <div id="page"> <div id="header"> <img src="images/logo.png" /> </div> </div> <div id="content"> <div id="container"> <div id="main"> <div id="menu"> <ul> <table width="100%" border="0"> <tr> <td width="15%"></td> <td width="30%%"><li><a href="index.php">Search Assets</a></li></td> <td width="30%"><li><a href="addAsset.php">Add Asset</a></li></td> <td width="25%"></td> </tr> </table> </ul> </div> <div id="text"> <ul> <li> <h1>View Asset</h1> </li> </ul> <br /> <?php switch ($type){ case "Server": $result = QueryServer($id); $ServerArray = GetServerData($result); PrintServerTable($ServerArray); break; case "Desktop"; break; case "Laptop"; break; } ?> </div> </div> </div> <div class="clear"></div> <div id="footer" align="center"> <p>&nbsp;</p> </div> </div> <div id="tagline"> Wagman Construction - Bridging Generations since 1902 </div> </body> </html>

    Read the article

  • Submitting a URL into a Form without "http://", with "www.", or with neither

    - by John
    (EDITED) Hello, In the form below, the filed for <div class="urlfield"><input name="url" type="url" id="url" maxlength="500"></div> fine when a URL is submitted that has a "http://" at the beginning of it. However, it doesn't work if a URL is submitted with only a "www." in front of it, or with neither a "http://" nor a "www." How can I make it work in all if the submitted URL has any or none of the following at the beginning of it: http:// www. http://www. Thanks in advance, John Form: echo '<div class="submittitle">Submit an item.</div>'; echo '<form action="http://www...com/.../submit2.php" method="post"> <input type="hidden" value="'.$_SESSION['loginid'].'" name="uid"> <div class="submissiontitle"><label for="title">Story Title:</label></div> <div class="submissionfield"><input name="title" type="title" id="title" maxlength="1000"></div> <div class="urltitle"><label for="url">Link:</label></div> <div class="urlfield"><input name="url" type="url" id="url" maxlength="500"></div> <div class="submissionbutton"><input name="submit" type="submit" value="Submit"></div> </form> '; submit2.php: <?php if($_SERVER['REQUEST_METHOD'] == "POST"){header('Location: http://www...com/.../submit2.php');} require_once "header.php"; if (isLoggedIn() == true) { $remove_array = array('http://www.', 'http://', 'https://', 'https://www.', 'www.'); $cleanurl = str_replace($remove_array, "", $_POST['url']); $cleanurl = strtolower($cleanurl); $cleanurl = preg_replace('/\/$/','',$cleanurl); $cleanurl = stripslashes($cleanurl); $title = $_POST['title']; $uid = $_POST['uid']; $title = mysql_real_escape_string($title); $title = stripslashes($title); $cleanurl = mysql_real_escape_string($cleanurl); $site1 = 'http://' . $cleanurl; $displayurl = parse_url($site1, PHP_URL_HOST); function isURL($url1 = NULL) { if($url1==NULL) return false; $protocol = '(http://|https://)'; $allowed = '[-a-z0-9]{1,63}'; $regex = "^". $protocol . // must include the protocol '(' . $allowed . '\.)'. // 1 or several sub domains with a max of 63 chars '[a-z]' . '{2,6}'; // followed by a TLD if(eregi($regex, $url1)==true) return true; else return false; } if(isURL($site1)==true) mysql_query("INSERT INTO submission VALUES (NULL, '$uid', '$title', '$cleanurl', '$displayurl', NULL)"); else echo "<p class=\"topicu\">Not a valid URL.</p>\n"; } else { show_loginform(); } if (!isLoggedIn()) { if (isset($_POST['cmdlogin'])) { if (checkLogin($_POST['username'], $_POST['password'])) { show_userbox(); } else { echo "Incorrect Login information !"; show_loginform(); } } else { show_loginform(); } } else { show_userbox(); } require_once "footer.php"; ?>

    Read the article

  • Get the last checked checkboxes...

    - by Sara
    Hi everyone, I'm not sure how to accomplish this issue which has been confusing me for a few days. I have a form that updates a user record in MySQL when a checkbox is checked. Now, this is how my form does this: if (isset($_POST['Update'])) { $paymentr = $_POST['paymentr']; //put checkboxes array into variable $paymentr2 = implode(', ', $paymentr); //implode array for mysql $query = "UPDATE transactions SET paymentreceived=NULL"; $result = mysql_query($query); $query = "UPDATE transactions SET paymentdate='0000-00-00'"; $result = mysql_query($query); $query = "UPDATE transactions SET paymentreceived='Yes' WHERE id IN ($paymentr2)"; $result = mysql_query($query); $query = "UPDATE transactions SET paymentdate=NOW() WHERE id IN ($paymentr2)"; $result = mysql_query($query); foreach ($paymentr as $v) { //should collect last updated records and put them into variable for emailing. $query = "SELECT id, refid, affid FROM transactions WHERE id = '$v'"; $result = mysql_query($query) or die("Query Failed: ".mysql_errno()." - ".mysql_error()."<BR>\n$query<BR>\n"); $trans = mysql_fetch_array($result, MYSQL_ASSOC); $transactions .= '<br>User ID:'.$trans['id'].' -- '.$trans['refid'].' -- '.$trans['affid'].'<br>'; } } Unfortunately, it then updates ALL the user records with the latest date which is not what I want it to do. The alternative I thought of was, via Javascript, giving the checkbox a value that would be dynamically updated when the user selected it. Then, only THOSE checkboxes would be put into the array. Is this possible? Is there a better solution? I'm not even sure I could wrap my brain around how to do that WITH Javascript. Does the answer perhaps lie in how my mysql code is written? Thanks - I sincerely appreciate it!!!

    Read the article

  • Linux apache developing configuration

    - by Jeffrey Vandenborne
    Recenly reinstalled my system, and came to a point where I need apache and php. I've been searching a long time, but I can't figure out how to configure apache the best way for a developer computer. The plan is simple, I want to install apache 2 + mysql server so I can develop some php website. I don't want to install lamp though, just the apache2, php5 and mysql. The problem that I've been looking an answer for is the permissions on the /var/www/ folder. I've tried making it my folder using the chown command, followed by a chmod -R 755 /var/www. Most things work then, but fwrite for example won't work, because I need to give write permissions to everyone, unless I change my global umask to 000 I'm not sure what I can do. In short: I want to install apache2, php5, mysql-server without using lamp, but configured in a way so I can open up netbeans, start a project with root in /var/www/, and run every single function without permission faults. Does anyone have experiences or workarounds to this? Extra: OS: Ubuntu 10.04 ARCH: x86_64

    Read the article

  • Appending Comment Number Anchors to Comments

    - by John
    Hello, I am using a PHP file called comments.php that has a query that enters values into a mySQL table called "comment." As the query does this, it auto-generates a field called "commentid", which is set to auto_increment in MySQL. The file also contains a loop what echoes out all comments for a given submission. It all works fine and dandy, but I want to simultaneously pull this "commentid" and turn it into a hashtag / anchor that when appended to the end of the URL makes that comment at the top of the user's browser. Someone said on another question that in order to do this one thing I should do is create an anchor on the row where the comment is being printed out. How can I do this? Thanks in advance, John The query that inserts comments into the MySQL table "comment": $query = sprintf("INSERT INTO comment VALUES (NULL, %d, %d, '%s', NULL)", $uid, $subid, $comment); mysql_query($query) or die(mysql_error()); The fields in the table "comment": commentid loginid submissionid comment datecommented The row in a loop where the comments are echoed out: echo '<td rowspan="3" class="commentname1">'.stripslashes($row["comment"]).'</td>';

    Read the article

  • Process data BEFORE a 301 Redirect?

    - by Jesse
    So, I've been working on a PHP link shortener (I know, just what the world needs). Basically when the page loads, php determines where it needs to go and sends a 301 Header to redirect the browser, like so... Header( "HTTP/1.1 301 Moved Permanently" ); header("Location: http://newsite.com"; Now, I'm trying to add some tracking to my redirects and insert some custom analytics data into a MySQL table before the redirect happen. It works perfectly if I don't specify the a redirect type and just use: header("Location: http://newsite.com"; But, of course as soon as you add in the 301 header, nothing else gets processed. Actually, on the first request, it sends the data to MySQL, but on any subsequent requests there's no communication with the database. I assume it's a browser caching issue, once it's seen the 301 it decides they're no reason to parse anything on future requests. But, does anyone know if there's any way to get around this? I'd really like to keep it as a 301 for SEO purposes (I believe if you don't specify it sends a 404 by default?). I thought about using .htaccess to prepend a file to the page that will do the MySQL work, but with the 301, wouldn't that just get ignored as well? Anyway, I'm not sure if there's any solution other than using a different type of redirect, but I'm ready to give up just yet. So, any suggestions would be much appreciated. Thanks!

    Read the article

  • Password security; Is this safe?

    - by Camran
    I asked a question yesterday about password safety... I am new at security... I am using a mysql db, and need to store users passwords there. I have been told in answers that hashing and THEN saving the HASHED value of the password is the correct way of doing this. So basically I want to verify with you guys this is correct now. It is a classifieds website, and for each classified the user puts, he has to enter a password so that he/she can remove the classified using that password later on (when product is sold for example). In a file called "put_ad.php" I use the $_POST method to fetch the pass from a form. Then I hash it and put it into a mysql table. Then whenever the users wants to delete the ad, I check the entered password by hashing it and comparing the hashed value of the entered passw against the hashed value in the mysql db, right? BUT, what if I as an admin want to delete a classified, is there a method to "Unhash" the password easily? sha1 is used currently btw. some code is very much appreciated. Thanks

    Read the article

  • Why does PDO print my password when the connection fails?

    - by Joe Hopfgartner
    I have a simple website where I establish a connection to a Mysql server using PDO. $dbh = new PDO('mysql:host=localhost;dbname=DB;port=3306', 'USER', 'SECRET',array(PDO::MYSQL_ATTR_INIT_COMMAND => "SET NAMES utf8")); I had some traffic on my site and the servers connection limit was reached, and the website throw this error, with my PLAIN password in it! Fatal error: Uncaught exception 'PDOException' with message 'SQLSTATE[08004] [1040] Too many connections' in /home/premiumize-me/html/index.php:64 Stack trace: #0 /home/premiumize-me/html/index.php(64): PDO-__construct('mysql:host=loca...', 'USER', 'SECRET', Array) #1 {main} thrown in /home/premiumize-me/html/index.php on line 64 Ironically I switched to PDO for security reasons, this really shocked me. Because this exact error is something you can provoke very easily on most sites using simple http flooding. I now wrapped my conenction into a try/catch clause, but still. I think this is catastrophic! So I am new to PDO and my questino is: What do I have to consider to be safe! How to I establish a connection in a secure way? Are there other known security holes like this one that I have to be aware of?

    Read the article

  • Need method to seek next/previous records id without cycling through all records.

    - by dqhendricks
    I am using MySQL and PHP. I have a MySQL blog post result set with id fields, and publish_date fields. I display one blog post per page, and the script knows which blog post to display based on $_GET['id'], which correlates to each blog entry's id field. I would like to reference them by id in the url, because I would like each blog post to have a perminant url. I would like to order the blog posts by publish date (descending). Now, on each page there will be next and previous links, which contain the $_GET['id'] value for the next and previous blog posts. How can I figure out what the id of the next and previous blog posts (determined by it's publish_date order) without cycling through each mysql result row? I can't mysql_data_seek(), because I do not know the row index of the current blog post id. I do not want to store a row index in a GET variable because the urls would no longer be perminant. I obviously cannot store the row index in a SESSION variable because then direct links to specific blog posts would have broken next and previous links. Any suggestions would be greatly appreciated.

    Read the article

  • How to randomly assign a partner?

    - by David
    I asked a question some time ago about creating a random circular partner assignment using php and mysql. This is a related issue. I am working from the following code to try to give two users new, randomly selected partners: $q = "SELECT user_id FROM users WHERE partner='$quit_partner' AND status='1'"; $r = mysqli_query ($dbc, $q) or trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc)); while ($row = mysqli_fetch_array($r)) { $users[] = $row[0]; } $current = end($users); $partners = array(); foreach ($users as $user) { $partners[$user] = $current; $current = $user; $q = "UPDATE users SET partner='{$partners[$user]}' WHERE user_id='{$user}'"; mysqli_query ($dbc, $q) or trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc)); } Basically, a particular user (lets say user #4) quits the activity, leaving multiple other users without a partner (hypothetically, users # 5,6,7). I need to find out who those users are, hence the first query. Once I find them, I throw them into an array. Then comes the difficult part. I want those newly partnerless users (5,6,7) to be randomly assigned new partners from everyone in the table. The current code is flawed in that it only assigns the newly partnerless users eachother. Thanks for your help.

    Read the article

  • After mysql_query, no result output

    - by Jerry
    I have a simple mysql_query() update command to update mysql. When a user submits my form, it will jump to an update page to update the data. The problem is that there's supposed to be some data shown after the update, but it comes out blank. My form <form id="form1" method="POST" action="scheduleUpdate.php" > <select name=std1> <option>AA</option> <option>BB</option> <option>CC</option> </select> <select name=std2> <option>DD</option> <option>EE</option> <option>FF</option> </select> .......//more drop down menu but the name is std3..std4..etc... ....... </form> scheduleUpdate.php //$i is the value posted from my main app to tell me how many std we have for($k=0;$k<$i;$k++){ $std=$_POST['std'.$k]; //if i remove the updateQuery, the html will output.I know the query is the problem but i //couldn't fix it.. $updateQuery=mysql_query("UPDATE board SET student='$std' WHERE badStudent='$std' or goodStudent='$std'",$connection); //no output below this line at all if($updateQuery){ DIE('mysql Error:'+mysql_error()); } } // I have bunch of HTML here....but no output at all!!!! MySQL will be updated after I hit submit, but it doesn't shown any HTML.

    Read the article

  • Test a database conection without codeigniter throwing a fit, can it be done?

    - by RobertWHurst
    I'm just about finished my first release of automailer, a program I've been working on for a while now. I've just got to finish writing the installer. Its job is to rewrite the codigniter configs from templates. I've got the read/write stuff working, but I'd like to be able to test the server credentials given by the user without codingiter throwing a system error if they're wrong. Is there a function other than mysql_connect that I can use to test a connection that will return true or false and won't make codeigniter have a fit? This is what I have function _test_connection(){ if(mysql_connect($_POST['host'], $_POST['username'], $_POST['password'], TRUE)) return TRUE; else return FALSE; } Codigniter doesn't like this and throws a system error. <div style="border:1px solid #990000;padding-left:20px;margin:0 0 10px 0;"> <h4>A PHP Error was encountered</h4> <p>Severity: Warning</p> <p>Message: mysql_connect() [<a href='function.mysql-connect'>function.mysql-connect</a>]: Unknown MySQL server host 'x' (1)</p> <p>Filename: controllers/install.php</p> <p>Line Number: 57</p> </div> I'd rather not turn off error reporting.

    Read the article

  • multiple mysql_real_query() in while loop

    - by Steve
    It seems that when I have one mysql_real_query() function in a continuous while loop, the query will get executed OK. However, if multiple mysql_real_query() are inside the while loop, one right after the other. Depending on the query, sometimes neither the first query nor second query will execute properly. This seems like a threading issue to me. I'm wondering if the mysql c api has a way of dealing with this? Does anyone know how to deal with this? mysql_free_result() doesn't work since I am not even storing the results. //keep polling as long as stop character '-' is not read while(szRxChar != '-') { // Check if a read is outstanding if (HasOverlappedIoCompleted(&ovRead)) { // Issue a serial port read if (!ReadFile(hSerial,&szRxChar,1, &dwBytesRead,&ovRead)) { DWORD dwErr = GetLastError(); if (dwErr!=ERROR_IO_PENDING) return dwErr; } } // Wait 5 seconds for serial input if (!(HasOverlappedIoCompleted(&ovRead))) { WaitForSingleObject(hReadEvent,RESET_TIME); } // Check if serial input has arrived if (GetOverlappedResult(hSerial,&ovRead, &dwBytesRead,FALSE)) { // Wait for the write GetOverlappedResult(hSerial,&ovWrite, &dwBytesWritten,TRUE); //load tagBuffer with byte stream tagBuffer[i] = szRxChar; i++; tagBuffer[i] = 0; //char arrays are \0 terminated //run query with tagBuffer if( strlen(tagBuffer)==PACKET_LENGTH ) { sprintf(query,"insert into scan (rfidnum) values ('"); strcat(query, tagBuffer); strcat(query, "')"); mysql_real_query(&mysql,query,(unsigned int)strlen(query)); i=0; } mysql_real_query(&mysql,"insert into scan (rfidnum) values ('2nd query')",(unsigned int)strlen("insert into scan (rfid) values ('2nd query')")); mysql_free_result(res); } }

    Read the article

  • Is it best to make fewer calls to the database and output the results in an array?

    - by Jonathan
    I'm trying to create a more succinct way to make hundreds of db calls. Instead of writing the whole query out every time I wanted to output a single field, I tried to port the code into a class that did all the query work. This is the class I have so far: class Listing { /* Connect to the database */ private $mysql; function __construct() { $this->mysql = new mysqli(DB_LOC, DB_USER, DB_PASS, DB) or die('Could not connect'); } function getListingInfo($l_id = "", $category = "", $subcategory = "", $username = "", $status = "active") { $condition = "`status` = '$status'"; if (!empty($l_id)) $condition .= "AND `L_ID` = '$l_id'"; if (!empty($category)) $condition .= "AND `category` = '$category'"; if (!empty($subcategory)) $condition .= "AND `subcategory` = '$subcategory'"; if (!empty($username)) $condition .= "AND `username` = '$username'"; $result = $this->mysql->query("SELECT * FROM listing WHERE $condition") or die('Error fetching values'); $info = $result->fetch_object() or die('Could not create object'); return $info; } } This makes it easy to access any info I want from a single row. $listing = new Listing; echo $listing->getListingInfo('','Books')->title; This outputs the title of the first listing in the category "Books". But if I want to output the price of that listing, I have to make another call to getListingInfo(). This makes another query on the db and again returns only the first row. This is much more succinct than writing the entire query each time, but I feel like I may be calling the db too often. Is there a better way to output the data from my class and still be succinct in accessing it (maybe outputting all the rows to an array and returning the array)? If yes, How?

    Read the article

  • MySQL Interview Questions

    - by Campbell
    Hi, I've been asked to screen some candidates for a MySQL DBA / Developer position for a role that requires an enterprise level skill set. I myself am a SQL Server person so I know what I would be looking for from that point of view with regards to scalability / design etc but is there anything specific I should be asking with regards to MySQL? I would ideally like to ask them about enterprise level features of MySQL that they would typically only use when working on a big database. Need to separate out the enterprise developers from the home / small website kind of guys. Thanks.

    Read the article

  • Can I force MySql table name case sensitivity on file systems that aren't case sensitive

    - by Brian Deacon
    So our target environment is linux, making mysql case-sensitive by default. I am aware that we can make our linux environment not case sensitive with the lower_case_table_names variable, but we would rather not. We have a few times been bitten with a case mismatch because our dev rigs are OSX, and mysql is not case sensitive there. Is there a way we can force table names to be case sensitive on my OSX install of MySql (5.0.83 if that matters) so that we catch a table name case mismatch prior to deploying to the integration servers running on linux?

    Read the article

  • Why are pieces of my HTML showing up on the page and breaking it? Is it PHP related?

    - by Jason Rhodes
    I've been building a site in PHP, HTML, CSS, and using a healthy dose of jQuery javascript. The site looks absolutely fine on my Mac browsers, but for some reason, when my client uses PC Safari, she's seeing strange bits of my HTML show up on the page. Here are some (small) screenshot examples: Figure 1: This one is just a closing </li> tag that should've been on the Media li element. Not much harm done, but strange. Figure 2: Here this was part of <div class='submenu'> and since the div tag didn't render properly, the entire contents of that div don't get styled correctly by CSS. Figure 3: This last example shows what should have been <a class='top current' href=... but for some reason half of the HTML tag stops being rendered and just gets printed out. So the rest of that list menu is completely broken. Here's the code from the header.php file itself. The main navigation section (seen in the screenshots) is further down, marked by a line of asterisks if you want to skip there. <?php // Setting up location variables if(isset($_GET['page'])) { $page = Page::find_by_slug($_GET['page']); } elseif(isset($_GET['post'])) { $page = Page::find_by_id(4); } else { $page = Page::find_by_id(1); } $post = isset($_GET['post']) ? Blogpost::find_by_slug($_GET['post']) : false; $front = $page->id == 1 ? true : false; $buildblog = $page->id == 4 ? true : false; $eventpage = $page->id == 42 ? true : false; // Setting up content edit variables $edit = isset($_GET['edit']) ? true : false; $preview = isset($_GET['preview']) ? true : false; // Finding page slug value $pageslug = $page->get_slug($loggedIn); ?> <!DOCTYPE html> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> <title> <?php if(!$post) { if($page->id != 1) { echo $page->title." | "; } echo $database->site_name(); } elseif($post) { echo "BuildBlog | ".$post->title; } ?> </title> <link href="<?php echo SITE_URL; ?>/styles/style.css" media="all" rel="stylesheet" /> <?php include(SITE_ROOT."/scripts/myJS.php"); ?> </head> <body class=" <?php if($loggedIn) { echo "logged"; } else { echo "public"; } if($front) { echo " front"; } ?>"> <?php $previewslug = str_replace("&edit", "", $pageslug); ?> <?php if($edit) { echo "<form id='editPageForm' action='?page={$previewslug}&preview' method='post'>"; } ?> <?php if($edit && !$preview) : // Edit original ?> <div id="admin_meta_nav" class="admin_meta_nav"> <ul class="center nolist"> <li class="title">Edit</li> <li class="cancel"><a class="cancel" href="?page=<?php echo $pageslug; ?>&cancel">Cancel</a></li> <li class="save"><input style='position: relative; z-index: 500' class='save' type="submit" name="newpreview" value="Preview" /></li> <li class="publish"><input style='position: relative; z-index: 500' class='publish button' type="submit" name="publishPreview" value="Publish" /></li> </ul> </div> <?php elseif($preview && !$edit) : // Preview your edits ?> <div id="admin_meta_nav" class="admin_meta_nav"> <ul class="center nolist"> <li class="title">Preview</li> <li class="cancel"><a class="cancel" href="?page=<?php echo $pageslug; ?>&cancel">Cancel</a></li> <li class="save"><a class="newpreview" href="?page=<?php echo $pageslug; ?>&preview&edit">Continue Editing</a></li> <li class="publish"><a class="publish" href="?page=<?php echo $pageslug; ?>&publishLastPreview">Publish</a></li> </ul> </div> <?php elseif($preview && $edit) : // Return to preview and continue editing ?> <div id="admin_meta_nav" class="admin_meta_nav"> <ul class="center nolist"> <li class="title">Edit Again</li> <li class="cancel"><a class="cancel" href="?page=<?php echo $pageslug; ?>&cancel">Cancel</a></li> <li class="save"><input style='position: relative; z-index: 500' class='save button' type="submit" name="newpreview" value="Preview" /></li> <li class="publish"><input style='position: relative; z-index: 500' class='publish button' type="submit" name="publishPreview" value="Publish" /></li> </ul> </div> <?php else : ?> <div id="meta_nav" class="meta_nav"> <ul class="center nolist"> <li><a href="login.php?logout">Logout</a></li> <li><a href="<?php echo SITE_URL; ?>/admin">Admin</a></li> <li><a href="<?php if($front) { echo "admin/?admin=frontpage"; } elseif($event || $eventpage) { echo "admin/?admin=events"; } elseif($buildblog) { if($post) { echo "admin/editpost.php?post={$post->id}"; } else { echo "admin/?admin=blog"; } } else { echo "?page=".$pageslug."&edit"; } ?>">Edit Mode</a></li> <li><a href="<?php echo SITE_URL; ?>/?page=donate">Donate</a></li> <li><a href="<?php echo SITE_URL; ?>/?page=calendar">Calendar</a></li> </ul> <div class="clear"></div> </div> <?php endif; ?> <div id="public_meta_nav" class="public_meta_nav"> <div class="center"> <ul class="nolist"> <li><a href="<?php echo SITE_URL; ?>/?page=donate">Donate</a></li> <li><a href="<?php echo SITE_URL; ?>/?page=calendar">Calendar</a></li> </ul> <div class="clear"></div> </div> </div> * Main Navigation Section, as seen in screenshots above, starts here ** <div class="header"> <div class="center"> <a class="front_logo" href="<?php echo SITE_URL; ?>"><?php echo $database->site_name(); ?></a> <ul class="nolist main_nav"> <?php $tops = Page::get_top_pages(); $topcount = 1; foreach($tops as $top) { $current = $top->id == $topID ? true : false; $title = $top->title == "Front Page" ? "Home" : ucwords($top->title); $url = ($top->title == "Front Page" || !$top->get_slug($loggedIn)) ? SITE_URL : SITE_URL . "/?page=".$top->get_slug($loggedIn); if(isset($_GET['post']) && $top->id == 1) { $current = false; } if(isset($_GET['post']) && $top->id == 4) { $current = true; } echo "<li"; if($topcount > 3) { echo " class='right'"; } echo "><a class='top"; if($current) { echo " current"; } echo "' href='{$url}'>{$title}</a>"; if($children = Page::get_children($top->id)) { echo "<div class='submenu'>"; echo "<div class='corner-helper'></div>"; foreach($children as $child) { echo "<ul class='nolist level1"; if(!$subchildren = Page::get_children($child->id)) { echo " nochildren"; } echo "'>"; $title = ucwords($child->title); $url = !$child->get_slug($loggedIn) ? SITE_URL : SITE_URL . "/?page=".$child->get_slug($loggedIn); if($child->has_published() || $loggedIn) { echo "<li><a class='title' href='{$url}'>{$title}</a>"; if($subchildren = Page::get_children($child->id)) { echo "<ul class='nolist level2'>"; foreach($subchildren as $subchild) { if($subchild->has_published() || $loggedIn) { $title = ucwords($subchild->title); $url = !$subchild->get_slug($loggedIn) ? SITE_URL : SITE_URL . "/?page=".$subchild->get_slug($loggedIn); echo "<li><a href='{$url}'>{$title}</a>"; } } echo "</ul>"; } echo "</li>"; } echo "</ul>"; } echo "</div>"; } echo "</li>"; $topcount++; } ?> </ul> <div class="clear"></div> </div> </div> <div id="mediaLibraryPopup" class="mediaLibraryPopup"> <h3>Media Library</h3> <ul class="box nolist"></ul> <div class="clear"></div> <a href="#" class="cancel">Cancel</a> </div> <div class="main_content"> Does anyone have any idea why the PC Safari browser would be breaking things up like this? I'm assuming it's PHP related but I cannot figure out why it would do that.

    Read the article

  • Mysql gem troubles with Snow Leopard upgrade

    - by so1o
    I have done everything that is on the web (i think) i have the new 64 bit xcode that came with snow leopard installed completely removed mysql, removed gems compeltely, removed rails installed mysql 64 bit, installed gems, installed mysql gem with the env ARCHFLAGS set I still get this nasty NameError: uninitialized constant MysqlCompat::MysqlRes from /Users/Navara/Sites/tuosystems/vendor/rails/activesupport/lib/active_support/dependencies.rb:440:in `load_missing_constant' from /Users/Navara/Sites/tuosystems/vendor/rails/activesupport/lib/active_support/dependencies.rb:80:in `rake_original_const_missing' from /usr/local/lib/ruby/gems/1.8/gems/rake-0.8.7/lib/rake.rb:2503:in `const_missing' Im not sure how to debug this.. any pointers will be greatly appreciated!

    Read the article

  • Connecting Named SQL Server Express 2005 from MySQL Migration Toolkit 1.1.10

    - by KoolKabin
    Hi guys, I am trying to migrate SQL Server Express 2005 database to mysql. I came across the mysql migration toolkit. When i started with the tool it asked for my sql server express information. I provided all the information of the sql express but it still can't connect. My machine has got 1.) SQL Server 2000 [Default instance eg computername ] 2.) SQL Server Express 2005 [Default Named Instance eg computername$SQLExpress ] *I can make easy connection from Microsoft SQL Server Management Studio. I am getting problem only while connecting from MySQl Migration toolkit 1.1.10

    Read the article

  • ClassNotFoundException in MySQL Connector/J

    - by Ephraim
    This has been asked before but I cannot find the answer I need. 1) Using Class.forName("com.mysql.java.Driver") in the eclipse IDE all works well. I load the correct jar (mysql-connector-java-5.1.20-bin.jar), no exception. When I create a jar for my app a1.jar and double click the jar, I get the ClassnotFoundException. I created a .bat file in Windows XP with java -classpath c:\temp\mysql-connector-java-5.1.20-bin.jar -jar c:\temp\a1.jar the app statrs with the same exception. Furthermore using System.getProperty ("java.class.path") shows c:\temp\a1.jar whilst in the IDE I can see several directories

    Read the article

< Previous Page | 214 215 216 217 218 219 220 221 222 223 224 225  | Next Page >