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  • How can I get all children from a parent row in the same table?

    - by Johnny Freeman
    Let's say I have a table called my_table that looks like this: id | name | parent_id 1 | Row 1 | NULL 2 | Row 2 | NULL 3 | Row 3 | 1 4 | Row 4 | 1 5 | Row 5 | NULL 6 | Row 6 | NULL 7 | Row 7 | 8 8 | Row 8 | NULL 9 | Row 9 | 4 10 | Row 10 | 4 Basically I want my final array in PHP to look like this: Array ( [0] => Array ( [name] => Row 1 [children] => Array ( [0] => Array ( [name] => Row 3 [children] => ) [1] => Array ( [name] => Row 4 [children] => Array ( [0] => Array ( [name] => Row 9 [children] => ) [1] => Array ( [name] => Row 10 [children] => ) ) ) ) ) [1] => Array ( [name] => Row 2 [children] => ) [2] => Array ( [name] => Row 5 [children] => ) [3] => Array ( [name] => Row 6 [children] => ) [4] => Array ( [name] => Row 8 [children] => Array ( [0] => Array ( [name] => Row 7 [children] => ) ) ) ) So, I want it to get all of the rows where parent_id is null, then find all nested children recursively. Now here's the part that I'm having trouble with: How can this be done with 1 call to the database? I'm sure I could do it with a simple select statement and then have PHP make the array look like this but I'm hoping this can be done with some kind of fancy db joining or something like that. Any takers?

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  • Yes another ON DUPLICATE KEY UPDATE query

    - by Andy Gee
    I've been reading all the questions on here but I still don't get it I have two identical tables of considerable size. I would like to update table packages_sorted with data from packages_sorted_temp without destroying the existing data on packages_sorted Table packages_sorted_temp contains data on only 2 columns db_id and quality_rank Table packages_sorted contains data on all 35 columns but quality_rank is 0 The primary key on each table is db_id and this is what I want to trigger the ON DUPLICATE KEY UPDATE with. In essence how do I merge these two tables by and change packages_sorted.quality_rank of 0 to the quality_rank stored in packages_sorted_temp under the same primary key Here's what's not working INSERT INTO `packages_sorted` ( `db_id` , `quality_rank` ) SELECT `db_id` , `quality_rank` FROM `packages_sorted_temp` ON DUPLICATE KEY UPDATE `packages_sorted`.`db_id` = `packages_sorted`.`db_id`

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  • Is there a way to echo this only once and not have it repeat?

    - by Fernando
    I have the following query: $select = mysql_query("SELECT * FROM posts WHERE id = $postIds"); while ($return = mysql_fetch_assoc($select)) { $postUrl = $return['url']; $postTitle = $return['title']; echo "<h1><a href='$postUrl'>".$postTitle."</a></h1>"; } Now the problem is, the variable $postIds often times contain the same id multiple times. So the title of the post echos itself multiple times. Is there a way to have it echo only once?

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  • undefined offset error in php

    - by user225269
    I don't know why but the code below is working when I have a different query: $result = mysql_query("SELECT * FROM student WHERE IDNO='".$_GET['id']."'") ?> <?php while ( $row = mysql_fetch_array($result) ) { ?> <?php list($year,$month,$day)=explode("-", $row['BIRTHDAY']); ?> <tr> <td width="30" height="35"><font size="2">Month:</td> <td width="30"><input name="mm" type="text" id="mm" onkeypress="return handleEnter(this, event)" value="<?php echo $month;?>"> <td width="30" height="35"><font size="2">Day:</td> <td width="30"><input name="dd" type="text" id="dd" maxlength="25" onkeypress="return handleEnter(this, event)" value="<?php echo $day;?>"> <td width="30" height="35"><font size="2">Year:</td> <td width="30"><input name="yyyy" type="text" id="yyyy" maxlength="25" onkeypress="return handleEnter(this, event)" value="<?php echo $year;?>"> And it works when this is my query: $idnum = mysql_real_escape_string($_POST['idnum']); mysql_select_db("school", $con); $result = mysql_query("SELECT * FROM student WHERE IDNO='$idnum'"); Please help, why do I get the undefined offset error when I use this query: $result = mysql_query("SELECT * FROM student WHERE IDNO='".$_GET['id']."'") I assume that the query is the problem because its the only thing that's different between the two.

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  • Secure php code for copyright

    - by cosy
    I have an eCommerce platform, and i wand to secure it. How can it be possible ? I don't want to somebody copy my code. Like a license for 1 year with a code for activation. Or somethings like this. Sorry my English, thanks a lot!

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  • PHP: Building A Stock Index Using Yahoo Finance [on hold]

    - by Jeremy
    I have the following code which is pulling data but it is not outputting properly. <?php class YahooStock { public function getQuotes(){ $stocks = array(); $result = array(); $s = file_get_contents("http://finance.yahoo.com/d/quotes.csv?s=AMZN+CRM+CNQR+CTL+CTXS+DWRE+EMC+GOOG+HP+IBM+JIVE+LNKD+MKTO+MSFT+N+NFLX+NOW+ORCL+RAX+SAP+T+VEEV+VMW+VZ+WDAY&f=npf6&e=.csv"); $data = explode( ',', $s); $result = $data; return $result; } } $objYahooStock = new YahooStock; foreach( $objYahooStock->getQuotes() as $code => $result){ echo 'Name:' . $result[0] . '<br />'; echo 'Price:' . $result[1] . '<br />'; echo 'Float:' . $result[2] . '<br />'; } ?> The output looks like it is separating every character with a comma instead of each column: Name:" Price:A Float:m Name: Price:I Float:n Name:3 Price:3 Float:2 Name: Price: Float: Any help is appreciated!

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  • SQL where clasue to work with Group by clasue after performing a count()

    - by Matt
    Tried my usual references at w3schools and google. No luck I'm trying to produce the following results. QTY is a derived column | Position | QTY -------------------- 1 Clerk 2 2 Mgr 2 Here's what I'm not having luck with: SELECT Position, Count(position) AS 'QTY' FROM tblemployee Where ('QTY' != 1) GROUP BY Position I know that my Position is set up as varchar(255) Count produces a integer data and my where clasue is accurate so that leads me to believe that that Count() is jamming me up. Please throw up an example so I can reference later. Thanks for the help!

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  • Get previous and next row from current id

    - by Hukr
    How can I do to get the next row in a table? `image_id` int(11) NOT NULL auto_increment `image_title` varchar(255) NOT NULL `image_text` mediumtext NOT NULL `image_date` datetime NOT NULL `image_filename` varchar(255) NOT NULL If the current image is 3 for example and the next one is 7 etc. this won’t work: $query = mysql_query("SELECT * FROM images WHERE image_id = ".intval($_GET['id'])); echo $_GET['id']+1; How should I do? thanks

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  • running a stored procedure inside a sql trigger

    - by Ying
    Hi all, the business logic in my application requires me to insert a row in table X when a row is inserted in table Y. Furthermore, it should only do it between specific times(which I have to query another table for). I have considered running a script every 5 minutes or so to check this, but I stumbled upon triggers and figured this might be a better way to do it. But I find the syntax for procedures a little bewildering and I keep getting an error I have no idea how to fix. Here is where I start: CREATE TRIGGER reservation_auto_reply AFTER INSERT ON reservation FOR EACH ROW BEGIN IF NEW.sent_type = 1 /* In-App */ THEN INSERT INTO `messagehistory` (`trip`, `fk`, `sent_time`, `status`, `message_type`, `message`) VALUES (NEW.trip, NEW.psk, 'NOW()', 'submitted', 4, 'This is an automated reply to reservation'); END; I get an error in the VALUES part of the statmenet but im not sure where. I still have to query the other table for the information I need, but I can't even get past this part. Any help is appreciated, including links to many examples..Thanks

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  • I want to add and remove items from a menu with PHP.

    - by CDeanMartin
    This menu will need to be updated daily. <html><head></head><body> <h1> Welcome to Burgerama </h1> <?php include("menuBuilder.php"); showBurgerMenu(); ?> </body></html> Menu items are stored in the database. Items have a display field; if it is on, the item should be displayed on the menu. The menu only displays 4 or 5 "specials" at a time, and the manager needs to change menu items easily. I want to make a menu editing page like this: <?php include("burger_queries.php"); dbconnect("burger_database"); foreach($menuItem in burger_database) { echo createToggleButton($menuItem); } ?> .. with a toggle button for each menu item. Ideally the button will be labeled with the menu item, in blue if the item is "on", and red if the item is "off." Clicking the button switches between "on" and "off" I am stuck trying to a get a button to execute an UPDATE query on its corresponding menu item.

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  • Input html tag not parsing properly in php

    - by Akaash
    I have a database with names that I would like displayed in the form of a table with checkboxes. Everything works until I try to place the html tag into my php code. When I put the input tag in it gives me the error: Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' I can't see where I would put a comma or semi colon. <form> <?php $name = $_POST['name']; $host = "mysql16.000webhost.com"; $user_name = "a1611480_akaash"; $pwd = "****"; $database_name = "a1611480_akaash"; $db = mysql_connect($host, $user_name, $pwd); mysql_select_db($database_name); $result = mysql_query("SELECT name FROM Sort"); $var = array(); while ($row = mysql_fetch_array($result)) { $var[] = $row['name']; } $unique = array_unique($var); foreach ($unique as $value) { echo "<p class = Body_text><label>$value</label> <input type="checkbox" name="name" value="$value" /> </p>\n"; } ?> </form>

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  • Android : Connecting to MySQL using PHP

    - by user1771128
    I followed the following article http://blog.sptechnolab.com/2011/02/10/android/android-connecting-to-mysql-using-php/ I am able to execute my php file. I executed it individually and its working fine. The problem is in the android execution part. Am posting the Log Cat for the error am facing. Tried putting in a List View with id "list" but the error stil 10-28 16:08:27.201: E/AndroidRuntime(664): **FATAL EXCEPTION: main** 10-28 16:08:27.201: E/AndroidRuntime(664): java.lang.RuntimeException: Unable to start activity ComponentInfo{com.example.city/com.example.city.City}: java.lang.RuntimeException: Your content must have a ListView whose id attribute is 'android.R.id.list' 10-28 16:08:27.201: E/AndroidRuntime(664): at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:1956) 10-28 16:08:27.201: E/AndroidRuntime(664): at android.app.ActivityThread.handleLaunchActivity(ActivityThread.java:1981) 10-28 16:08:27.201: E/AndroidRuntime(664): at android.app.ActivityThread.access$600(ActivityThread.java:123) 10-28 16:08:27.201: E/AndroidRuntime(664): at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1147) 10-28 16:08:27.201: E/AndroidRuntime(664): at android.os.Handler.dispatchMessage(Handler.java:99) 10-28 16:08:27.201: E/AndroidRuntime(664): at android.os.Looper.loop(Looper.java:137) 10-28 16:08:27.201: E/AndroidRuntime(664): at android.app.ActivityThread.main(ActivityThread.java:4424) 10-28 16:08:27.201: E/AndroidRuntime(664): at java.lang.reflect.Method.invokeNative(Native Method) 10-28 16:08:27.201: E/AndroidRuntime(664): at java.lang.reflect.Method.invoke(Method.java:511) 10-28 16:08:27.201: E/AndroidRuntime(664): at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:784) 10-28 16:08:27.201: E/AndroidRuntime(664): at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:551) 10-28 16:08:27.201: E/AndroidRuntime(664): at dalvik.system.NativeStart.main(Native Method) 10-28 16:08:27.201: E/AndroidRuntime(664): Caused by: java.lang.RuntimeException: Your content must have a ListView whose id attribute is 'android.R.id.list' 10-28 16:08:27.201: E/AndroidRuntime(664): at android.app.ListActivity.onContentChanged(ListActivity.java:243) 10-28 16:08:27.201: E/AndroidRuntime(664): at com.android.internal.policy.impl.PhoneWindow.setContentView(PhoneWindow.java:254) 10-28 16:08:27.201: E/AndroidRuntime(664): at android.app.Activity.setContentView(Activity.java:1835) 10-28 16:08:27.201: E/AndroidRuntime(664): at com.example.city.City.onCreate(City.java:35) 10-28 16:08:27.201: E/AndroidRuntime(664): at android.app.Activity.performCreate(Activity.java:4465) 10-28 16:08:27.201: E/AndroidRuntime(664): at android.app.Instrumentation.callActivityOnCreate(Instrumentation.java:1049) 10-28 16:08:27.201: E/AndroidRuntime(664): at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:1920) 10-28 16:08:27.201: E/AndroidRuntime(664): ... 11 more

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  • Securing database keys for client-side processing

    - by danp
    I have a tree of information which is sent to the client in a JSON object. In that object, I don't want to have raw IDs which are coming from the database. I thought of making a hash of the id and a field in the object (title, for example) or a salt, but I'm worried that this might have a serious effect on processing overhead. SELECT * FROM `things` where md5(concat(id,'some salt')) = md5('1some salt'); Is there a standard practice for obscuring IDs in this kind of situation?

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  • SQL Query That Should Return Least two days record

    - by Aryans
    I have a table "abc" where i store timestamp having multiple records let suppose 1334034000 Date:10-April-2012 1334126289 Date:11-April-2012 1334291399 Date:13-April-2012 I want to build a sql query where I can find at first attempt the records having last two day values and so second time the next two days . . . Example: Select *,dayofmonth(FROM_UNIXTIME(i_created)) from notes where dayofmonth(FROM_UNIXTIME(i_created)) > dayofmonth(FROM_UNIXTIME(i_created)) -2 order by dayofmonth(FROM_UNIXTIME(i_created)) this query returns all the records date wise but we need very most two days record. Please suggest accordingly. Thanks in advance

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  • Help converting subquery to query with joins

    - by Tim
    I'm stuck on a query with a join. The client's site is running mysql4, so a subquery isn't an option. My attempts to rewrite using a join aren't going too well. I need to select all of the contractors listed in the contractors table who are not in the contractors2label table with a given label ID & county ID. Yet, they might be listed in contractors2label with other label and county IDs. Table: contractors cID (primary, autonumber) company (varchar) ...etc... Table: contractors2label cID labelID countyID psID This query with a subquery works: SELECT company, contractors.cID FROM contractors WHERE contractors.complete = 1 AND contractors.archived = 0 AND contractors.cID NOT IN ( SELECT contractors2label.cID FROM contractors2label WHERE labelID <> 1 AND countyID <> 1 ) I thought this query with a join would be the equivalent, but it returns no results. A manual scan of the data shows I should get 34 rows, which is what the subquery above returns. SELECT company, contractors.cID FROM contractors LEFT OUTER JOIN contractors2label ON contractors.cID = contractors2label.cID WHERE contractors.complete = 1 AND contractors.archived = 0 AND contractors2label.labelID <> 1 AND contractors2label.countyID <> 1 AND contractors2label.cID IS NULL

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  • Confused as to why my PHP include isn't working

    - by Sam
    I had a prototype of my website working correctly, meaning it connected to the database correctly. This was done with just one file called "connect.php" which had mysql_connect() and such inside it. I then separated the connect information into to separate files, one containing the account information (account.php) and one containing the connect function (connect.php), with correct information (I triple checked) and it isn't connecting properly. All I can think of is that I'm not including it the right way. This is what I have in a file: <?php include('account.php'); include('connect.php'); include('functions.php'); ..... ?>

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  • check for several conditions when a user logs in

    - by paul
    I would like to accomplish the following: If a username or password field is null, notify the user. If user name already exists, do not insert into the database and notify user to create a different name. if the username is unique and password is not null, return the username to the user. As of now it always returns "Please enter a different user name." I believe the issue has to do with the database query but I am not sure. If anyone can have a look and see if I am making an error, I greatly appreciate it, thanks. if ($userName or $userPassword = null) { echo "Please enter a user name and password or return to the homepage."; } elseif (mysql_num_rows(mysql_query("SELECT count(userName) FROM logininfo WHERE userName = '$userName'")) ==1) { echo "Please enter a different user name."; } elseif ($userName and $userPassword != null) { echo "Your login name is: $userName"; }

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  • SQL query joining rows with same value

    - by user1285737
    I need to write a query that creates a view that calculates the total cost of each sale, by considering quantity and price of each bought item. The view should return the debit and total cost. In the answer each debit-number should only occur once. Thanks in advance Table ITEM: ID NAME PRICE 118 Jeans 100 120 Towel 20 127 Shirt 55 Table DEBIT: DEBIT ITEM Quantity 100581 118 5 100581 120 1 100586 127 5

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  • laravel multiple where clauses within a loop

    - by user1424508
    Pretty much I want the query to select all records of users that are 25 years old AND are either between 150-170cm OR 190-200cm. I have this query written down below. However the problem is it keeps getting 25 year olds OR people who are 190-200cm instead of 25 year olds that are 150-170 OR 25 year olds that 190-200cm tall. How can I fix this? thanks $heightarray=array(array(150,170),array(190,200)); $user->where('age',25); for($i=0;$i<count($heightarray);i++){ if($i==0){ $user->whereBetween('height',$heightarray[$i]) }else{ $user->orWhereBetween('height',$heightarray[$i]) } } $user->get(); Edit: I tried advanced wheres (http://laravel.com/docs/queries#advanced-wheres) and it doesn't work for me as I cannot pass the $heightarray parameter into the closure. from laravel documentation DB::table('users') ->where('name', '=', 'John') ->orWhere(function($query) { $query->where('votes', '>', 100) ->where('title', '<>', 'Admin'); }) ->get();

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  • how do I copy value from one table and inserted to another in the same database??

    - by mathew
    I am having a tough time to do this... I have created two table say table-1 and table-2 in same database.what I want is I need to copy some values from table-1 and insert the same to table-2. I have tried many ways but it does not seems work. below is my code can any one tell me where I am missing?? $db = mysql_connect("localhost", "user", "pass") or die("Could not connect."); mysql_select_db("comdata",$db)or die(mysql_error()); $resultb = mysql_query("SELECT * FROM table-2")or die(mysql_error()); $row = mysql_fetch_array($resultb); $days = (strtotime(date("Y-m-d")) - strtotime($row['regtime'])) / (60 * 60 * 24); if($row > 0 && $days < 1){ $person = $row['person']; $catogr = $row['catog']; $position = $row['position']; $location = $row['location']; $rank = $row['rank']; mysql_close($db); }else{ $db = mysql_connect("localhost", "user", "pass") or die("Could not connect."); mysql_select_db("comdata",$db)or die(mysql_error()); $result = mysql_query("SELECT * FROM table-1 WHERE regtime = DATE(NOW()) ORDER BY rank ASC LIMIT 1;")or die(mysql_error()); $row = mysql_fetch_array($result); $person = $row['person']; $catogr = $row['catog']; $position = $row['position']; $location = $row['location']; $rank = $row['rank']; mysql_query("INSERT INTO table-2 (regtime,person,catog,position,location,rank) VALUES(NOW(),'$person','$catogr','$position','$location','$rank')"); mysql_close($db); }

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  • problem with logout script in php

    - by user225269
    I'm a beginner in php, and I am trying to create a login and logout. But I am having problems in logging out. My logout just calls for the login form which is this: <? session_start(); session_destroy(); ?> <table width="300" border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#CCCCCC"> <tr> <form name="form1" method="post" action="checklogin.php"> <td> <table width="100%" border="0" cellpadding="3" cellspacing="1" bgcolor="#FFFFFF"> <tr> <td colspan="3"><strong>Member Login </strong></td> </tr> <tr> <td width="78">Username</td> <td width="6">:</td> <td width="294"><input name="myusername" type="text" id="myusername"></td> </tr> <tr> <td>Password</td> <td>:</td> <td><input name="mypassword" type="text" id="mypassword"></td> </tr> <tr> <td>&nbsp;</td> <td>&nbsp;</td> <td><input type="submit" name="Submit" value="Login"></td> </tr> </table> </td> </form> </tr> </table> My problem is, when I try to press the back button in the browser. Whoever user is using it can still access what is not supposed to be accessed when a user hasn't logged in. Do I need to add a code on the user page? I have this code on the user page: <? session_start(); if(!session_is_registered(myusername)){ header("location:main_login.php"); } ?> What can you recommend that I would do so that a script will prompt to enter the username and password again when a user clicks on the back button.

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