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  • sql statement question. Need to query 3 tables in one go!

    - by Stefan
    Hey there, I have an sql database. In this database is 3 tables I need to query. The first table has all the item info called item and the other two tables has data for votes and comments called userComment and the third for votes called userItem I currently have a function which uses this sql query to get the latest more popular (in terms of both votes and comments): $sql = "SELECT itemID, COUNT(*) AS cnt FROM ( SELECT `itemID` FROM `userItem` WHERE FROM_UNIXTIME( `time` ) >= NOW() - INTERVAL 1 DAY UNION ALL SELECT `itemID` FROM `userComment` WHERE FROM_UNIXTIME( `time` ) >= NOW() - INTERVAL 1 DAY AND `itemID` > 0 ) q GROUP BY `itemID` ORDER BY cnt DESC"; I know how to change this for either by votes alone or comments.... HOWEVER - I need to query the database to only return the itemID's of the ones which have specific conditions in only the item table these are WHERE categoryID = 'xx' AND typeID = 'xx' If the sql ninja could please help me on this one? Do I have to first return the results from the above query and the for each in the array fetched then check each against the item table and see if it fits the conditions to build a new array - or is that overkill? Thanks, Stefan

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  • What are the best practices for storing PHP session data in a database?

    - by undefined
    I have developed a web application that uses a web server and database hosted by a web host (on the ground) and a server running on Amazon Web Services EC2. Both servers may be used by a user during a session and both will need to know some session information about a user. I don't want to POST the information that is needed by both servers because I dont want it to be visible to browsers / Firebug etc. So I need my session data to persist across servers. And I think that this means that the best option is to store all / some of the data that I need in the database rather than in a session. The easiest thing to do seems to be to keep the sessions but to POST the session_id between servers and use this as the key to lookup the data I need from a 'user_session_data' table in the database. I have looked at Tony Marston's article "Saving PHP Session Data to a database" - should I use this or will a table with the session data that I need and session_id as key suffice? What would be the downside of creating my own table and set of methods for storing the data I need in the database?

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  • Resetting AUTO_INCREMENT on myISAM without rebuilding the table

    - by Artem
    Please help I am in major trouble with our production database. I had accidentally inserted a key with a very large value into an autoincrement column, and now I can't seem to change this value without a huge rebuild time. "ALTER TABLE tracks_copy AUTO_INCREMENT = 661482981" Is super-slow. How can I fix this in production? I can't get this to work either (has no effect): myisamchk tracks.MYI --set-auto-increment=661482982 Any ideas? Basically, no matter what I do I get an overflow: SHOW CREATE TABLE tracks CREATE TABLE tracks ( ... ) ENGINE=MYISAM AUTO_INCREMENT=2147483648 DEFAULT CHARSET=latin1

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  • Database Design for One to One relationships

    - by Greelmo
    I'm trying to finalize my design of the data model for my project, and am having difficulty figuring out which way to go with it. I have a table of users, and an undetermined number of attributes that apply to that user. The attributes are in almost every case optional, so null values are allowed. Each of these attributes are one to one for the user. Should I put them on the same table, and keep adding columns when attributes are added (making the user table quite wide), or should I put each attribute on a separate table with a foreign key to the user table. I have decided against using the EAV model. Thanks!

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  • Form validation

    - by kielie
    Hi guys, I need to create a form that has many of the same fields, that have to be inserted into a database, but the problem I have is that if a user only fills in one or two of the rows, the form will still submit the blank data of the empty fields along with the one or two fields the user has filled in. How can I check for the rows that have not been filled in and leave them out of the query? or check for those that have been filled in and add them to the query. . . The thank_you.php file will capture the $_POST variables and add them to the database. <form method="post" action="thank_you.php"> Name: <input type="text" size="28" name="name1" /> E-mail: <input type="text" size="28" name="email1" /> <br /> Name: <input type="text" size="28" name="name2" /> E-mail: <input type="text" size="28" name="email2" /> <br /> Name: <input type="text" size="28" name="name3" /> E-mail: <input type="text" size="28" name="email3" /> <br /> Name: <input type="text" size="28" name="name4" /> E-mail: <input type="text" size="28" name="email4" /> <input type="image" src="images/btn_s.jpg" /> </form> I am assuming that I could use javascript or jQuery to accomplish this, how would I go about doing this? Thanx in advance for the help.

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  • Doctrine-CLI database creation issue.

    - by gokujou
    I have Doctrine setup in my Zend Framework application and I built my schema YAML file. But when I tell Doctrine to build the tables it says it does but it doesn't actually make them. It creates the models, and will create the DB but it will not populate the DB with the tables and throws no errors. Does anyone have a guess or know why this is not working? Thank you.

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  • PHP MySQLi isn't letting me alter a table (adding a new column)

    - by asdasd
    Well thats pretty much it. This is my query: $query = 'ALTER TABLE permissions ADD '.$name.' INT NOT NULL DEFAULT \'0\''; Where $name is already checked to exist with only lower case alpha letters, and not more than 20 length. Im just starting this out with very simple names. The next 4 lines of code after that one are: if($stmt = $db -> prepare($query)) { $success = $stmt -> execute(); $stmt -> close(); if(!$success) echo 'ERROR: Unsuccessful query: ',$db->error,PHP_EOL; } And I get back, every time ERROR: Unsuccessful query: And no error message. Is there a way to get more error messages so I can see what is failing? I can add new columns through phpmyadmin, but that really doesnt help me at all. The $db is fine, i do lots of stuff before and after this one section. It is only adding new column to the table that fails. side question: prepare() rejected my query every time when i tried to make those 2 variables, the $name and the 0 value as ? ? prepared statement values. Thats why they are in the real query and not bound later. If i could change that too I would like that.

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  • JOIN two tables to show already purchased items

    - by Norbert
    I have a table where I keep all my templates: templates template_id template_name template_price These templates can be purchased by a registered user and then are inserted in the payments table: payments payment_id template_id user_id Is there a way to join these two tables and get not just a list of templates that have been purchased by a certain user, but all the templates? And then figure out from there which ones have already been purchased? I used this SELECT, but only the ones that the user bought showed up. I would like to have all the rows from templates, but empty in case the user_id doesn't match. SELECT * FROM templates LEFT JOIN payments ON templates.template_id = payments.template_id WHERE user_id = 2 GROUP BY templates.template_id

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  • How to add condition on multiple-join table

    - by Jean-Philippe
    Hi, I have those two tables: client: id (int) #PK name (varchar) client_category: id (int) #PK client_id (int) category (int) Let's say I have those datas: client: {(1, "JP"), (2, "Simon")} client_category: {(1, 1, 1), (2, 1, 2), (3, 1, 3), (4,2,2)} tl;dr client #1 has category 1, 2, 3 and client #2 has only category 2 I am trying to build a query that would allow me to search multiple categories. For example, I would like to search every clients that has at least category 1 and 2 (would return client #1). How can I achieve that? Thanks!

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  • change password code error

    - by ejah85
    I've created a code to change a password. Now it seem contain an error. When I fill in the form to change password, and click save the error message: Warning: mysql_real_escape_string() expects parameter 2 to be resource, null given in C:\Program Files\xampp\htdocs\e-Complaint(FYP)\userChangePass.php on line 103 Warning: mysql_real_escape_string() expects parameter 2 to be resource, null given in C:\Program Files\xampp\htdocs\e-Complaint(FYP)\userChangePass.php on line 103 I really don’t know what the error message means. Please guys. Help me fix it. Here's is the code: <?php session_start(); ?> <?php # change password.php //set the page title and include the html header. $page_title = 'Change Your Password'; //include('templates/header.inc'); if(isset($_POST['submit'])){//handle the form require_once('connectioncomplaint.php');//connect to the db. //include "connectioncomplaint.php"; //create a function for escaping the data. function escape_data($data){ global $dbc;//need the connection. if(ini_get('magic_quotes_gpc')){ $data=stripslashes($data); } return mysql_real_escape_string($data, $dbc); }//end function $message=NULL;//create the empty new variable. //check for a username if(empty($_POST['userid'])){ $u=FALSE; $message .='<p> You forgot enter your userid!</p>'; }else{ $u=escape_data($_POST['userid']); } //check for existing password if(empty($_POST['password'])){ $p=FALSE; $message .='<p>You forgot to enter your existing password!</p>'; }else{ $p=escape_data($_POST['password']); } //check for a password and match againts the comfirmed password. if(empty($_POST['password1'])) { $np=FALSE; $message .='<p> you forgot to enter your new password!</p>'; }else{ if($_POST['password1'] == $_POST['password2']){ $np=escape_data($_POST['password1']); }else{ $np=FALSE; $message .='<p> your new password did not match the confirmed new password!</p>'; } } if($u && $p && $np){//if everything's ok. $query="SELECT userid FROM access WHERE (userid='$u' AND password=PASSWORD('$p'))"; $result=@mysql_query($query); $num=mysql_num_rows($result); if($num == 1){ $row=mysql_fetch_array($result, MYSQL_NUM); //make the query $query="UPDATE access SET password=PASSWORD('$np') WHERE userid=$row[0]"; $result=@mysql_query($query);//run the query. if(mysql_affected_rows() == 1) {//if it run ok. //send an email,if desired. echo '<p><b>your password has been changed.</b></p>'; include('templates/footer.inc');//include the HTML footer. exit();//quit the script. }else{//if it did not run OK. $message= '<p>Your password could not be change due to a system error.We apolpgize for any inconvenience.</p><p>' .mysql_error() .'</p>'; } }else{ $message= '<p> Your username and password do not match our records.</p>'; } mysql_close();//close the database connection. }else{ $message .='<p>Please try again.</p>'; } }//end oh=f the submit conditional. //print the error message if there is one. if(isset($message)){ echo'<font color="red">' , $message, '</font>'; } ?> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <body> <script language="JavaScript1.2">mmLoadMenus();</script> <table width="604" height="599" border="0" align="center" cellpadding="0" cellspacing="0"> <tr> <td height="130" colspan="7"><img src="images/banner(E-Complaint)-.jpg" width="759" height="130" /></td> </tr> <tr> <td width="100" height="30" bgcolor="#ABD519"></td> <td width="100" bgcolor="#ABD519"></td> <td width="100" bgcolor="#ABD519"></td> <td width="100" bgcolor="#ABD519"></td> <td width="100" bgcolor="#ABD519"></td> <td width="160" bgcolor="#ABD519"> <?php include "header.php"; ?>&nbsp;</td> </tr> <tr> <td colspan="7" bgcolor="#FFFFFF"> <fieldset><legend> Enter your information in the form below:</legend> <p><b>User ID:</b> <input type="text" name="username" size="10" maxlength="20" value="<?php if(isset($_POST['userid'])) echo $_POST['userid']; ?>" /></p> <p><b>Current Password:</b> <input type="password" name="password" size="20" maxlength="20" /></p> <p><b>New Password:</b> <input type="password" name="password1" size="20" maxlength="20" /></p> <p><b>Confirm New Password:</b> <input type="password" name="password2" size="20" maxlength="20" /></p> </fieldset> <div align="center"> <input type="submit" name="submit" value="Change My Password" /></div> </form><!--End Form--> </td> </tr> </table> </body> </html>

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  • change password code error.....

    - by shimaTun
    I've created a code to change a password. Now it seem contain an error.before i fill the form. the page display the error message: Parse error: parse error, unexpected $end in C:\Program Files\xampp\htdocs\e-Complaint(FYP)\userChangePass.php on line 222 this the code: <?php # userChangePass.php //this page allows logged in user to change their password. $page_title='Change Your Password'; //if no first_name variable exists, redirect the user if(!isset($_SESSION['nameuser'])){ header("Location: http://" .$_SERVER['HTTP_HOST']. dirname($_SERVER['PHP_SELF'])."/index.php"); ob_end_clean(); exit(); }else{ if(isset($_POST['submit'])) {//handle form. require_once('connectioncomplaint.php'); //connec to the database //check for a new password and match againts the confirmed password. if(eregi ("^[[:alnum:]]{4,20}$", stripslashes(trim($_POST['password1'])))){ if($_POST['password1'] == $_POST['password2']){ $p =escape_data($_POST['password1']); }else{ $p=FALSE; echo'<p><font color="red" size="+1"> Your password did not match the confirmed password!</font></p>'; } }else{ $p=FALSE; echo'<p><font color="red" size="+1"> Please Enter a valid password!</font></p>'; } if($p){ //if everything OK. //make the query $query="UPDATE access SET password=PASSWORD('$p') WHERE userid={$_SESSION['userid']}"; $result=@mysql_query($query);//run the query. if(mysql_affected_rows() == 1) {//if it run ok. //send an email,if desired. echo '<p><b>your password has been changed.</b></p>'; //include('templates/footer.inc');//include the HTML footer. exit(); }else{//if it did not run ok $message= '<p>Your password could not be change due to a system error.We apolpgize for any inconvenience.</p><p>' .mysql_error() .'</p>'; } mysql_close();//close the database connection. }else{//failed the validation test. echo '<p><font color="red" size="+1"> Please try again.</font></p>'; } }//end of the main Submit conditional. ?> And code for form: <h1>Change Your Password</h1> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <fieldset> <p><b>New Password:</b><input type="password" name="password1" size="20" maxlength="20" /> <small>Use only letters and numbers.Must be between 4 and 20 characters long.</small></p> <p><b>Confirm New Password:</b><input type="password" name="password2" size="20" maxlength="20" /></p> </fieldset> <div align="center"> <input type="submit" name="submit" value="Change My Password" /></div> </form><!--End Form-->

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  • Does a SELECT happen all at once, or progressively

    - by AmbroseChapel
    I have a process which finds a list of files to be deleted using a SELECT wheredelete= 'Y'. I set this process running the other day but it takes a while because it actually does the file deletions too. And in the middle of its long operation, I was using the application and deleted one more file. At this point I realised I didn't know if that file would be deleted, because I didn't know if the SELECT would have found all the files at the start, or if it was finding them progressively and would get to my newly-deleted file eventually.

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  • Having a Link Only Appear If a Logged-In User Appears on a Dynamic List

    - by John
    Hello, For the function below, I would like the link <div class="footervote"><a href="http://www...com/.../footervote.php">Vote</a></div> to only appear if the logged in user currently appears on editorlist.php. (I. e. if the loginid in the function corresponds to any of the usernames that currently appear in editorlist.php.) Appearing on editorlist.php is something that is dynamic. How can I do this? Thanks in advance, John function show_userbox() { // retrieve the session information $u = $_SESSION['username']; $uid = $_SESSION['loginid']; // display the user box echo '<div id="userbox"> <div class="username">'.$u.'</div> <div class="submit"><a href="http://www...com/.../submit.php">Submit an item.</a></div> <div class="changepassword"><a href="http://www...com/.../changepassword.php">Change Password</a></div> <div class="logout"><a href="http://www...com/.../logout.php">Logout</a></div> <div class="footervote"><a href="http://www...com/.../footervote.php">Vote</a></div> </div>'; } On editorlist.php: $sqlStr = "SELECT l.loginid, l.username, l.created, DATEDIFF(NOW(), l.created) AS days, COALESCE(s.total, 0) AS countSubmissions, COALESCE(c.total, 0) AS countComments, COALESCE(s.total, 0) * 10 + COALESCE(c.total, 0) AS totalScore, DATEDIFF(NOW(), l.created) + COALESCE(s.total, 0) * 10 + COALESCE(c.total, 0) AS totalScore2 FROM login l LEFT JOIN ( SELECT loginid, COUNT(1) AS total FROM submission GROUP BY loginid ) s ON l.loginid = s.loginid LEFT JOIN ( SELECT loginid, COUNT(1) AS total FROM comment GROUP BY loginid ) c ON l.loginid = c.loginid GROUP BY l.loginid ORDER BY totalScore2 DESC LIMIT 10"; $result = mysql_query($sqlStr); $arr = array(); echo "<table class=\"samplesrec1edit\">"; while ($row = mysql_fetch_array($result)) { echo '<tr>'; echo '<td class="sitename1edit1"><a href="http://www...com/.../members/index.php?profile='.$row["username"].'">'.stripslashes($row["username"]).'</a></td>'; echo '<td class="sitename1edit2">'.($row["countSubmissions"]).'</td>'; echo '<td class="sitename1edit2">'.($row["countComments"]).'</td>'; echo '<td class="sitename1edit2">'.($row["days"]).'</td>'; echo '<td class="sitename1edit2">'.($row["totalScore2"]).'</td>'; echo '</tr>'; } echo "</table>";

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  • Rails find :conditions

    - by Sam
    I have a Reservation model that I'm searching for with three fields. The container_id must always be self.id but as confirmed and auto_confirmed only one needs to be true. I have the following but it doesn't preform what I need: Reservation.find(:all, :conditions => ['container_id = ? AND confirmed = ? OR auto_confirm = ?', self.id, true, true,]) How should I change this?

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  • How do I get the median/mode/range of a column in SQL using Java?

    - by Derek
    I have to get the median, mode and range of test scores from one column in a table but I am unsure how to go about doing that. When you connect to the database using java, you are normally returned a ResultSet that you can make a table or something out of but how do you get particular numbers or digits? Is there an SQL command to get the median/mode/range or will I have to calculate this myself, and how do you pull out numbers from the table in order to be able to calculate the mode/median/range? Thanks.

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  • Accessing data entered into multiple Django forms and generating them onto a new URL

    - by pedjk
    I have a projects page where users can start up new projects. Each project has two forms. The two forms are: class ProjectForm(forms.Form): Title = forms.CharField(max_length=100, widget=_hfill) class SsdForm(forms.Form): Status = forms.ModelChoiceField(queryset=P.ProjectStatus.objects.all()) With their respective models as follows: class Project(DeleteFlagModel): Title = models.CharField(max_length=100) class Ssd(models.Model): Status = models.ForeignKey(ProjectStatus) Now when a user fills out these two forms, the data is saved into the database. What I want to do is access this data and generate it onto a new URL. So I want to get the "Title" and the "Status" from these two forms and then show them on a new page for that one project. I don't want the "Title" and "Status" from all the projects to show up, just for one project at a time. If this makes sense, how would I do this? I'm very new to Django and Python (though I've read the Django tutorials) so I need as much help as possible. Thanks in advance Edit: The ProjectStatus code is (under models): class ProjectStatus(models.Model): Name = models.CharField(max_length=30) def __unicode__(self): return self.Name

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  • PHP form validation submit problem

    - by TaG
    Every time I try to submit the form and I have not entered nothing in the year field I get Incorrect year! how can I still submit the form without having to enter a year. In other words leaving the year field blank and not getting a warning? Here is the PHP code. if(preg_match('/^\d{4,}$/', $_POST['year'])) { $year = mysqli_real_escape_string($mysqli, $_POST['year']); } else { $year = NULL; } if($year == NULL) { echo '<p class="error">Incorrect year!</p>'; } else { //do something }

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  • Add all lines multiplied by another line in another table

    - by russell
    Hi, I hope I can explain this good enough. I have 3 tables. wo_parts, workorders and part2vendor. I am trying to get the cost price of all parts sold in a month. I have this script. $scoreCostQuery = "SELECT SUM(part2vendor.cost*wo_parts.qty) as total_score FROM part2vendor INNER JOIN wo_parts ON (wo_parts.pn=part2vendor.pn) WHERE workorder=$workorder"; What I am trying to do is each part is in wo_parts (under partnumber [pn]). The cost of that item is in part2vendor (under part number[pn]). I need each part price in part2vendor to be multiplied by the quantity sold in wo_parts. The way all 3 tie up is workorders.ident=wo_parts.workorder and part2vendor.pn=wo_parts.pn. I hope someone can assist. The above script does not give me the same total as when added by calculator.

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  • How to retrieve column total when rows are paginated?

    - by Rick
    Hey guys I have a column "price" in a table and I used a pagination script to generate the data displayed. Now the pagination is working perfectly however I am trying to have a final row in my HTML table to show the total of all the price. So I wrote a script to do just that with a foreach loop and it sort of works where it does give me the total of all the price summed up together however it is the sum of all the rows, even the ones that are on following pages. How can I retrieve just the sum of the rows displayed within the pagination? Thank you! Here is the query.. SELECT purchase_log.id, purchase_log.date_purchased, purchase_log.total_cost, purchase_log.payment_status, cart_contents.product_name, members.first_name, members.last_name, members.email FROM purchase_log LEFT JOIN cart_contents ON purchase_log.id = cart_contents.purchase_id LEFT JOIN members ON purchase_log.member_id = members.id GROUP BY id ORDER BY id DESC LIMIT 0,30";

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  • How to make an add friend/defriend function in PHP?

    - by user300371
    I have created a site where people can create a profile. But I am trying to figure out how to start on making an add friend button so users can have friends. In my user table, i have user_id, first_name, last_name, email, etc. Should I somehow relate the user_id of the user and the friend in a friend table? I am a novice to programming, so these things are still new to me. Thanks!

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