Search Results

Search found 14310 results on 573 pages for 'mysql sock'.

Page 355/573 | < Previous Page | 351 352 353 354 355 356 357 358 359 360 361 362  | Next Page >

  • How do I code a loop for my echo statements?

    - by ggg
    <?php defined('_JEXEC') or die('Restricted access'); $db =& JFactory::getDBO(); $query0 = "SELECT * FROM `#__chesspositions` WHERE . . . . ."; //echo $query0; $db->setQuery($query0); $ginfo = $db->loadObjectList(); //echo $ginfo[0]; echo $db->getErrorMsg(); if(empty($ginfo)){ echo "<center><h2 style='color:navy'>No game found, we apologize</h2></center>"; }else{ $query1= "SELECT * FROM `#__chessmoves` WHERE Id='".$ginfo[0]->MoveDataId."'"; $db->setQuery($query1); echo $db->getErrorMsg(); $gmove = $db->loadObjectList(); } //define array; //how do I code a foreach loop (or any other type of loop) here? //I'm having trouble properly defining the array and structuring the syntax. echo "[Event \"".$ginfo[0]->Event."\"]\n"; echo "[Site \"".$ginfo[0]->Site."\"]\n"; echo "[Date \"".$ginfo[0]->Date."\"]\n"; echo "[Round \"".$ginfo[0]->Round."\"]\n"; echo "[White \"".$ginfo[0]->White."\"]\n"; echo "[Black \"".$ginfo[0]->Black."\"]\n"; echo "[Result \"".$ginfo[0]->Result."\"]\n"; echo "[ECO \"".$ginfo[0]->ECO."\"]\n"; echo "[WhiteElo \"".$ginfo[0]->WhiteElo."\"]\n"; echo "[BlackElo \"".$ginfo[0]->BlackElo."\"]\n"; echo "[Annotator \"".$ginfo[0]->Annotator."\"]\n"; echo "[SetUp \"".$ginfo[0]->SetUp."\"]\n"; echo $gmove[0]->MoveData; ?>

    Read the article

  • Killing the mysqld process

    - by Josh K
    I have a table with ~800k rows. I ran an update users set hash = SHA1(CONCAT({about eight fields})) where 1; Now I have a hung Sequel Pro process and I'm not sure about the mysqld process. This is two questions: What harm can possibly come from killing these programs? I'm working on a separate database, so no damage should come to other databases on the system, right? Assume you had to update a table like this. What would be a quicker / more reliable method of updating without writing a separate script. I just checked with phpMyAdmin and it appears as though the query is complete. I still have Sequel Pro using 100% of both my cores though...

    Read the article

  • SQL query to retrieve financial year data grouped by the year

    - by mlevit
    Hi, I have a database with lets assume two columns (service_date & invoice_amount). I would like to create an SQL query that would retrieve and group the data for each financial year (July to June). I have two years of data so that is two financial years (i.e. 2 results). I know I can do this manually by creating an SQL query to group by month then run the data through PHP to create the financial year data but I'd rather have an SQL query. All ideas welcome. Thanks

    Read the article

  • PHP Multiple User Login Form - Navigation to Different Pages Based on Login Credentials

    - by Zulu Irminger
    I am trying to create a login page that will send the user to a different index.php page based on their login credentials. For example, should a user with the "IT Technician" role log in, they will be sent to "index.php", and if a user with the "Student" role log in, they will be sent to the "student/index.php" page. I can't see what's wrong with my code, but it's not working... I'm getting the "wrong login credentials" message every time I press the login button. My code for the user login page is here: <?php session_start(); if (isset($_SESSION["manager"])) { header("location: http://www.zuluirminger.com/SchoolAdmin/index.php"); exit(); } ?> <?php if (isset($_POST["username"]) && isset($_POST["password"]) && isset($_POST["role"])) { $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["username"]); $password = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["password"]); $role = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["role"]); include "adminscripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM Users WHERE username='$manager' AND password='$password' AND role='$role' LIMIT 1"); $existCount = mysql_num_rows($sql); if (($existCount == 1) && ($role == 'IT Technician')) { while ($row = mysql_fetch_array($sql)) { $id = $row["id"]; } $_SESSION["id"] = $id; $_SESSION["manager"] = $manager; $_SESSION["password"] = $password; $_SESSION["role"] = $role; header("location: http://www.zuluirminger.com/SchoolAdmin/index.php"); } else { echo 'Your login details were incorrect. Please try again <a href="http://www.zuluirminger.com/SchoolAdmin/index.php">here</a>'; exit(); } } ?> <?php if (isset($_POST["username"]) && isset($_POST["password"]) && isset($_POST["role"])) { $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["username"]); $password = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["password"]); $role = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["role"]); include "adminscripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM Users WHERE username='$manager' AND password='$password' AND role='$role' LIMIT 1"); $existCount = mysql_num_rows($sql); if (($existCount == 1) && ($role == 'Student')) { while ($row = mysql_fetch_array($sql)) { $id = $row["id"]; } $_SESSION["id"] = $id; $_SESSION["manager"] = $manager; $_SESSION["password"] = $password; $_SESSION["role"] = $role; header("location: http://www.zuluirminger.com/SchoolAdmin/student/index.php"); } else { echo 'Your login details were incorrect. Please try again <a href="http://www.zuluirminger.com/SchoolAdmin/index.php">here</a>'; exit(); } } ?> And the form that the data is pulled from is shown here: <form id="LoginForm" name="LoginForm" method="post" action="http://www.zuluirminger.com/SchoolAdmin/user_login.php"> User Name:<br /> <input type="text" name="username" id="username" size="50" /><br /> <br /> Password:<br /> <input type="password" name="password" id="password" size="50" /><br /> <br /> Log in as: <select name="role" id="role"> <option value="">...</option> <option value="Head">Head</option> <option value="Deputy Head">Deputy Head</option> <option value="IT Technician">IT Technician</option> <option value="Pastoral Care">Pastoral Care</option> <option value="Bursar">Bursar</option> <option value="Secretary">Secretary</option> <option value="Housemaster">Housemaster</option> <option value="Teacher">Teacher</option> <option value="Tutor">Tutor</option> <option value="Sanatorium Staff">Sanatorium Staff</option> <option value="Kitchen Staff">Kitchen Staff</option> <option value="Parent">Parent</option> <option value="Student">Student</option> </select><br /> <br /> <input type="submit" name = "button" id="button" value="Log In" onclick="javascript:return validateLoginForm();" /> </h3> </form> Once logged in (and should the correct page be loaded, the validation code I have at the top of the script looks like this: <?php session_start(); if (!isset($_SESSION["manager"])) { header("location: http://www.zuluirminger.com/SchoolAdmin/user_login.php"); exit(); } $managerID = preg_replace('#[^0-9]#i', '', $_SESSION["id"]); $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["manager"]); $password = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["password"]); $role = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["role"]); include "adminscripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM Users WHERE username='$manager' AND password='$password' AND role='$role' LIMIT 1"); $existCount = mysql_num_rows($sql); if ($existCount == 0) { header("location: http://www.zuluirminger.com/SchoolAdmin/index.php"); exit(); } ?> Just so you're aware, the database table has the following fields: id, username, password and role. Any help would be greatly appreciated! Many thanks, Zulu

    Read the article

  • SQL Query Update is not working

    - by Brett Powell
    Hey guys, I am using pawn script for something, and everything works great except for one of my queries. For some reason, it will not work, and I am hoping it is simple enough someone can spot my mistake as I have been banging my head on it for days. http://ampaste.net/m6a887d30 The two highlighted lines are the queries that are not working. The other one works fine, but the values for 'class1kills' and 'class2kills' remain at 0. Here is a screenshot from phpmyadmin incase I did something silly. http://brutalservers.net/sql.png

    Read the article

  • mysql_fetch_array() not displaying all results

    - by user1666995
    I have a database with a calendar table (each row represents one day) with 4 years of rows (2012, 2013, 2014, 2015). I use the column name calyear for the year. I use the following code to find values for distinct years then display it: $year = mysql_query("SELECT DISTINCT calyear FROM calendar"); while($yeararray = mysql_fetch_array($year)) { echo($yeararray['calyear']."<br />"); } The problem is it only displays the years 2013, 2014, 2015 even though when I use echo(mysql_num_rows($year); it displays the value 4 which I take to mean all 4 years are there. I'm not quite sure where I'm going wrong with this.

    Read the article

  • Function for putting all database table to an array

    - by jasmine
    I have written a function to print database table to an array like this $db_array= Array( ID=>1, PARENTID =>1, TITLE => LIPSUM, TEXT =>LIPSUM ) My function is: function dbToArray($table) { $allArrays =array(); $query = mysql_query("SELECT * FROM $table"); $dbRow = mysql_fetch_array($query); for ($i=0; $i<count($dbRow) ; $i++) { $allArrays[$i] = $dbRow; } $txt .='<pre>'; $txt .= print_r($allArrays); $txt .= '</pre>'; return $txt; } Anything wrong in my function. Any help is appreciated about my problem. Thanks in advance

    Read the article

  • hibernate not throwing stale state exception nor it is overwriting data

    - by Reddy
    Our application do the following. 1. Start the transaction. 2. Execute a query using prepared statement 3. Check a condition to see the number of rows updated are equal to the required number. 4. It commits on success of above condition otherwise it will roll back However the problem is that when two threads are simultaneously enter this code. Thread-1 is updating a row in step 2. It checked the condition and committed successfully since the condition is successful. Thread-2 started execution somewhere between steps 1 & 4, and it is failing on at condition checking at step 3 (as it is getting number of updated rows as 0). I expected second thread to throw an exception but it is not. What could be the problem?

    Read the article

  • Magento Search returns All Products

    - by user338844
    I have set up a website (www.autobodypartsnow.com) and the search function has gone haywire. Whenever you search for anything, it returns every product in the system making it useless. I have fiddled around with it with no prevail. I have done the Like and FullText thing in the Configuration - nothing. I have reset the Search and FS cache. I have actually gone in the product attributes and disabled all the attributes from being searchable (just to see) and Rebuilt the Search Cache, cleared the FS cache and it still does the same thing - which tells me that something it pretty wrong. Any Ideas? Thanks in Advance!

    Read the article

  • Improving performance in this query

    - by Luiz Gustavo F. Gama
    I have 3 tables with user logins: sis_login = administrators tb_rb_estrutura = coordinators tb_usuario = clients I created a VIEW to unite all these users by separating them by levels, as follows: create view `login_names` as select `n1`.`cod_login` as `id`, '1' as `level`, `n1`.`nom_user` as `name` from `dados`.`sis_login` `n1` union all select `n2`.`id` as `id`, '2' as `level`, `n2`.`nom_funcionario` as `name` from `tb_rb_estrutura` `n2` union all select `n3`.`cod_usuario` as `id`, '3' as `level`, `n3`.`dsc_nome` as `name` from `tb_usuario` `n3`; So, can occur up to three ids repeated for different users, which is why I separated by levels. This VIEW is just to return me user name, according to his id and level. considering it has about 500,000 registered users, this view takes about 1 second to load. too much time, but is becomes very small when I need to return the latest posts on the forum of my website. The tables of the forums return the user id and level, then look for a name in this VIEW. I have registered 18 forums. When I run the query, it takes one second for each forum = 18 seconds. OMG. This page loads every time somebody enter my website. This is my query: select `x`.`forum_id`, `x`.`topic_id`, `l`.`nome` from ( select `t`.`forum_id`, `t`.`topic_id`, `t`.`data`, `t`.`user_id`, `t`.`user_level` from `tb_forum_topics` `t` union all select `a`.`forum_id`, `a`.`topic_id`, `a`.`data`, `a`.`user_id`, `a`.`user_level` from `tb_forum_answers` `a` ) `x` left outer join `login_names` `l` on `l`.`id` = `x`.`user_id` and `l`.`level` = `x`.`user_level` group by `x`.`forum_id` asc USING EXPLAIN: id select_type table type possible_keys key key_len ref rows Extra 1 PRIMARY <derived2> ALL NULL NULL NULL NULL 6 Using temporary; Using filesort 1 PRIMARY <derived4> ALL NULL NULL NULL NULL 530415 4 DERIVED n1 ALL NULL NULL NULL NULL 114 5 UNION n2 ALL NULL NULL NULL NULL 2 6 UNION n3 ALL NULL NULL NULL NULL 530299 NULL UNION RESULT ALL NULL NULL NULL NULL NULL 2 DERIVED t ALL NULL NULL NULL NULL 3 3 UNION r ALL NULL NULL NULL NULL 3 NULL UNION RESULT ALL NULL NULL NULL NULL NULL Somebody can help me or give a suggestion?

    Read the article

  • getting sql records

    - by droidus
    when i run this code, it returns the topic fine... $query = mysql_query("SELECT topic FROM question WHERE id = '$id'"); if(mysql_num_rows($query) > 0) { $row = mysql_fetch_array($query) or die(mysql_error()); $topic = $row['topic']; } but when I change it to this, it doesn't run at all. why is this happening? $query = mysql_query("SELECT topic, lock FROM question WHERE id = '$id'"); if(mysql_num_rows($query) > 0) { $row = mysql_fetch_array($query) or die(mysql_error()); $topic = $row['topic']; $lockedThread = $row['lock']; echo "here: " . $lockedThread; }

    Read the article

  • Top x rows and group by (again)

    - by Tibor Szasz
    Hello, I know it's a frequent question but I just can't figure it out and the examples I found didn't helped. What I learned, the best strategy is to try to find the top and bottom values of the top range and then select the rest, but implementing is a bit tricky. Example table: id | title | group_id | votes I'd like to get the top 3 voted rows from the table, for each group. I'm expecting this result: 91 | hello1 | 1 | 10 28 | hello2 | 1 | 9 73 | hello3 | 1 | 8 84 | hello4 | 2 | 456 58 | hello5 | 2 | 11 56 | hello6 | 2 | 0 17 | hello7 | 3 | 50 78 | hello8 | 3 | 9 99 | hello9 | 3 | 1 I've fond complex queries and examples, but they didn't really helped.

    Read the article

  • Socket ReceiveAll

    - by rielz
    I am trying to capture ip packets in c#. Everything is working fine, except that i only get outgoing packets. My Code: using (Socket sock = new Socket(AddressFamily.InterNetwork, SocketType.Raw, ProtocolType.IP)) { sock.Bind(new IPEndPoint(MYADDRESS, 0)); sock.SetSocketOption(SocketOptionLevel.IP, SocketOptionName.HeaderIncluded, true); sock.IOControl(IOControlCode.ReceiveAll, BitConverter.GetBytes(1), null); while (true) { byte[] buffer = new byte[sock.ReceiveBufferSize]; int count = sock.Receive(buffer); // ... } } The problem is definitely my pc! But maybe there is a workaround ...

    Read the article

  • 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-->

    Read the article

  • How to join nearly identical several queries into one?

    - by Devyn
    Hi, Assume I have an order_dummy table where order_dummy_id, order_id, user_id, book_id, author_id are stored. You may complain the logic of my table but I somehow need to do it that way. I want to execute following queries. SELECT * FROM order_dummy WHERE order_id = 1 AND user_id = 1 AND book_id = 1 ORDER BY `order_dummy_id` DESC LIMIT 1 SELECT * FROM order_dummy WHERE order_id = 1 AND user_id = 1 AND book_id = 2 ORDER BY `order_dummy_id` DESC LIMIT 1 SELECT * FROM order_dummy WHERE order_id = 1 AND user_id = 1 AND book_id = 3 ORDER BY `order_dummy_id` DESC LIMIT 1 Please keep in mind that several numbers of same book is included in one order. Therefore, I list order_dummy_id by descending and limit 1 so only LATEST ORDER of A BOOK is shown. But my goal is to show other books in that way in one table. I used group by like this ... SELECT * FROM order_dummy WHERE order_id = 1 AND user_id = 1 GROUP BY book_id but it only shows order_dummy_id with ascending result. I have no idea anymore. Looking forward your kindness help!

    Read the article

  • selecting data from table based on date .

    - by mehdi
    i have database table like this +-------+--------------+----------+ | id | ip | date | +-------+--------------+----------+ | 505 |192.168.100.1 |2010-04-03| | 252 |192.168.100.5 |2010-03-03| | 426 |192.168.100.6 |2010-03-03| | 201 |192.168.100.7 |2010-04-03| | 211 |192.168.100.10|2010-04-03| +-------+--------------+----------+ how can i retirive data from this table where month=03 how to write sql to do that . select * from table where month=03 something like that .

    Read the article

  • 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>

    Read the article

  • Product Name Print Several times, How to fix.?

    - by mans
    i had added the following Opencart module for my order report list... http://www.opencart.com/index.php?route=extension/extension/info&extension_id=3597&filter_search=order%20list%20filter%20model&page=4 I have problems with the column "Products". If there are more than one option the products name prints several times. So if I got a product with three options the product name prints three times. Is there any way to fix this problem? i want print product name and model number only once, any idea.? i will attach the results what i got now... this is my sql query... public function getOrders($data = array()) { $sql = "select o.order_id,o.email,o.telephone,CONCAT(o.shipping_address_1, ' ', o.shipping_address_2) AS address,CONCAT(o.firstname, ' ', o.lastname) AS customer,o.payment_zone AS state,o.payment_address_2 AS block, o.payment_address_1 AS address,o.payment_postcode AS postcode,(SELECT os.name FROM " . DB_PREFIX . "order_status os WHERE os.order_status_id = o.order_status_id AND os.language_id = '" . (int)$this->config->get('config_language_id') . "') AS status,o.payment_city AS city,GROUP_CONCAT(pd.name) AS pdtname,GROUP_CONCAT(op.model) AS model,o.date_added,sum(op.quantity) AS quantity,GROUP_CONCAT(opt.value ) AS options, GROUP_CONCAT(opt.order_product_id ) AS ordprdid,GROUP_CONCAT(op.order_product_id ) AS optprdid, GROUP_CONCAT(op.quantity) AS opquantity from `" . DB_PREFIX . "order` o LEFT JOIN " . DB_PREFIX . "order_product op ON (op.order_id = o.order_id) LEFT JOIN " . DB_PREFIX . "product_description pd ON (pd.product_id = op.product_id and pd.language_id = '" . (int)$this->config->get('config_language_id') . "') LEFT JOIN " . DB_PREFIX . "order_option opt ON (opt.order_product_id = op.order_product_id) "; Product Name = GROUP_CONCAT(pd.name) AS pdtname,

    Read the article

  • a query is inserted from PHPMYAdmin but not from PHP

    - by iyad al aqel
    i'm writing a php code to insert form values in a forum values $dbServer = mysql_connect("localhost" , "root", "") ; if(!$dbServer) die ("Unable to connect"); mysql_select_db("kfumWonder"); $name= $_POST['name'] ; $password= md5($_POST['password']); $email= $_POST['email'] ; $major= $_POST['major'] ; $dateOfBirth=$_POST['dateOfBirth'] ; $webSite = $_POST['website']; $joinDate= date("Y m d") ; $query = "INSERT INTO user (name, password, email, major, dob, website, join_date) Values ('$name', '$password', '$email', '$major', '$dateOfBirth', '$webSite' , '$joinDate')" ; //echo $query ; $result = mysql_query($query) ; if (! $result ) echo " no results " ; this works perfectly fine when i took the printed query and run it in PHPMyAdmin but when i run this code nothing happens , any ideas !?

    Read the article

  • Problem with PHP/Java bridge.

    - by Jack
    I am using Tomcat 6. I am running a php script using the JavaBridge. I get the following error when I run my code. Fatal error: Call to undefined function mysqli_connect() in C:\Program Files\apache-tomcat-6.0.26\webapps\JavaBridge\xxxx\xxxxx.php on line 534 Please help.

    Read the article

< Previous Page | 351 352 353 354 355 356 357 358 359 360 361 362  | Next Page >