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  • 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,

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  • 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

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  • User has many computers, computers have many attributes in different tables, best way to JOIN?

    - by krismeld
    I have a table for users: USERS: ID | NAME | ---------------- 1 | JOHN | 2 | STEVE | a table for computers: COMPUTERS: ID | USER_ID | ------------------ 13 | 1 | 14 | 1 | a table for processors: PROCESSORS: ID | NAME | --------------------------- 27 | PROCESSOR TYPE 1 | 28 | PROCESSOR TYPE 2 | and a table for harddrives: HARDDRIVES: ID | NAME | ---------------------------| 35 | HARDDRIVE TYPE 25 | 36 | HARDDRIVE TYPE 90 | Each computer can have many attributes from the different attributes tables (processors, harddrives etc), so I have intersection tables like this, to link the attributes to the computers: COMPUTER_PROCESSORS: C_ID | P_ID | --------------| 13 | 27 | 13 | 28 | 14 | 27 | COMPUTER_HARDDRIVES: C_ID | H_ID | --------------| 13 | 35 | So user JOHN, with id 1 owns computer 13 and 14. Computer 13 has processor 27 and 28, and computer 13 has harddrive 35. Computer 14 has processor 27 and no harddrive. Given a user's id, I would like to retrieve a list of that user's computers with each computers attributes. I have figured out a query that gives me a somewhat of a result: SELECT computers.id, processors.id AS p_id, processors.name AS p_name, harddrives.id AS h_id, harddrives.name AS h_name, FROM computers JOIN computer_processors ON (computer_processors.c_id = computers.id) JOIN processors ON (processors.id = computer_processors.p_id) JOIN computer_harddrives ON (computer_harddrives.c_id = computers.id) JOIN harddrives ON (harddrives.id = computer_harddrives.h_id) WHERE computers.user_id = 1 Result: ID | P_ID | P_NAME | H_ID | H_NAME | ----------------------------------------------------------- 13 | 27 | PROCESSOR TYPE 1 | 35 | HARDDRIVE TYPE 25 | 13 | 28 | PROCESSOR TYPE 2 | 35 | HARDDRIVE TYPE 25 | But this has several problems... Computer 14 doesnt show up, because it has no harddrive. Can I somehow make an OUTER JOIN to make sure that all computers show up, even if there a some attributes they don't have? Computer 13 shows up twice, with the same harddrive listet for both. When more attributes are added to a computer (like 3 blocks of ram), the number of rows returned for that computer gets pretty big, and it makes it had to sort the result out in application code. Can I somehow make a query, that groups the two returned rows together? Or a query that returns NULL in the h_name column in the second row, so that all values returned are unique? EDIT: What I would like to return is something like this: ID | P_ID | P_NAME | H_ID | H_NAME | ----------------------------------------------------------- 13 | 27 | PROCESSOR TYPE 1 | 35 | HARDDRIVE TYPE 25 | 13 | 28 | PROCESSOR TYPE 2 | 35 | NULL | 14 | 27 | PROCESSOR TYPE 1 | NULL | NULL | Or whatever result that make it easy to turn it into an array like this [13] => [P_NAME] => [0] => PROCESSOR TYPE 1 [1] => PROCESSOR TYPE 2 [H_NAME] => [0] => HARDDRIVE TYPE 25 [14] => [P_NAME] => [0] => PROCESSOR TYPE 1

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  • Can't use where clause on correlated columns.

    - by Keyo
    I want to add a where clause to make sure video_count is greater than zero. Only categories which are referenced once or more in video_category.video_id should be returned. Because video_count is not a field in any table I cannot do this. Here is the query. SELECT category . * , ( SELECT COUNT( * ) FROM video_category WHERE video_category.category_id = category.category_id ) AS 'video_count' FROM category WHERE category.status = 1 AND video_count > '0' AND publish_date < NOW() ORDER BY updated DESC; Thanks for the help.

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  • 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; }

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  • Define keys in temporary table creation

    - by imperium2335
    How do I define the keys for a temporary table that is being created from a SELECT statement? I have: CREATE temporary TABLE _temp_unique_parts_trading engine=memory AS (SELECT parts_trading.enquiryref, sellingcurrency, jobs.id AS jobID FROM parts_trading, jobs WHERE jobs.enquiryref = parts_trading.enquiryref GROUP BY parts_trading.enquiryref) But where do I define the keys?

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  • 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

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  • 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

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  • 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 !?

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  • 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.

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  • mod rewrite, title slugs and htaccess

    - by chris
    I have been taken in to provide some seo guidance on a website which has been running since 2005. My problem is i want to use clean urls. The code that handles the url is hidden away in some class file.. and with over a few thousand lines of code its a struggle to rewrite it. So I'm think, I have gone through all the products and created a slug for them as a field in the product table. Is it possible to do something like an intermediate file for htaccess. Some thing like 1./clean-slug-comes-in/ 2.htaccess catches this and uses slug.php to find the relevant product id for the slug. 3.Then product.php?id=(ID.found.from.2) is loaded?

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  • status update error (null field)

    - by ejah85
    hai guys... i .ve the problem that i cannot be recovered yet... i have one form where admin need to approve or reject the booking request... i've set the b_status field in table usage IN PROCESS default value... i want to update the b_status value BOOKING APPROVED when user click APPROVE button.. otherwise, the b_status will update the value as BOOKING REJECTED when user click on the REJECT button here's is the form code: <?php $db = mysql_connect('localhost','root') or die ("unable to connect"); mysql_select_db('fyp',$db) or die ("able to select"); $sql="SELECT * FROM vehicle WHERE v_status='READY'"; $result = mysql_query($sql) or die ("Query failed!"); ?> <tr><td>&nbsp;</td></tr> <tr> <tr> <td width="200"><font face="Arial" size="2" font color="#000000">Registration Number </font></td> <td><select name="regno"> <option value="" selected>--Registration No--</option> <?php while($row = mysql_fetch_array($result)){?> <option value="<?php echo $row['regno']; ?>"><?php echo $row['regno']; ?></option> <?php } ?> </select></td> <td><font face="Arial" size="2" font color="#000000">Reason</font></td> <td><textarea name="reason" rows="3" cols="50 "value = ""></textarea></td> </tr> <?php $db = mysql_connect('localhost','root') or die ("unable to connect"); mysql_select_db('fyp',$db) or die ("able to select"); $sql="SELECT * FROM driver WHERE d_status='READY'"; $result = mysql_query($sql) or die ("Query failed!"); ?> <tr> <td><font face="Arial" size="2" font color="#000000">Driver</font></td> <td><select id = "d_name" name="d_name"> <option value="" selected>--Driver Name--</option> <?php while($row = mysql_fetch_array($result)){?> <option value="<?php echo $row['d_name']; ?>"><?php echo $row['d_name']; ?></option> <?php } ?> </select></td> </tr> <tr> <?php mysql_close($db); ?> </table> <p></p> <center><input name="APPROVED" type="submit" id="APPROVED" value="APPROVED"> <input name="REJECT" type="submit" id="REJECT" value="REJECT"> </center> </div> </center> and this is the process page code: <?php $db = mysql_connect('localhost','root') or die ("unable to connect"); mysql_select_db('fyp',$db) or die ("able to select"); $bookingno=mysql_real_escape_string($_POST['bookingno']); $username=mysql_real_escape_string($_POST['username']); $name=mysql_real_escape_string($_POST['name']); $department=mysql_real_escape_string($_POST['department']); $g_date=mysql_real_escape_string($_POST['g_date']); $g_time=mysql_real_escape_string($_POST['g_time']); $r_date=mysql_real_escape_string($_POST['r_date']); $r_time=mysql_real_escape_string($_POST['r_time']); $destination=mysql_real_escape_string($_POST['destination']); $pass_num=mysql_real_escape_string($_POST['pass_num']); $trip_purpose=mysql_real_escape_string($_POST['trip_purpose']); $regno=mysql_real_escape_string($_POST['regno']); $d_name=mysql_real_escape_string($_POST['d_name']); $reason=mysql_real_escape_string($_POST['reason']); $b_status=mysql_real_escape_string($_POST['b_status']); $sql = "INSERT INTO `usage` VALUES('$bookingno','$username','$name','$department','$g_date','$g_time','$r_date','$r_time','$destination', '$pass_num','$trip_purpose','$regno','$d_name','$reason','$b_status')"; $query = "INSERT INTO `usage` VALUES b_status ='BOOKING APPROVED'"; $result = @mysql_query($query); $query1 = "UPDATE driver SET d_status ='OUT' WHERE '$d_name'=d_name"; $result1 = @mysql_query($query1); if(isset($_POST['APPROVED'])) { $query2 = "UPDATE `usage` SET b_status ='BOOKING APPROVED' WHERE '$b_status'='IN PROCESS'"; $result2 = @mysql_query($query2); } if (isset($_POST['REJECT'])) { $query3 = "UPDATE `usage` SET b_status ='BOOKING REJECTED' WHERE '$b_status'='IN PROCESS'"; $result3 = @mysql_query($query3); } //$result = mysql_query($sql) or die ("error!"); $result = mysql_query($sql) or trigger_error (mysql_error().' in '.$sql); i.ve the problem on the b_status field.. plz guys... help me ya :-)

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  • cakePHP and GROUP BY

    - by Lizard
    I am trying to solve a hopefully simple problem here is the query I am trying produce: SELECT `categories`.*, COUNT(`entities`.id) FROM `categories` LEFT JOIN `entities` ON (`categories`.`id` = `entities`.`category_id`) GROUP BY `categories`.`id` I am really struggling to do this is in cakePHP 1.2 How would/should I go about doing this... (I am using 'Containable' if that helps) Thanks in advance

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  • Search a string to find which records in table are inside said string

    - by Improfane
    Hello, Say I have a string. Then I have a number of unique tokens or keywords, potentially a large number in a database. I want to search and find out which of these database strings are inside the string I provide (and get the IDs of them). Is there a way of using a query to search the provided string or must it be taken to application space? Am I right in thinking that this is not a 'full text search'? Would the best method be to insert it into the database to make it a full text search?

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  • Correct Sql Script for Formula

    - by Madan Madan
    Can anyone help me write SQL script for the following formula? If DEP = 1 If DROP 1 PLV = 334.86 * exp(0.3541 * ACTIVE_DAYS) + 0.25 * DROP + 20 * DEP Else If DROP < 0 PLV = DROP + 70 * ACTIVE_DAYS Else PLV = 0.25 * DROP + 70 * ACTIVE_DAYS The SQL script which I have is the following SELECT IF(dep=1, if(dep=1, (334.86 * exp(0.3541 * act_days)) + (0.25 * 'drop') + (20 * dep), if('drop'<0, 'drop' + (70 * act_days), (0.25 * 'drop') + (70 * act_days))),'0') as PLV But the above query is not right as something is missing where the formula says Else PLV = 0.26 * DROP Thanks,

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  • SQL Querying for Threaded Messages

    - by Harper
    My site has a messaging feature where one user may message another. The messages support threading - a parent message may have any number of children but only one level deep. The messages table looks like this: Messages - Id (PK, Auto-increment int) - UserId (FK, Users.Id) - FromUserId (FK, Users.Id) - ParentMessageId (FK to Messages.Id) - MessageText (varchar 200) I'd like to show messages on a page with each 'parent' message followed by a collapsed view of the children messages. Can I use the GROUP BY clause or similar construct to retrieve parent messages and children messages all in one query? Right now I am retrieving parent messages only, then looping through them and performing another query for each to get all related children messages. I'd like to get messages like this: Parent1 Child1 Child2 Child3 Parent2 Child1 Parent3 Child1 Child2

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  • How phpmyadmin, php and everything else works?

    - by Tom
    Ok, I guess I got crazy, but really. How phpmyadmin works? Does it have his own phpmyadmin or what? And how php works? Why writing echo 'hello'; it returns hello in the browser? I am really interested on how these things really works, maybe you know any books or smth to figure it out? Thank you.

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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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  • 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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  • 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.

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  • How to compare a string with an option value

    - by user225269
    I have this html form which has options: <tr> <td width="30" height="35"><font size="3">*List:</td> <td width="30"><input name="specific" type="text" id="specific" maxlength="25" value=""> </td> <td><font size="3">*By:</td> <td> <select name="general" id="general"> <font size="3"> <option value="YEAR">Year</option> <option value="ADDRESS">Address</option> </select></td></td> </tr> And I'm trying to have this as the form action: if ('{$_POST["ADDRESS"]}'="ADDRESS") Which will compare if the value in the option in the html form matches the word "Address". If it matches then it will execute this query: $saddress= mysql_real_escape_string($_POST['specific']);<--this is the input form where the user will put the specific address to search. mysql_query("SELECT * FROM student WHERE ADDRESS='$saddress'"); Please I need help in here, I thinks its wrong: if ('{$_POST["ADDRESS"]}'="ADDRESS")

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  • [Ruby on Rails] Data Structure

    - by siulamvictor
    I am building a online form, with about 20 multiple choice checkboxes. I can get the nested data with this command. raise params.to_yaml I need to store these data and call them again later. I want to sort out which user chose which specific checkbox, i.e. who chose checkbox no.2? What's the best way to store these data in database?

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  • PHP MySQLi and MySQLi_STMT: Which insert_id to use?

    - by Carvell Fenton
    Hello all, Both the MySQLi and MySQLi_STMT classes have an $insert_id property. If I am connected to my database using a MySQLi object (say $db), and then I perform an INSERT with a MySQLi_STMT object (say $stmt), to get the id of the last INSERT, should I use: $last_id = $db->insert_id; or $last_id = $stmt->insert_id; Or would they be the same, in which case it doesn't matter? I thought this might be a quick answer for someone, and save me the time of writing the test code to check it. Thanks in advance as always.

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