Search Results

Search found 36756 results on 1471 pages for 'mysql real query'.

Page 511/1471 | < Previous Page | 507 508 509 510 511 512 513 514 515 516 517 518  | Next Page >

  • How to get rank based on SUM's?

    - by Kenan
    I have comments table where everything is stored, and i have to SUM everything and add BEST ANSWER*10. I need rank for whole list, and how to show rank for specified user/ID. Here is the SQL: SELECT m.member_id AS member_id, (SUM(c.vote_value) + SUM(c.best)*10) AS total FROM comments c LEFT JOIN members m ON c.author_id = m.member_id GROUP BY c.author_id ORDER BY total DESC LIMIT {$sql_start}, 20

    Read the article

  • problem parsing JSON Strings

    - by blacktooth
    var records = JSON.parse(JsonString); for(var x=0;x<records.result.length;x++) { var record = records.result[x]; ht_text+="<b><p>"+(x+1)+" " +record.EMPID+" " +record.LOCNAME+" " +record.DEPTNAME+" " +record.CUSTNAME +"<br/><br/><div class='slide'>" +record.REPORT +"</div></p></b><br/>"; } The above code works fine when the JsonString contains an array of entities but fails for single entity. result is not identified as an array! Whats wrong with it? http://pastebin.com/hgyWw5hd

    Read the article

  • Storing unique values into an array and comparing against a loop - PHP

    - by Aphex22
    I'm writing a PHP report which is designed to be exported purely as a CSV file, using commma delimiters. There are three columns relating to product_id, these three columns are as follows: SKU Parent / Child Parent SKU 12345 parent 12345 12345_1 child 12345 12345_2 child 12345 12345_3 child 12345 12345_4 child 12345 18099 parent 18099 18099_1 child 18099 Here's a link to the full CSV file: http://i.imgur.com/XELufRd.png At the moment the code looks like this: $sql = "select * from product WHERE on_amazon = 'on' AND active = 'on'"; $result = mysql_query($sql) or die ( mysql_error() );?> <? // set headers echo " Type, SKU, Parent / Child, Parent SKU, Product name, Manufacturer name, Gender, Product_description, Product price, Discount price, Quantity, Category, Photo 1, Photo 2, Photo 3, Photo 4, Photo 5, Photo 6, Photo 7, Photo 8, Color id, Color name, Size name <br> "; // load all stock while ($line = mysql_fetch_assoc($result) ) { ?> <?php // Loop through each possible size variation to see whether any of the quantity column has stock > 0 $con_size = array (35,355,36,37,375,38,385,39,395,40,405,41,415,42,425,43,435,44,445,45,455,46,465,47,475,48,485); $arrlength=count($con_size); for($x=0;$x<$arrlength;$x++) { // check if size is available if($line['quantity_c_size_'.$con_size[$x].'_chain'] > 0 ) { ?> <? echo 'Shoes'; ?>, <?=$line['product_id']?>, , , <?=$line['title']?>, <? $brand = $line['jys_brand']; echo ucfirst($brand); ?>, <? $gender = $line['category']; if ($gender == 'Mens') { echo 'H'; } else{ echo 'F'; } ?>, <?=preg_replace('/[^\da-z]/i', ' ', $line['amazon_desc']) ?>, <?=$line['price']?>, <?=$line['price']?>, <?=$line['quantity_c_size_'.$con_size[$x].'_chain']?>, <? $category = $line['style1']; switch ($category) { case "ankle-boots": echo "10013"; break; case "knee-high-boots": echo "10011"; break; case "high-heel-boots": echo "10033"; break; case "low-heel-boots": echo "10014"; break; case "wedge-boots": echo "10014"; break; case "western-boots": echo "10032"; break; case "flat-shoes": echo "10034"; break; case "high-heel-shoes": echo "10039"; break; case "low-heel-shoes": echo "10039"; break; case "wedge-shoes": echo "10035"; break; case "ballerina-shoes": echo "10008"; break; case "boat-shoes": echo "10018"; break; case "loafer-shoes": echo "10037"; break; case "work-shoes": echo "10039"; break; case "flat-sandals": echo "10041"; break; case "low-heel-sandals": echo "10042"; break; case "high-heel-sandals": echo "10042"; break; case "wedge-sandals": echo "10042"; break; case "mule-sandals": echo "10038"; break; case "mary-jane-shoes": echo "10039"; break; case "sports-shoes": echo "10026"; break; case "court-shoes": echo "10035"; break; case "peep-toe-shoes": echo "10035"; break; case "flat-boots": echo "10609"; break; case "mid-calf-boots": echo "10014"; break; case "trainer-shoes": echo "10009"; break; case "wellington-boots": echo "10012"; break; case "lace-up-boots": echo "10609"; break; case "chelsea-and-jodphur-boots": echo "10609"; break; case "desert-and-chukka-boots": echo "10032"; break; case "lace-up-shoes": echo "10034"; break; case "slip-on-shoes": echo "10043"; break; case "gibson-and-derby-shoes": echo "10039"; break; case "oxford-shoes": echo "10039"; break; case "brogue-shoes": echo "10039"; break; case "winter-boots": echo "10021"; break; case "slipper-shoes": echo "10016"; break; case "mid-heel-shoes": echo "10039"; break; case "sandals-and-beach-shoes": echo "10044"; break; case "mid-heel-sandals": echo "10042"; break; case "mid-heel-boots": echo "10014"; break; default: echo ""; } ?>, http://www.getashoe.co.uk/full/<?=$line['product_id']?>_1.jpg, http://www.getashoe.co.uk/full/<?=$line['product_id']?>_2.jpg, http://www.getashoe.co.uk/full/<?=$line['product_id']?>_3.jpg, http://www.getashoe.co.uk/full/<?=$line['product_id']?>_4.jpg, , , , , <? $colour = preg_replace('/[^\da-z]/i', ' ', $line['colour']); if( preg_match( '/white.*/i', $colour)) { echo '1'; } elseif( preg_match( '/yellow.*/i', $colour)) { echo '4'; } elseif( preg_match( '/orange.*/i', $colour)) { echo '7'; } elseif( preg_match( '/red.*/i', $colour)) { echo '8'; } elseif( preg_match( '/pink.*/i', $colour)) { echo '13'; } elseif( preg_match( '/purple.*/i', $colour)) { echo '15'; } elseif( preg_match( '/blue.*/i', $colour)) { echo '19'; } elseif( preg_match( '/green.*/i', $colour)) { echo '25'; } elseif( preg_match( '/brown.*/i', $colour)) { echo '28'; } elseif( preg_match( '/grey.*/i', $colour)) { echo '35'; } elseif( preg_match( '/black.*/i', $colour)) { echo '38'; } elseif( preg_match( '/gold.*/i', $colour)) { echo '41'; } elseif( preg_match( '/silver.*/i', $colour)) { echo '46'; } elseif( preg_match( '/multi.*/i', $colour)) { echo '594'; } elseif( preg_match( '/beige.*/i', $colour)) { echo '6887'; } elseif( preg_match( '/nude.*/i', $colour)) { echo '6887'; } else { echo '534'; } ?>, <?=$line['colour']?>, <?=$con_size[$x]?> <br> <? // finish checking if size is available } } ?> So at the moment this is simply echoing out the product_ID into the SKU column. The code would need to enter the product_id into an array and check whether it is unique. If the product_id is unique to the array, then the product_id is echoed out unaltered, and parent is echoed out to the 'Parent/Child' column and then the product_id is repeated to the 'Parent SKU' column. However, if the array is checked and the product_id already exists in the array, then the product_id is echoed out to the 'SKU' column with a suffix i.e. _1. Then child is echoed to the 'Parent / Child' column and the original parent product_id echoed to the 'Parent SKU' column. HOWEVER - the same SKU cannot be repeated with the same suffix i.e. 12345_1, 12345_1 - so presumably there would be to be another array for the suffixed SKUs to be checked against. If anybody could help, it would be great. Thanks --- UPDATE ANSWER --- I managed to solved this myself and thought I would share my solution for future reference. /* * Array to collect product_ids and check whether unique. * If unique product_id becomes parent SKU * If not product_id becomes child of previous parent and suffixed with _1, _2 etc... */ if (!in_array($line['product_id'], $SKU)) { $SKU[] = $line['product_id']; $parent = $line['product_id']; $a = 0; ?> <? echo 'Shoes'; ?>, <? echo $parent; ?>, <? echo "Parent"; ?>, <? echo $parent; ?>, <? } else { $child = $line['product_id'] . "_" . $a; ?> <? echo 'Shoes'; ?>, <? echo $child; ?>, <? echo "Child"; ?>, <? echo $child; <? // increment suffix value for child SKU $a++; ?>

    Read the article

  • PHP Login, Store Session Variables.

    - by Andreas Carlbom
    Yo. I'm trying to make a simple login system in PHP and my problem is this: I don't really understand sessions. Now, when I log a user in, I run session_register("user"); but I don't really understand what I'm up to. Does that session variable contain any identifiable information, so that I for example can get it out via $_SESSION["user"] or will I have to store the username in a separate variable? Thanks.

    Read the article

  • One to two relationship in Doctrine with YAML

    - by Jeremy DeGroot
    I'm working on my first Symfony project with Doctrine, and I've run into a hitch. I'm trying to express a game with two players. The relationship I want to have is PlayerOne and PlayerTwo each being keyed to an ID in the Users table. This is part of what I've got so far: Game: actAs: { Timestampable:- } columns: id: { type: integer, notnull: true, unique: true } startDate: { type: timestamp, notnull: true } playerOne: { type: integer, notnull: true } playerTwo: { type: integer, notnull: true } winner: { type: integer, notnull:true, default:0 } relations: User: { onUpdate: cascade, local: playerOne, foreign: id} User: { onUpdate: cascade, local: playerTwo, foreign: id} That doesn't work. It builds fine, but the SQL it generates only includes a constraint for playerTwo. I've tried a few other things: User: { onUpdate: cascade, local: [playerOne, playerTwo], foreign: id} Also: User: [{ onUpdate: cascade, local: playerOne, foreign: id}, { onUpdate: cascade, local: playerTwo, foreign: id}] Those last two throw errors when I try to build. Is there anyone out there who understands what I'm trying to do and can help me achieve it?

    Read the article

  • how to specify a BIGINT in a ruby scaffold?

    - by webdestroya
    I am trying to create a model in ruby that uses a BIGINT datatype (as opposed to the INT done by :integer). I have search all over Google, but all I seem to find is "run an SQL statement to alter the table to a BIGINT" - This seems a bit hack-ish to me, so I wanted to know if there was a way to specify a bigint in the ruby system like :big_int or something Any ideas?

    Read the article

  • problem during data modification

    - by nectar
    here my code - if($pin == '105') { $sqltree = "INSERT INTO tbltree (`userId`, `level`, `superId`, `rootId`, `childcount`) VALUES ('$child1', '1', '$newid', '$myroot', '0');"; mysql_query($sqltree); update_level($newid); } function update_level() { //for 1st level $newid = $_SESSION['newid']; //getting senior's level 1 and to increase by 1 $sqlgetlevel = "SELECT superId,level1 FROM tbltree WHERE userID='$newid'"; echo "<br>test:".$sqlgetlevel; $result = mysql_query($sqlgetlevel,$link)or die(mysql_error()); //line 340 $row = mysql_fetch_array($result, MYSQL_ASSOC); $level1 = $row["level1"]; $level1 = $level1 + 1; //update increased level $sqlupdate = "UPDATE tbltree SET level1='$level1' WHERE userId='$newid';"; mysql_query($sqlupdate,$link)or die(mysql_error()); //change superId for new level $superid = $row["superId"]; } ERROR - test:SELECT superId,level1 FROM tbltree WHERE userID='29277640' Warning: mysql_query() expects parameter 2 to be resource, null given in C:\xampp\htdocs\303\levelupdate.php on line 340

    Read the article

  • Php INNER JOING jqGrid help

    - by yanike
    I'm trying to get INNER JOIN to work with JQGRID, but I can't get it working. I want the code to get the first_name and last_name from members using the "efrom" from messages that matches the "id" from members. $col = array(); $col["title"] = "From"; $col["name"] = "messages.efrom"; $col["width"] = "70"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $col = array(); $col["title"] = "First Name"; $col["name"] = "members.first_name"; $col["width"] = "80"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $col = array(); $col["title"] = "Last Name"; $col["name"] = "members.last_name"; $col["width"] = "80"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $col = array(); $col["title"] = "Subject"; $col["name"] = "messages.esubject"; $col["width"] = "300"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $col = array(); $col["title"] = "Date"; $col["name"] = "messages.edatetime"; $col["width"] = "150"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $g = new jqgrid(); $grid["sortname"] = 'messages.edatetime'; $g->select_command = "SELECT messages.efrom, messages.esubject, messages.edatetime, members.first_name, members.last_name FROM messages INNER JOIN members ON messages.efrom = members.id";

    Read the article

  • Why is str_replace not replacing this string?

    - by Niall
    I have the following PHP code which should load the data from a CSS file into a variable, search for the old body background colour, replace it with the colour from a submitted form, resave the CSS file and finally update the colour in the database. The problem is, str_replace does not appear to be replacing anything. Here is my PHP code (stored in "processors/save_program_settings.php"): <?php require("../security.php"); $institution_name = mysql_real_escape_string($_POST['institution_name']); $staff_role_title = mysql_real_escape_string($_POST['staff_role_title']); $program_location = mysql_real_escape_string($_POST['program_location']); $background_colour = mysql_real_escape_string($_POST['background_colour']); $bar_border_colour = mysql_real_escape_string($_POST['bar_border_colour']); $title_colour = mysql_real_escape_string($_POST['title_colour']); $url = $global_variables['program_location']; $data_background = mysql_query("SELECT * FROM sents_global_variables WHERE name='background_colour'") or die(mysql_error()); $background_output = mysql_fetch_array($data_background); $css = file_get_contents($url.'/default.css'); $str = "body { background-color: #".$background_output['data']."; }"; $str2 = "body { background-color: #".$background_colour."; }"; $css2 = str_replace($str, $str2, $css); unlink('../default.css'); file_put_contents('../default.css', $css2); mysql_query("UPDATE sents_global_variables SET data='{$institution_name}' WHERE name='institution_name'") or die(mysql_error()); mysql_query("UPDATE sents_global_variables SET data='{$staff_role_title}' WHERE name='role_title'") or die(mysql_error()); mysql_query("UPDATE sents_global_variables SET data='{$program_location}' WHERE name='program_location'") or die(mysql_error()); mysql_query("UPDATE sents_global_variables SET data='{$background_colour}' WHERE name='background_colour'") or die(mysql_error()); mysql_query("UPDATE sents_global_variables SET data='{$bar_border_colour}' WHERE name='bar_border_colour'") or die(mysql_error()); mysql_query("UPDATE sents_global_variables SET data='{$title_colour}' WHERE name='title_colour'") or die(mysql_error()); header('Location: '.$url.'/pages/start.php?message=program_settings_saved'); ?> Here is my CSS (stored in "default.css"): @charset "utf-8"; /* CSS Document */ body,td,th { font-family: Arial, Helvetica, sans-serif; font-size: 14px; color: #000; } body { background-color: #CCCCFF; } .main_table th { background:#003399; font-size:24px; color:#FFFFFF; } .main_table { background:#FFF; border:#003399 solid 1px; } .subtitle { font-size:20px; } input#login_username, input#login_password { height:30px; width:300px; font-size:24px; } input#login_submit { height:30px; width:150px; font-size:16px; } .timetable_cell_lesson { width:100px; font-size:10px; } .timetable_cell_tutorial_a, .timetable_cell_tutorial_b, .timetable_cell_break, .timetable_cell_lunch { width:100px; background:#999; font-size:10px; } I've run some checks using the following code in the PHP file: echo $css . "<br><br>" . $str . "<br><br>" . $str2 . "<br><br>" . $css2; exit; And it outputs (as you can see it's not changing anything in the CSS): @charset "utf-8"; /* CSS Document */ body,td,th { font-family: Arial, Helvetica, sans-serif; font-size: 14px; color: #000; } body { background-color: #CCCCFF; } .main_table th { background:#003399; font-size:24px; color:#FFFFFF; } .main_table { background:#FFF; border:#003399 solid 1px; } .subtitle { font-size:20px; } input#login_username, input#login_password { height:30px; width:300px; font-size:24px; } input#login_submit { height:30px; width:150px; font-size:16px; } .timetable_cell_lesson { width:100px; font-size:10px; } .timetable_cell_tutorial_a, .timetable_cell_tutorial_b, .timetable_cell_break, .timetable_cell_lunch { width:100px; background:#999; font-size:10px; } body { background-color: #CCCCFF; } body { background-color: #FF5719; } @charset "utf-8"; /* CSS Document */ body,td,th { font-family: Arial, Helvetica, sans-serif; font-size: 14px; color: #000; } body { background-color: #CCCCFF; } .main_table th { background:#003399; font-size:24px; color:#FFFFFF; } .main_table { background:#FFF; border:#003399 solid 1px; } .subtitle { font-size:20px; } input#login_username, input#login_password { height:30px; width:300px; font-size:24px; } input#login_submit { height:30px; width:150px; font-size:16px; } .timetable_cell_lesson { width:100px; font-size:10px; } .timetable_cell_tutorial_a, .timetable_cell_tutorial_b, .timetable_cell_break, .timetable_cell_lunch { width:100px; background:#999; font-size:10px; }

    Read the article

  • Connecting to 3rd party databse in Joomla!?

    - by Michael
    I need to connect to another database in Joomla! that's on another server. This is for a plugin and I need to pull some data from a table. Now what I don't want is to use this database to run Joomla!, I already have Joomla! installed and running on its own database on its server but I want to connect to another database (ON TOP of the current one) to pull some data, then disconnect from that 3rd party database - all while keeping the original Joomla database connection in tact.

    Read the article

  • Fiscal year, quarters, student table, and faculty table... How do I relate these?!

    - by yuudachi
    I have a student and faculty table. The primary key for student is studendID (SID) and faculty's primary key is facultyID, naturally. Student has an advisor column and a requested advisor column, which are foreign key to faculty. That's simple enough, right? However, now I have to throw in dates. I want to be able to view who their advisor was for a certain quarter (such as 2009 Winter) and who they had requested. The result will be a table like this: Year | Term | SID | Current | Requested ------------------------------------------------ 2009 | Winter | 860123456 | 1 | NULL 2009 | Winter | 860445566 | 3 | NULL 2009 | Winter | 860369147 | 5 | 1 And then if I feel like it, I could also go ahead and view a different year and a different term. I am not sure how these new table(s) will look like. Will there be a year table with three columns that are Fall, Spring and Winter? And what will the Fall, Spring, Winter table have? I am new to the art of tables, so this is baffling me... Also, I feel I should clarify how the site works so far now. Admin can approve student requests, and what happens is that the student's current advisor gets overwritten with their request. However, I think I should not do that anymore, right?

    Read the article

  • Remove duplicate records/objects uniquely identified by multiple attributes

    - by keruilin
    I have a model called HeroStatus with the following attributes: id user_id recordable_type hero_type (can be NULL!) recordable_id created_at There are over 100 hero_statuses, and a user can have many hero_statuses, but can't have the same hero_status more than once. A user's hero_status is uniquely identified by the combination of recordable_type + hero_type + recordable_id. What I'm trying to say essentially is that there can't be a duplicate hero_status for a specific user. Unfortunately, I didn't have a validation in place to assure this, so I got some duplicate hero_statuses for users after I made some code changes. For example: user_id = 18 recordable_type = 'Evil' hero_type = 'Halitosis' recordable_id = 1 created_at = '2010-05-03 18:30:30' user_id = 18 recordable_type = 'Evil' hero_type = 'Halitosis' recordable_id = 1 created_at = '2009-03-03 15:30:00' user_id = 18 recordable_type = 'Good' hero_type = 'Hugs' recordable_id = 1 created_at = '2009-02-03 12:30:00' user_id = 18 recordable_type = 'Good' hero_type = NULL recordable_id = 2 created_at = '2009-012-03 08:30:00' (Last two are not a dups obviously. First two are.) So what I want to do is get rid of the duplicate hero_status. Which one? The one with the most-recent date. I have three questions: How do I remove the duplicates using a SQL-only approach? How do I remove the duplicates using a pure Ruby solution? Something similar to this: http://stackoverflow.com/questions/2790004/removing-duplicate-objects. How do I put a validation in place to prevent duplicate entries in the future?

    Read the article

  • Multitenant shared user account?

    - by jpartogi
    Dear all, Based on your experience, which is the route to go for a multi-tenant user login? One user login per account. Which means if there is one user that has access to multiple account, there will be redundancy of record in the database One user login for all account that she has privileges to. Which means one user record has access to multiple account if she has privileges to that account. From your experience, which one is better and why? I was thinking to choose the latter, but I don't know whether it will cause security issue or less flexibility. Thank you for sharing your experience.

    Read the article

  • Having a problem displaying data from last inserted data

    - by Gideon
    I'm designing a staff rota planner....have three tables Staff (Staff details), Event (Event details), and Job (JobId, JobDate, EventId (fk), StaffId (fk)). I need to display the last inserted job detail with the staff name. I've been at it for couple of hours and getting nowhere. Thanks for the help in advance. My code is the following: $eventId = $_POST['eventid']; $selectBox = $_POST['selectbox']; $timePeriod = $_POST['time']; $selectedDate = $_POST['date']; $count = count($selectBox); //constructing the staff selection if (empty($selectBox)) { echo "<p>You didn't select any member of staff to be assigned."; echo "<p><input type='button' value='Go Back' onClick='history.go(-1)'>"; } else { echo "<p> You selected ".$count. " staff for this show."; for ($i=0;$i<$count;$i++) { $selectId = $selectBox[$i]; //insert the details into the Job table in the database $insertJob = "INSERT INTO Job (JobDate, TimePeriod, EventId, StaffId) VALUES ('".$selectedDate."', '".$timePeriod."', ".$eventId.", ".$selectId.")"; $exeinsertJob = mysql_query($insertJob) or die (mysql_error()); } } //display the inserted job details $insertedlist = "SELECT Job.JobId, Staff.LastName, Staff.FirstName, Job.JobDate, Job.TimePeriod FROM Staff, Job WHERE Job.StaffId = Staff.StaffId AND Job.EventId = $eventId AND Job.JobDate = ".$selectedDate; $exeinsertlist = mysql_query($insertedlist) or die (mysql_error()); if ($exeinsertlist) { echo "<p><table cellspacing='1' cellpadding='3'>"; echo "<tr><th colspan=5> ".$eventname."</th></tr>"; echo "<tr><th>Job Id</th><th>Last Name</th> <th>First Name </th><th>Date</th><th>Hours</th></tr>"; while ($joblistarray = mysql_fetch_array($exeinsertlist)) { echo "<tr><td align=center>".$joblistarray['JobId']." </td><td align=center>".$joblistarray['LastName']."</td><td align=center>".$joblistarray['FirstName']." </td><td align=center>".$joblistarray['JobDate']." </td><td align=center>".$joblistarray['TimePeriod']."</td></tr>"; } echo "</table>"; echo "<h3><a href=AssignStaff.php>Add More Staff?</a></h3>"; } else { echo "The Job list can not be displayed at this time. Try again."; echo "<p><input type='button' value='Go Back' onClick='history.go(-1)'>"; }

    Read the article

  • Advanced count and join in Rails

    - by trobrock
    I am try to find the top n number of categories as they relate to articles, there is a habtm relationship set up between the two. This is the SQL I want to execute, but am unsure of how to do this with ActiveRecord, aside from using the find_by_sql method. is there any way of doing this with ActiveRecord methods: SELECT "categories".id, "categories".name, count("articles".id) as counter FROM "categories" JOIN "articles_categories" ON "articles_categories".category_id = "categories".id JOIN "articles" ON "articles".id = "articles_categories".article_id GROUP BY "categories".id ORDER BY counter DESC LIMIT 5;

    Read the article

  • php paypal ipn membership error

    - by aleksander haugas
    hello I integrated the membership subscription in the website, but I fail to insert data to the database. checking out all the function I make a txt file with php with data and generates me correctly. but I can not verify any data and less insert. <?php // read the post from PayPal system and add 'cmd' $req = 'cmd=_notify-validate'; foreach ($_POST as $key => $value) { $value = urlencode(stripslashes($value)); $req .= "&$key=$value"; } // Post back to PayPal system to validate $header .= "POST /cgi-bin/webscr HTTP/1.0\r\n"; $header .= "Content-Type: application/x-www-form-urlencoded\r\n"; $header .= "Content-Length: " . strlen($req) . "\r\n\r\n"; $fp = fsockopen ('ssl://www.paypal.com', 443, $errno, $errstr, 30); // Assign posted variables to local variables $item_name = $_POST['item_name']; $item_number = $_POST['item_number']; $payment_status = $_POST['payment_status']; $payment_amount = $_POST['mc_gross']; $payment_currency = $_POST['mc_currency']; $txn_id = $_POST['txn_id']; $receiver_email = $_POST['receiver_email']; $payer_email = $_POST['payer_email']; $user_id = $_POST['custom']; //Here i make the txt file and it works correctly //but after this does not work or make a txt file //Insert and update data if (!$fp) { // HTTP ERROR } else { fputs ($fp, $header . $req); while (!feof($fp)) { $res = fgets ($fp, 1024); if (strcmp ($res, "VERIFIED") == 0) { //Verificamos el estado del pedido if ($payment_status == 'Completed') { $txn_id_check = mysql_query("SELECT Transaction_ID FROM ".$db_table_prefix."Users_Payments WHERE Transaction_ID='".$txn_id."'"); if (mysql_num_rows($txn_id_check) !=1) { //Si el pago se ha realizado al dueño especificado y coincide con el dueño if ($receiver_email == $website_Payment_Email) { //Tipo de cuenta segun pagado if ($payment_amount == '10.00' && $payment_currency == 'EUR') { //add txn_id to database $log_query = mysql_query("INSERT INTO `".$db_table_prefix."Users_Payments` VALUES ('','".$txn_id."','".$payer_email."')"); //update premium to 3 $update_premium = mysql_query("UPDATE ".$db_table_prefix."Users SET Group_ID='3' WHERE User_ID='".$user_id."'"); } } } } } else if (strcmp ($res, "INVALID") == 0) { // log for manual investigation } } fclose ($fp); } ?> I call this file with require_once () with the connection to the database any ideas? thanks

    Read the article

  • Calculate time from timezones in php

    - by Ramya
    Hai I have the system with employees having different timezones in their profile. I would like to show the date according to their timezones specified. The GMT time zone values are placed in the database. could you guys help me

    Read the article

  • Need to map classes to different databases at runtime in Hibernate

    - by serg555
    I have MainDB database and unknown number (at compile time) of UserDB_1, ..., UserDB_N databases. MainDB contains names of those UserDB databases in some table (new UserDB can be created at runtime). All UserDB have exactly the same table names and fields. How to handle such situation in Hibernate? (database structure cannot be changed). Currently I am planning to create generic User classes not mapped to anything and just use native SQL for all queries: session.createSQLQuery("select * from " + db + ".user where id=1") .setResultTransformer(Transformers.aliasToBean(User.class)); Is there anything better I can do? Ideally I would want to have mappings for UserDB tables and relations and use HQL on required database.

    Read the article

  • While loop problems

    - by Luke
    I have put together the following code, the problem is that each while loop is only returning one set of data. $result = mysql_query("SELECT date FROM ".TBL_FIXTURES." WHERE compname = '$comp_name' GROUP BY date"); $i = 1; echo "<table cellspacing=\"10\" style='border: 1px dotted' width=\"300\" bgcolor=\"#eeeeee\">"; while ($row = mysql_fetch_assoc($result)) { $date=date("F j Y", $row['date']); echo $date; echo " <tr> <td>Fixture $i - Deadline on $date</td> </tr> "; $result = mysql_query("SELECT * FROM ".TBL_FIXTURES." WHERE compname = '$comp_name' AND date = '$row[date]' ORDER BY date"); while ($row = mysql_fetch_assoc($result)) { extract ($row); echo " <tr> <td>$home_user - $home_team V $away_user - $away_team</td> </tr> "; } $i++; } echo "</table>"; I should get many dates, and then each set of fixture below. At the moment, the first row from the first while loop is present, along with the data from the second while loop. However, it doesn't continue? Any ideas where I can correct this? Thanks

    Read the article

  • SQL hidden techniques?

    - by AlexRednic
    What are those pro/subtle techniques that SQL provides and not many know about which also cut code and improve performance? eg: I have just learned how to use CASE statements inside aggregate functions and it totally changed my approach on things. Are there others?

    Read the article

  • Form submit capture into database not working

    - by kielie
    Hi guys, I have a form that has two buttons on it, one yes, one no, and at the moment I capture the clicked button into a database, but for some reason, it doesn't always capture the clicked button, I have gone through all my code and everything seems fine, here is the form <div id="mid_prompt"> <form action="refer.php" method="post" onsubmit="return submit_form()" > <div class="prompt_container" style="float: left;"> <span class="prompt_item"><input type="image" src="images/yes.jpg" alt="submit" name="refer" value="yes" /></span> </div> </form> <form action="thank_you.php" method="post" onsubmit="return submit_form()" > <div class="prompt_container" style="float: right;"> <span class="prompt_item"><input type="image" src="images/no.jpg" alt="submit" name="refer" value="no" /></span> </div> </form> </div> I am using sessions to carry the variables all the way to the end of the forms where I then write all the data to a database, I have checked my sessions and they seem to be working fine, the only one that is giving problems is the yes/no. So basically I need it to capture that yes/no everytime. Thanx in advance!

    Read the article

  • How to secure phpMyAdmin

    - by Andrei
    Hi, I have noticed that there are strange requests to my website trying to find phpmyadmin, like /phpmyadmin/ /pma/ etc. Now I have installed PMA on Ubuntu via apt and would like to access it via webaddress different from /phpmyadmin/. What can I do to change it? Thanks

    Read the article

< Previous Page | 507 508 509 510 511 512 513 514 515 516 517 518  | Next Page >