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  • Broken count(*) after adding LEFT JOIN

    - by Iain Urquhart
    Since adding the LEFT JOIN to the query below, the count(*) has been returning some strange values, it seems to have added the total rows returned in the query to the 'level': SELECT `n`.*, exp_channel_titles.*, round((`n`.`rgt` - `n`.`lft` - 1) / 2, 0) AS childs, count(*) - 1 + (`n`.`lft` > 1) + 1 AS level, ((min(`p`.`rgt`) - `n`.`rgt` - (`n`.`lft` > 1)) / 2) > 0 AS lower, (((`n`.`lft` - max(`p`.`lft`) > 1))) AS upper FROM `exp_node_tree_6` `n` LEFT JOIN `exp_channel_titles` ON (`n`.`entry_id`=`exp_channel_titles`.`entry_id`), `exp_node_tree_6` `p`, `exp_node_tree_6` WHERE `n`.`lft` BETWEEN `p`.`lft` AND `p`.`rgt` AND ( `p`.`node_id` != `n`.`node_id` OR `n`.`lft` = 1 ) GROUP BY `n`.`node_id` ORDER BY `n`.`lft` I'm totally stumped... Thank you!

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  • Perl Script to search and replace in .SQL query file with user inputs

    - by T.Mount
    I have a .SQL file containing a large number of queries. They are being run against a database containing data for multiple states over multiple years. The machine I am running this on can only handle running the queries for one state, in one year, at a time. I am trying to create a Perl script that takes user input for the state abbreviation, the state id number, and the year. It then creates a directory for that state and year. Then it opens the "base" .SQL file and searches and replaces the base state id and year with the user input, and saves this new .SQL file to the created directory. The current script I have (below) stops at open(IN,'<$infile') with "Can't open [filename]" It seems that it is having difficulty finding or opening the .SQL file. I have quadruple-checked to make sure the paths are correct, and I have even tried replacing the $path with an absolute path for the base file. If it was having trouble with creating the new file I'd have more direction, but since it can't find/open the base file I do not know how to proceed. #!/usr/local/bin/perl use Cwd; $path = getcwd(); #Cleans up the path $path =~ s/\\/\//sg; #User inputs print "What is the 2 letter state abbreviation for the state? Ex. 'GA'\n"; $stlet = <>; print "What is the 2 digit state abbreviation for the state? Ex. '13'\n"; $stdig = <>; print "What four-digit year are you doing the calculations for? Ex. '2008'\n"; $year = <>; chomp $stlet; chomp $stdig; chomp $year; #Creates the directory mkdir($stlet); $new = $path."\/".$stlet; mkdir("$new/$year"); $infile = '$path/Base/TABLE_1-26.sql'; $outfile = '$path/$stlet/$year/TABLE_1-26.sql'; open(IN,'<$infile') or die "Can't open $infile: $!\n"; open(OUT,">$infile2") or die "Can't open $outfile: $!\n"; print "Working..."; while (my $search = <IN>) { chomp $search; $search =~ s/WHERE pop.grp = 132008/WHERE pop.grp = $stdig$year/g; print OUT "$search\n"; } close(IN); close(OUT); I know I also probably need to tweak the regular expression some, but I'm trying to take things one at a time. This is my first Perl script, and I haven't really been able to find anything that handles .SQL files like this that I can understand. Thank you!

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  • [PHP] Local/Dev/Live deployment - best workflow

    - by Adam Kiss
    Hello, situation We our little company with 3 people, each has a localhost webserver and most projects (previous and current) are on one PC network shared disk. We have virtual server, where some of our clients' sites and our site. Our standard workflow is: Coder PC ? Programmer localhost ? dev domain (client.company.com) ? live version (client.com) It often happens, that there are two or three guys working on same projects at the same time - one is on dev version, two are on localhost. When finished, we try to synchronize the files on dev version and ideally not to mess up any files, which *knock knock * doesn't happen often. And then one of us deploys dev version on live webserver. question we are looking for a way to simplify this workflow while updating websites - ideally some sort of diff uploader or VCS probably (Git/SVN/VCS/...), but we are not completely sure where to begin or what way would be ideal, therefore I ask you, fellow stackoverflowers for your experience with website / application deployment and recommended workflow. We probably will also need to use Mac in process, so if it won't be a problem, that would be even better. Thank you

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

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

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  • Cannot add or update a child row: a foreign key constraint fails

    - by myaccount
    // Getting the id of the restaurant to which we are uploading the pictures $restaurant_id = intval($_GET['restaurant-id']); if(isset($_POST['submit'])) { $tmp_files = $_FILES['rest_pics']['tmp_name']; $target_files = $_FILES['rest_pics']['name']; $tmp_target = array_combine($tmp_files, $target_files); $upload_dir = $rest_pics_path; foreach($tmp_target as $tmp_file => $target_file) { if(move_uploaded_file($tmp_file, $upload_dir."/".$target_file)) { $sql = sprintf(" INSERT INTO rest_pics (branch_id, pic_name) VALUES ('%s', '%s')" , mysql_real_escape_string($restaurant_id) , mysql_real_escape_string(basename($target_file))); $result = mysql_query($sql) or die(mysql_error()); } I get the next error: Cannot add or update a child row: a foreign key constraint fails (rest_v2.rest_pics, CONSTRAINT rest_pics_ibfk_1 FOREIGN KEY (branch_id) REFERENCES rest_branches (branch_id) ON DELETE CASCADE ON UPDATE CASCADE However, this error totally disappears and everything goes well when I put directly the restaurant id (14 for example) instead of $restaurant_id variable in the sql query. The URL am getting the id from is: http://localhost/rest_v2/public_html/admin/add-delete-pics.php?restaurant-id=2 Any help please?

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  • Why does this SELECT ... JOIN statement return no results?

    - by Stephen
    I have two tables: 1. tableA is a list of records with many columns. There is a timestamp column called "created" 2. tableB is used to track users in my application that have locked a record in tableA for review. It consists of four columns: id, user_id, record_id, and another timestamp collumn. I'm trying to select up to 10 records from tableA that have not been locked by for review by anyone in tableB (I'm also filtering in the WHERE clause by a few other columns from tableA like record status). Here's what I've come up with so far: SELECT tableA.* FROM tableA LEFT OUTER JOIN tableB ON tableA.id = tableB.record_id WHERE tableB.id = NULL AND tableA.status = 'new' AND tableA.project != 'someproject' AND tableA.created BETWEEN '1999-01-01 00:00:00' AND '2010-05-06 23:59:59' ORDER BY tableA.created ASC LIMIT 0, 10; There are currently a few thousand records in tableA and zero records in tableB. There are definitely records that fall between those timestamps, and I've verified this with a simple SELECT * FROM tableA WHERE created BETWEEN '1999-01-01 00:00:00' AND '2010-05-06 23:59:59' The first statement above returns zero rows, and the second one returns over 2,000 rows.

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  • Force reload/refresh when pressing the back button

    - by FlyingCat
    I will try my best to explain this. I have an application that show the 50+ projects in my view page. The user can click the individual project and go to the update page to update the project information. Everything works fine except that after user finish updating the individual project information and hit 'back' button on the browser to the previous view page. The old project information (before update) is still there. The user has to hit refresh to see the updated information. It not that bad, but I wish to provide better user experience. Any idea to fix this? Thanks a lot.

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  • undefined offset error in php

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

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  • PHP Foreach statement issue. Multiple rows are returned

    - by Daniel Patilea
    I'm a PHP beginner and lately i've been having a problem with my source code. Here it is: <html> <head> <title> Bot </title> <link type="text/css" rel="stylesheet" href="main.css" /> </head> <body> <form action="bot.php "method="post"> <lable>You:<input type="text" name="intrebare"></lable> <input type="submit" name="introdu" value="Send"> </form> </body> </html> <?php //error_reporting(E_ALL & ~E_NOTICE); mysql_connect("localhost", "root", "") or die(mysql_error()); mysql_select_db("robo") or die(mysql_error()); $intrebare=$_POST['intrebare']; $query = "SELECT * FROM dialog where intrebare like '%$intrebare%'"; $result = mysql_query($query) or die(mysql_error()); $row = mysql_fetch_array($result) or die(mysql_error()); ?> <div id="history"> <?php foreach($row as $rows){ echo "<b>The robot says: </b><br />"; echo $row['raspuns']; } ?> </div> It returns me the result x6 times. This problem appeared when I've made that foreach because I wanted the results to stuck on the page one by one after every sql querry. Can you please tell me what seems to be the problem? Thanks!

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  • php search database for row

    - by Brenden Morley
    Okay I got code the code to pull data based on a users account number well here is what im using (And yes I know it isnt safe now that is the reason for my post) <?php include('config.php'); $user_info = fetch_user_info($_GET['AccountNumber']); ?> <html> <body> <div> <?php if ($user_info === false){ $Output = 'http://www.MyDomain.Com/'; echo '<META HTTP-EQUIV=Refresh CONTENT="0; URL='.$Output.'">'; }else{ ?> <center> <title><?php echo $user_info['FirstName'], ' ', $user_info['LastName'], ' - ', $user_info['City'], ', ', $user_info['State']; ?> - Name of site</title> So basically what this code is allowing me to do is have a file called Profile.php And when a user visits this this page it will return the data Like this http://MyDomain.com/Profile.php?AccountNumber=50b9c965b7c3b How can I do this securely cause right now its using a get method really unsafe to retive the account number from the url bar.

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  • Select one column from a row in hibernate

    - by ComfortablyNumb
    I am trying to do a simple query to get a unique result back in hibernate here is my code. public String getName(Integer id) { Session session = getSessionFactory().openSession(); String name = (String)session.createSQLQuery("SELECT name FROM users WHERE user_id = :userId").setParameter("userId", id).uniqueResult(); return name; } The name that is being returned is stored as HTML text that includes html syntacx language. I think this is what is causing the problem but it doesnt make sense I just want to return it as a string. It is only happening on this one field name, I can get every other field in the row but this one it gives me the error. I am getting an exception. The exception I am getting is No Dialect mapping for JDBC type: -1; nested exception is org.hibernate.HibernateException How do you query for a specific column on a row in hibernate?

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  • Referral System PHP

    - by Liam
    I have a membership based website and im planning on implementing a referral system. My website is credit based, the idea is that if User X refers User Y, then User X gets 100 bonus credits. Has anybody built a referral system before and if so what obstacles should I bear in mind? I've had a snoop round SO tonight but couldn't find any suitable answers. My theory is to give each user a random string which is generated and stored in the DB when they sign up, The user will then be presented with a URL incl. that string which when they pass to somebody (User Z), User Z is then sent to a page, the page then uses the GET method to gather the Random string and update the DB Row they currently occupy, does this sound feasible or could it easily be breached? Thanks

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  • NOT LIKE not working on comparison to a column

    - by rodling
    Data is fairly large and takes few minutes to run it every time, so its taking a lot of time debugging this problem. When I run like concat('%',T.item,'%') on smaller data it seems to identify items properly. However, when I run it on the main DB (the code shown), it still shows many(maybe even all) of the exceptions. EDIT: it seems when i add NOT it stops identifying items select distinct T.comment from (select comment, source, item from data, non_informative where ticker != "O" and source != 7 and source != 6) as T where T.comment not like concat('%',T.item,'%') order by T.comment; comment and source are in data, item is in non_informative Some items from T.item: 'Stock Analysis -', '#InsideTrades', 'IIROC Trade' Example comment which should be removed '#InsideTrades #4 | MACNAB CRAIG (Director,Officer,Chief Executive Officer): Filed Form 4 for $NNN (NATIONAL RETA' Can't seem to figure out it why shows all the items

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  • How to pass an array of objects trough a jquery $.post?

    - by majc
    Hi, I want to pass the result of a query trough a $.post. function GetAllTasks() { $sql = "select t.id as task_id, description, createdat, createdby, max_requests, max_duration, j.name as job_name from darkfuture.tasks t, darkfuture.jobs j where t.job_id = j.id"; $sqlresult = mysql_query($sql) or die("The list of works failed: ".mysql_error($this->con)); $result = array(); while($row = mysql_fetch_assoc($sqlresult)) { $task = new TasksResult(); $task->id = $row["task_id"]; $task->description = $row["description"]; $task->createdat = $row["createdat"]; $task->createdby = $row["createdby"]; $task->max_requests = $row["max_requests"]; $task->max_duration = $row["max_duration"]; $task->job_id = $row["job_name"]; array_push($result, $task); } mysql_free_result($sqlresult); return $result; } Here is how i call it: $tasksDB = new TasksDB(); $tasks = $tasksDB->GetAllTasks(); Now i want to pass $tasks through here: $.post("views/insert_tasks.php",{'tasks[]': $tasks}, function(data) { }); I know this {'tasks[]': $tasks} it's wrong but i don't know how to do it right. Some help will be appreciated. Thanks in advance!

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  • uploading images with the help of arrays and fetch errors

    - by bonny
    i use a script to upload a couple of images to a directory. the code works great in case of just one picture will be uploaded. if i like to upload two images or more and have an extension that is not accepted, the script will upload the one with the extension that is allowed to upload and shows the errormessage for the one who is not accepted. but the upload takes place. that's my first problem. second problem will be: in case of an errormessage i would like to display a message in which it is said, which of the images will be not allowed. i do not know how to fetch this one that has an unaccepted ending into a variable that i can echo to the errormessage. here is the code i use: if(!empty($_FILES['image']['tmp_name'])){ $allowed_extension = array('jpg', 'jpeg', 'png', 'bmp', 'tiff', 'gif'); foreach($_FILES['image']['name'] as $key => $array_value){ $file_name = $_FILES['image']['name'][$key]; $file_size = $_FILES['image']['size'][$key]; $file_tmp = $_FILES['image']['tmp_name'][$key]; $file_extension = strtolower(end(explode('.', $file_name))); if (in_array($file_extension, $allowed_extension) === false){ $errors[] = 'its an unaccepted format in picture $variable_that_count'; continue; } if ($file_size > 2097152){ $errors[] = 'reached maxsize of 2MB per file in picture $variable_that_count'; } if (count($errors) == 0){ $path = "a/b/c/"; $uploadfile = $path."/".basename($_FILES['image']['name'][$key]); if (move_uploaded_file($_FILES['image']['tmp_name'][$key], $uploadfile)){ echo "success."; } } } } hope it will be clear what i like to approach. if there is someone who could help out i really would appreciate. thanks a lot.

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  • PHP export to text file - Only saving first line.

    - by wertz8090
    I'm trying to export some extracted $_POST information into a text file, however my code is only capturing the first variable and ignoring the rest. I'm saving the information to the text file in this manner: $values = "First Name: $fName\r\n"; $values .= "Last Name: $lName\r\n"; $values .= "Address: $address\r\n"; etc. This is the code I use to write to the text file: $fp = @fopen("person.data", "w") or die("Couldn't open person.data for writing!"); $numBytes = @fwrite($fp, $values) or die("Couldn't write values to file!"); @fclose($fp); Any ideas on why it would only save the first $values ($fName) variable but not the rest of them? It actually saves the first part of the $values string for all of them (so I see Last Name:, Address:, etc. on separate lines in the text file) but the called variables $lName and $address do not appear.

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  • duplicate record

    - by user349953
    Insert into Attendancemst ( emp_code , name, date , timetable , on_duty,out_duty,clockin , clockout, late, early, mis_in , mis_out , absent , halfday, total_time ) values (pemp_code,pname,pdate,ptimetable,pon_duty,pout_duty ,pclockin,pclockout,plate, pearly, pmis_in,pmis_out,pabsent,phalfday,ptotal_time )ON DUPLICATE KEY UPDATE emp_code=pemp_code and date = pdate;

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  • Moving from dedicated to shared cpanel - any scripts to do all / some of the install tasks ?

    - by mbbcat
    Hi, I have a few hundred phpld sites to move - each has its own cpanel, ( & the target may have shared cpanel) & I can do a full cpanel backup on the original server, but I don't have whm on the current host - the backups are fairly easy to organize but the installs so far means picking through files & setting up db's & mail etc by hand - I am thinking there ought to be an easier ie scripted way to do the installs or at least some parts - can anyone please suggest something ? I would like to migrate the stats at the same time Thanks M

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  • Connecting PHP Server and Android?

    - by user3439988
    I am trying to create a simple test application to transfer data back and forth between my server and Android device. The following are the things I aim for: Ability to upload data and files to the server. To be able to view my files on the server. To be able to download the files from the server to my android device. Ability of the server to send me updates on the files or notifications to my phone. I need a safe and secure way to do these things. I have tried these: HTTPPost requests onto the server and echoing the output accordingly and capturing the HTTPresponse and parsing it. For files I have tried using MultipartEntity, but I think that has been deprecated.

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