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  • Root users and mysql: `sudo mysql` vs `/root/.my.cnf`

    - by user67641
    I have a /root/.my.cnf file which stores the mysql root user's password: [client] password = "my password" When I log in as system root and enter mysql, I get a passwordless login: myuser@local:$ sudo su root@local:$ mysql mysql> But when I try to do the same just using sudo, I get access denied: myuser@local:$ sudo mysql ERROR 1045 (28000): Access denied for user 'root'@'localhost' (using password: NO) How can I get sudo mysql to log me in as the mysql root user, without entering a password?

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  • procmail don't execute php script

    - by Phliplip
    Hi, I have setup a kannel SMS gateway on my FreeBSD 7.2 - the service works great. I'm now trying to setup a email2sms feature. For this i have created a system user called kannel and all mails are forwarded to this user. In the home dir of kannel i have the following files. -rw-r--r-- 1 kannel kannel 81B 17 jan 09:50 .procmailrc lrwxr-x--- 1 root kannel 58B 14 jan 13:24 email2sms.php @ -> some-what-some-where -rw-rw-rw- 1 root kannel 5,8K 17 jan 09:52 log.email2sms -rw------- 1 kannel kannel 1,3K 17 jan 09:50 procmail.log -rw-r----- 1 root kannel 606B 14 jan 13:28 rawmail.txt The file email2sms.php is a symlink to the a php script (ZendFramework Application) that takes the email from STDIN, and uses ZendFramework to parse that mail into an object. It then do a http request to the SMS gateway. The php-script works. Content of .procmailrc LOGFILE=$HOME/procmail.log VERBOSE=yes :0 | php email2sms.php >> log.email2sms From last sent email i have this in procmail.log procmail: [97744] Mon Jan 17 09:50:40 2011 procmail: [97744] Mon Jan 17 09:50:40 2011 procmail: Assigning "LASTFOLDER= php email2sms.php >> log.email2sms" procmail: Executing " php email2sms.php >> log.email2sms" procmail: Notified comsat: "kannel@:/home/user/kannel/ php email2sms.php >> log.email2sms" From [email protected] Mon Jan 17 09:50:40 2011 Subject: asdf as Folder: php email2sms.php >> log.email2sms 2600 But there is no new output to log.email2sms, and the script should output the subject of the email. If i sudo as the kannel user and pipe a file with raw email to the script, it executes just fine. [root@webserver /home/user/kannel]# /home/user/kannel/ sudo -u kannel cat rawmail.txt | php email2sms.php >> log.email2sms And the command outputs to log.email2sms as desired. Any ideas guys?

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  • How to start mysql server

    - by Vineeth
    I installed mysql using yum install mysql on fedora 12. Now how do I start the mysql server? [root@localhost init.d]# which mysql /usr/bin/mysql [root@localhost init.d]# mysql --version mysql Ver 14.14 Distrib 5.1.46, for redhat-linux-gnu (x86_64) using readline 5.1 Please, help

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  • What is the best way to recover from a mysql replication fail?

    - by Itai Ganot
    Today, the replication between our master mysql db server and the two replication servers dropped. I have a procedure here which was written a long time ago and i'm not sure it's the fastest method to recover for this issue. I'd like to share with you the procedure and I'd appreciate if you could give your thoughts about it and maybe even tell me how it can be done quicker. At the master: RESET MASTER; FLUSH TABLES WITH READ LOCK; SHOW MASTER STATUS; And copy the values of the result of the last command somewhere. Wihtout closing the connection to the client (because it would release the read lock) issue the command to get a dump of the master: mysqldump mysq Now you can release the lock, even if the dump hasn't end. To do it perform the following command in the mysql client: UNLOCK TABLES; Now copy the dump file to the slave using scp or your preferred tool. At the slave: Open a connection to mysql and type: STOP SLAVE; Load master's data dump with this console command: mysql -uroot -p < mysqldump.sql Sync slave and master logs: RESET SLAVE; CHANGE MASTER TO MASTER_LOG_FILE='mysql-bin.000001', MASTER_LOG_POS=98; Where the values of the above fields are the ones you copied before. Finally type START SLAVE; And to check that everything is working again, if you type SHOW SLAVE STATUS; you should see: Slave_IO_Running: Yes Slave_SQL_Running: Yes That's it! At the moment i'm in the stage of copying the db from the master to the other two replication servers and it takes more than 6 hours to that point, isn't it too slow? The servers are connected through a 1gb switch.

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  • How to connect with MySQL server if it won't connect via the socket?

    - by cwd
    I have an account on a shared server. I have jailshell access and also PhpMyAdmin. I want to run mysql commands via SSH but I'm getting an error: $ mysql -u mySqlUser -p mySqlPw Can't connect to local MySQL server through socket '/var/lib/mysql/mysql.sock' I can connect with PHP and phpMyAdmin, so would it be possible to call mysql from the shell and have it connect via an ip and port instead of the socket? The file /var/lib/mysql/mysql.sock does not exist - maybe that is intentional, and the only thing in /etc/my.cnf is [mysqld] skip-innodb More Info I don't have access to change system settings. I did a search in /var for mysql.sock but found nothing. However, phpMyAdmin might be connecting via a socket somehow: Really it would just be great if I could connect via IP. Also tried these two syntaxes: $ mysql -u mySqlUser -p mySqlPw -h localhost $ mysql -u mySqlUser -p mySqlPw -h localhost -P 3306 Both with the same result: ERROR 2002 (HY000): Can't connect to local MySQL server through socket '/var/lib/mysql/mysql.sock' (2)

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  • I could use some help with my SQL command

    - by SuperSpy
    I've got a database table called 'mesg' with the following structure: receiver_id | sender_id | message | timestamp | read Example: 2 *(«me)* | 6 *(«nice girl)* | 'I like you, more than ghoti' | yearsago | 1 *(«seen it)* 2 *(«me)* | 6 *(«nice girl)* | 'I like you, more than fish' | now | 1 *(«seen it)* 6 *(«nice girl)* | 2 *(«me)* | 'Hey, wanna go fish?' | yearsago+1sec | 0 *(«she hasn't seen it)* It's quite a tricky thing that I want to achieve. I want to get: the most recent message(=ORDER BY time DESC) + 'contact name' + time for each 'conversation'. Contact name = uname WHERE uid = 'contact ID' (the username is in another table) Contact ID = if(sessionID*(«me)*=sender_id){receiver_id}else{sender_id} Conversation is me = receiver OR me = sender For example: From: **Bas Kreuntjes** *(« The message from bas is the most recent)* Hey $py, How are you doing... From: **Sophie Naarden** *(« Second recent)* Well hello, would you like to buy my spam? ... *(«I'll work on that later >p)* To: **Melanie van Deijk** *(« My message to Melanie is 3th)* And? Did you kiss him? ... That is a rough output. QUESTION: Could someone please help me setup a good SQL command. This will be the while loup <?php $sql = "????"; $result = mysql_query($sql); while($fetch = mysql_fetch_assoc($result)){ ?> <div class="message-block"> <h1><?php echo($fetch['uname']); ?></h1> <p><?php echo($fetch['message']); ?></p> <p><?php echo($fetch['time']); ?></p> </div> <?php } ?> I hope my explanation is good enough, if not, please tell me. Please don't mind my English, and the Dutch names (I am Dutch myself) Feel free to correct my English UPDATE1: Best I've got until now: But I do not want more than one conversation to show up... u=user table m=message table SELECT u.uname, m.message, m.receiver_uid, m.sender_uid, m.time FROM m, u WHERE (m.receiver_uid = '$myID' AND u.uid = m.sender_uid) OR (m.sender_uid = '$myID' AND u.uid = m.receiver_uid) ORDER BY time DESC;

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  • PHP-How to Pass Multiple Value In Form Field

    - by Tall boY
    hi i have a php based sorting method with drop down menu to sort no of rows, it is working fine. i have another sorting links to sort id & title, it is also working fine. but together they are not working fine. what happens is that when i sort(say by title) using links, result gets sorted by title, then if i sort rows using drop down menu rows get sorted but result gets back to default of id sort. sorting codes for id & tite is if ($orderby == 'title' && $sortby == 'asc') {echo " <li id='scurrent'><a href='?rpp=$rowsperpage&order=title&sort=asc'>title-asc:</a></li> ";} else {echo " <li><a href='?rpp=$rowsperpage&order=title&sort=asc'>title-asc:</a></li> ";} if ($orderby == 'title' && $sortby == 'desc') {echo " <li id='scurrent'><a href='?rpp=$rowsperpage&order=title&sort=desc'>title-desc:</a></li> ";} else {echo " <li><a href='?rpp=$rowsperpage&order=title&sort=desc'>title-desc:</a></li> ";} if ($orderby == 'id' && $sortby == 'asc') {echo " <li id='scurrent'><a href='?rpp=$rowsperpage&order=id&sort=asc'>id-asc:</a></li> ";} else {echo " <li><a href='?rpp=$rowsperpage&order=id&sort=asc'>id-asc:</a></li> ";} if ($orderby == 'id' && $sortby == 'desc') {echo " <li id='scurrent'><a href='?rpp=$rowsperpage&order=id&sort=desc'>id-desc:</a></li> ";} else {echo " <li><a href='?rpp=$rowsperpage&order=id&sort=desc'>id-desc:</a></li> ";} ?> sorting codes for rows is <form action="is-test.php" method="get"> <select name="rpp" onchange="this.form.submit()"> <option value="10" <?php if ($rowsperpage == 10) echo 'selected="selected"' ?>>10</option> <option value="20" <?php if ($rowsperpage == 20) echo 'selected="selected"' ?>>20</option> <option value="30" <?php if ($rowsperpage == 30) echo 'selected="selected"' ?>>30</option> </select> </form> this method passes only rows per page(rpp) into url. i want it to pass order, sort& rpp. is there a way around to pass multiple values in form fields like this. <form action="is-test.php" method="get"> <select name="rpp, order, sort" onchange="this.form.submit()"> <option value="10, $orderby, $sortby" <?php if ($rowsperpage == 10) echo 'selected="selected"' ?>>10</option> <option value="20, $orderby, $sortby" <?php if ($rowsperpage == 20) echo 'selected="selected"' ?>>20</option> <option value="30, $orderby, $sortby" <?php if ($rowsperpage == 30) echo 'selected="selected"' ?>>30</option> </select> </form> this may seem silly but it just to give you an idea of what i am trying to implement,(i am very new to php) please suggest any way to make this work. thanks

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  • php includes that include another php file

    - by Emma
    I'm getting really muddled up now and my brain hurts! :( lol Root: index.php Includes: cat.php dog.php index includes dog: include("includes/dog.php"); dog includes cat: include("cat.php"); When I run index, for cat it says: A link to the server could not be established Access denied for user ... However, if I run dog, I get no problems... I'm guessing its the path, but i've tried ./includes/cat.php to no joy...

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  • Updating Checked Checkboxes using CodeIgniter + MySQL

    - by Tim
    Hello I have about 8 check boxes that are being generated dynamically from my database. This is the code in my controller //Start Get Processes Query $this->db->select('*'); $this->db->from('projects_processes'); $this->db->where('process_enabled', '1'); $data['getprocesses'] = $this->db->get(); //End Get Processes Query //Start Get Checked Processes Query $this->db->select('*'); $this->db->from('projects_processes_reg'); $this->db->where('project_id', $project_id); $data['getchecked'] = $this->db->get(); //End Get Processes Query This is the code in my view. <?php if($getprocesses->result_array()) { ?> <?php foreach($getprocesses->result_array() as $getprocessrow): ?> <tr> <td><input <?php if($getchecked->result_array()) { foreach($getchecked->result_array() as $getcheckedrow): if($getprocessrow['process_id'] == $getcheckedrow['process_id']) { echo 'checked'; } endforeach; }?> type="checkbox" name="progresscheck[]" value="<?php echo $getprocessrow['process_id']; ?>"><?php echo $getprocessrow['process_name']; ?><br> </td> </tr> <?php endforeach; ?> This generates the checkboxes into the form and also checks the appropriate ones as specified by the database. The problem is updating them. What I have been doing so far is simply deleting all checkbox entries for the project and then re-inserting all the values into the database. This is bad because 1. It's slow and horrible. 2. I lose all my meta data of when the check boxes were checked. So I guess my question is, how do I update only the checkboxes that have been changed? Thanks, Tim

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  • stored procedure in MYsql access in PHP

    - by xcodemaddy
    Hi.. I am creating stored procedure ,here code : CREATE PROCEDURE `dbnm`.`getlogin` ( IN uid INT, IN upass VARCHAR(45) ) BEGIN if exists(select uphno,pass from user_master where uphno=uid and pass=upass)then ***true else ***false end if; END $$ DELIMITER ; i want return value(**true or **false) in stored procedure from PHP by calling sp PHP code: $res = $mysqli->query('call getlogin("1","rashmi")'); how to acesss boolean value in PHP from sp? Thanks

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  • PHP, MySQL, and temporary tables

    - by dave
    Php novice. 1.Is there anything wrong with this PHP & MySQL code? include_once "db_login.php" ; $sql = "DROP TEMPORARY TABLE IF EXISTS temp_sap_id_select" ; mysql_query ( $sql ) or ( "Error " . mysql_error () ) ; $sql = " CREATE TEMPORARY TABLE temp_sap_id_select ( `current_page` INT NOT NULL, `total_pages` INT NOT NULL, `select_date` DATE NOT NULL, `select_schcode` CHAR(6) NOT NULL, `select_user` CHAR(30) NOT NULL, `select_id` CHAR(9) NOT NULL ) " ; mysql_query ( $sql ) or ( "Error " . mysql_error () ) ; 2.Admittedly, I'm a dull boy, but I'll ask anyway: If I'm using a MySQL GUI and have open the target database, will it be aware of the above temporary table created by PHP/MySQL (IF the temporary table is properly created)?

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  • Database INSERT does not take place

    - by reggie
    My code is as follows: <?php include("config.php"); $ip=$_SERVER['REMOTE_ADDR']; if($_POST['id']) { $id=$_POST['id']; $id = mysql_escape_String($id); $ip_sql=mysql_query("select ip_add from Voting_IP where mes_id_fk='$id' and ip_add='$ip'"); $count=mysql_num_rows($ip_sql); if($count==0) { $sql = "update Messages set up=up+1 where mes_id='$id'"; mysql_query($sql); $sql_in = "insert into Voting_IP (mes_id_fk,ip_add) values ('$id','$ip')"; mysql_query($sql_in) or die(mysql_error()); echo "<script>alert('Thanks for the vote');</script>"; } else { echo "<script>alert('You have already voted');</script>"; } $result=mysql_query("select up from Messages where mes_id='$id'"); $row=mysql_fetch_array($result); $up_value=$row['up']; echo "<img src='button.png' width='110' height='90'>"; echo $up_value; } ?> My problem is that the insert process does not take place at all. The script tags echos an alert box. Even the img tag is echoed to the web page. But the insert process does not take place. The config file is fine. Note: This code works on my local machine which has PHP 5.3 but it does not work on the server which has PHP 5.2. Any advice?

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  • PHP database simulation

    - by Emdiesse
    I have a PHP script that works by calling items from a database based upon the time they were placed in there and it deletes them if they are older than 5 minutes. Basically, I want to now simulate what would happen if this database was being updated regularly. So I was considering sticking in some code that loads an XML file then goes through and parses that into the database based upon the time data located within a node of the xml data... but the problem there is I want it to continually loop through an enter this data so it'll never actually run the other processes So I was thinking of having another PHP script do that that could do this independantly of the php script that is going to display this data... In theory: I am looking to have a button that I can press and it will then run some php code to load up an XML file from a directory on my web server and then iterate though the data sending the data, to a database, based upon the time within a node in the PHP script and when the script was first called So back to my page that displayed the data... if I continually hit refresh it will contain different results each time because data is being added by the other process and this php script removes the older data when it is refreshed Any information on this? Is there a way I can silently, and safely, run a php script without it being loaded into a browser... like a thread!?

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  • including php file from another server with php

    - by ermac2014
    hi I have 2 php files each file on different server. lets say the first file called includeThis.php and the second called main.php the first file is located in (http://)www.sample.com/includeThis.php and the second located in (http://)www.mysite.com/main.php so now what I want is to include the first file into my second file. the contents of the first file is like: <?php $foo = "this is data from file one"; ?> the second file like: <?php include "http://www.sample.com/includeThis.php"; echo $foo; ?> is there any way I can do this? thanks in advance

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  • Get current PHP executable from within script?

    - by benizi
    I want to run a PHP cli program from within PHP cli. On some machines where this will run, both php4 and php5 are installed. If I run the outer program as php5 outer.php I want the inner script to be run with the same php version. In Perl, I would use $^X to get the perl executable. It appears there's no such variable in PHP? Right now, I'm using $_SERVER['_'], because bash (and zsh) set the environment variable $_ to the last-run program. But, I'd rather not rely on a shell-specific idiom. UPDATE: Version differences are but one problem. If PHP isn't in PATH, for example, or isn't the first version found in PATH, the suggestions to find the version information won't help. Additionally, csh and variants appear to not set the $_ environment variable for their processes, so the workaround isn't applicable there. UPDATE 2: I was using $_SERVER['_'], until I discovered that it doesn't do the right thing under xargs (which makes sense... zsh sets it to the command it ran, which is xargs, not php5, and xargs doesn't change the variable). Falling back to using: $version = explode('.', phpversion()); $phpcli = "php{$version[0]}";

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  • Mysql stored procedures

    - by Richard M
    Hello, I have a unique issue that I need advice on. I've been writing asp.net apps with SQL Server back ends for the past 10 years. During that time, I have also written some PHP apps, but not many. I'm going to be porting some of my asp.net apps to PHP and have run into a bit of an issue. In the Asp.net world, it's generally understood that when accessing any databases, using views or stored procedures is the preferred way of doing so. I've been reading some PHP/MySQL books and I'm beginning to get the impression that utilizing stored procedures in mysql is not advisable. I hesitate in using that word, advisable, but that's just the felling I get. So, the advice I'm looking for is basically, am I right or wrong? Do PHP developers use stored procedures at all? Or, is it something that is shunned? Thanks in advance.

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  • Extremely basic PHP and Mysql

    - by fighella
    Background: I am more of a designer than a programmer, but have hacked templates for many open source CMS's (Drupal, Joomla, Wordpress) I want to start from scratch in regards to the relations of php and a mysql database. Lets assume I have a working database and php engine locally. What would be my first step to connecting to my database and creating a table... (im happy to be led to an appropriate tutorial...) Many of the tutorials I have seen start with basic php, but I would rather explore the connection between the db and the php.

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  • mysql_real_escape_string() just makes an empty string?

    - by James P
    I am using a jQuery AJAX request to a page called like.php that connects to my database and inserts a row. This is the like.php code: <?php // Some config stuff define(DB_HOST, 'localhost'); define(DB_USER, 'root'); define(DB_PASS, ''); define(DB_NAME, 'quicklike'); $link = mysql_connect(DB_HOST, DB_USER, DB_PASS) or die('ERROR: ' . mysql_error()); $sel = mysql_select_db(DB_NAME, $link) or die('ERROR: ' . mysql_error()); $likeMsg = mysql_real_escape_string(trim($_POST['likeMsg'])); $timeStamp = time(); if(empty($likeMsg)) die('ERROR: Message is empty'); $sql = "INSERT INTO `likes` (like_message, timestamp) VALUES ('$likeMsg', $timeStamp)"; $result = mysql_query($sql, $link) or die('ERROR: ' . mysql_error()); echo mysql_insert_id(); mysql_close($link); ?> The problematic line is $likeMsg = mysql_real_escape_string(trim($_POST['likeMsg']));. It seems to just return an empty string, and in my database under the like_message column all I see is blank entries. If I remove mysql_real_escape_string() though, it works fine. Here's my jQuery code if it helps. $('#like').bind('keydown', function(e) { if(e.keyCode == 13) { var likeMessage = $('#changer p').html(); if(likeMessage) { $.ajax({ cache: false, url: 'like.php', type: 'POST', data: { likeMsg: likeMessage }, success: function(data) { $('#like').unbind(); writeLikeButton(data); } }); } else { $('#button_container').html(''); } } }); All this jQuery code works fine, I've tested it myself independently. Any help is greatly appreciated, thanks.

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  • I have a problem with mysql and php

    - by neo skosana
    Hi I have a problem, this is my code: $db = new mysqli("localhost", "root", "", "blah"); $result1 = $db-query("select * from c_register where email = '$eml' and password = '$pass'"); if($result1-fetch_array()) { $auth->createSession(); $_SESSION['user'] = 'client'; promptUser("You have successfully logged in!!!","index.php"); } $db = new mysqli("localhost", "root", "", "blah"); $result2 = $db-query("select * from b_register where email = '$eml' and password = '$pass'"); if($result2-fetch_array()) { $auth->createSession(); $_SESSION['user'] = 'business'; promptUser("You have successfully logged in!!!","index.php"); } $db = new mysqli("localhost", "root", "", "blah"); $result3 = $db-query("select * from g_register where email = '$eml' and password = '$pass'"); if($result3-fetch_array()) { $auth->createSession(); $_SESSION['user'] = 'employee'; promptUser("You have successfully logged in!!!","index.php"); } $db = new mysqli("localhost", "root", "", "blah"); $result4 = $db-query("select * from k_register where email = '$eml' and password = '$pass'"); if($result4-fetch_array()) { $auth->createSession(); $_SESSION['user'] = 'super'; promptUser("You have successfully logged in!!!","index.php"); } else { promptUser("Username/Password do not match. Please try again!!!",""); } Funny enough this code works, but I no that I went about it the wrong way. I am new with php and mysql, so please help. I also tried e.gresult4->free(); for all the variable that save the data, and I got this error: Fatal error: Call to a member function free() on a non-object in...

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  • MySQL thousands of updates, slowing down.

    - by noryb009
    I need to run a PHP loop for a total of 100, 000 times (about 10, 000 each script-run), and each loop has about 5 MySQL UPDATES to it. When I run the loop 50 times, it takes 3 sec. When I run the loop 1000 times, it takes about 1300 sec. As you can see, MySQL is slowing down ALOT with more UPDATEs. This is an example update: mysql_query("UPDATE table SET `row1`=`row1` +1 WHERE `UniqueValue`='5'"); This is generated randomly from PHP, but I can store it in a variable and run it every n loops. Is there any way to either make MySQL and PHP run at a consistent speed (is PHP storing hidden variables?), or split up the script so they do? Note: I am running this for a development purposes, not for production, so there will only be 1 computer accessing the data.

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  • Php + mysql transactions examples

    - by Donator
    I really haven't found normal example of php file where mysql transactions are being used. Can you show me simple example of that? And one more question. I've already created a lot of programming and didn't use transaction, maybe I can put any php function or smth to header.php that if one mysql_query fails, then others too? Thank you.

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  • mysql php problem: no error message despite error_reporting(E_ALL) line

    - by herrturtur
    index.php <html> <head> <title>Josh's Online Playground</title> </head> <body> <form method="POST" action="action.php"> <table> <tr> <td>"data for stuff"</td> <td><input type="text" ?></td> </tr> <tr> <td><input type="submit"></td> </tr> </table> </form> </body> </html> action.php <?php error_reporting(E_ALL); ini_sit("display_errors", 1); $mysqli = new mysqli('localhost', 'root', 'password', 'website'); $result = $mysqli->query("insert into stuff (data) values (' .$_POST['data'] ."'); echo $mysqli->error(); if($result = $mysqli->query("select data from stuff")){ echo 'There are '.$result->num_rows.' results.'; } while ($row = $result->fetch_object()){ echo 'stuff' . $row->data; } ?> Despite the first two lines in action.php, I get no error or warning messages. Instead I get a blank page after clicking the submit button. Do I have to do something differently to insert data?

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  • PHP+MYSQL Server Config

    - by Matias
    Hi guys, I am parsing an XML file with PHP and inserting the rows in a MYSQL database. I am using PHP simplexml_load_files to load the XML and a foreach to loop through the array and insert the rows into my database. It works perfectly fine with small files i am testing, but it comes to reality I need to parse a large 500mb XML file and nothing happens. I was wondering what was the right Php.ini config for this case ? I have a VPS Linux Cent OS, with 256 mb of dedicated Memory and MYSQL 5.0.5. I have also set php memory_limit = 256M (maximum of my server) Any suggestions, similar experiences will be greatly appreciated Thanks

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  • Form Submitting Incorrect Information to MySQL Database

    - by ThatMacLad
    I've created a form that submits data to a MySQL database but the Date, Time, Year and Month fields constantly revert to the exact same date (1st January 1970) despite the fact that when I submit the information to the database the form displays the current date, time etc to me. I've already set it so that the time and date fields automatically display the current time and date. Could someone please help me with this. Form: <html> <head> <title>Blog | New Post</title> <link rel="stylesheet" href="css/newposts.css" type="text/css" /> </head> <body> <div class="new-form"> <div class="header"> <a href="edit.php"><img src="images/edit-home-button.png"></a> </div> <div class="form-bg"> <?php if (isset($_POST['submit'])) { $month = htmlspecialchars(strip_tags($_POST['month'])); $date = htmlspecialchars(strip_tags($_POST['date'])); $year = htmlspecialchars(strip_tags($_POST['year'])); $time = htmlspecialchars(strip_tags($_POST['time'])); $title = htmlspecialchars(strip_tags($_POST['title'])); $entry = $_POST['entry']; $timestamp = strtotime($month . " " . $date . " " . $year . " " . $time); $entry = nl2br($entry); if (!get_magic_quotes_gpc()) { $title = addslashes($title); $entry = addslashes($entry); } mysql_connect ('localhost', 'root', 'root') ; mysql_select_db ('tmlblog'); $sql = "INSERT INTO php_blog (timestamp,title,entry) VALUES ('$timestamp','$title','$entry')"; $result = mysql_query($sql) or print("Can't insert into table php_blog.<br />" . $sql . "<br />" . mysql_error()); if ($result != false) { print "<p class=\"success\">Your entry has successfully been entered into the blog. </p>"; } mysql_close(); } ?> <?php $current_month = date("F"); $current_date = date("d"); $current_year = date("Y"); $current_time = date("H:i"); ?> <form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> <input class="field" type="text" name="date" id="date" size="2" value="<?php echo $current_month; ?>" /> <input class="field" type="text" name="date" id="date" size="2" value="<?php echo $current_date; ?>" /> <input class="field" type="text" name="date" id="date" size="2" value="<?php echo $current_year; ?>" /> <input type="text" name="time" id="time" size="5"value="<?php echo $current_time; ?>" /> <input class="field2" type="text" id="title" value="Title Goes Here." name="title" size="40" /> <textarea class="textarea" cols="80" rows="20" name="entry" id="entry" class="field2"></textarea> <input class="field" type="submit" name="submit" id="submit" value="Submit"> </form> </div> </div> </div> <div class="bottom"></div> </body> </html>

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  • MySQL Database Query - Codeigniter

    - by user2450349
    I am building an application with Codeigniter and need some help with a DB query. I have a table called users with the following fields: user_id, user_name, user_password, user_email, user_role, user_manager_id In my app, I pull all records from the user table using the following: function get_clients() { $this->db->select('*'); $this->db->where('user_role', 'client'); $this->db->order_by("user_name", "Asc"); $query = $this->db->get("users"); return $query->result_array(); } This works as expected, however when I display the results in the view, I also want to display a new column called Manager which will display the managers user_name field. The user_manager_id is the id of the user from the same table. Im guessing you can create an outer join on the same table but not sure. In the view, I am displaying the returned info as follows: <table class="table table-striped" id="zero-configuration"> <thead> <tr> <th>Name</th> <th>Email</th> <th>Manager</th> </tr> </thead> <tbody> <?php foreach($clients as $row) { ?> <tr> <td><?php echo $row['user_name']; ?> (<?php echo $row['user_username']; ?>)</td> <td><?php echo $row['user_email']; ?></td> <td><?php echo $row['???']; ?></td> </tr> <?php } ?> </tbody> </table> Any idea of how I can form the query and display the manager name is the view? Example: user_id user_name user_password user_email user_role user_manager_id 1 Ollie adjjk34jcd [email protected] client null 2 James djklsdfsdjk [email protected] client 1 When i query the database, i want to display results like this: Ollie [email protected] James [email protected] Ollie

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