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  • Returning mySQL error with jQuery & AJAX

    - by kel
    I've got a form inserting data into mySQL. It works but I'm trying to add error handling in case something happens. If I break the Insert statements mySQL dies but I'm still getting a success message on the front end. What am I doing wrong? AJAX function postData(){ var employeeName = jQuery('#employeeName').val(); var hireDate = jQuery('#hireDate').val(); var position = jQuery('#position').val(); var location = jQuery('#location').val(); var interveiwer = jQuery('#interviewersID').val(); var q01 = jQuery('#q01').val(); var q02 = jQuery('#q02').val(); var q03 = jQuery('#q03').val(); var q04 = jQuery('#q04').val(); var q05 = jQuery('#q05').val(); var summary = jQuery('#summary').val(); jQuery.ajax({ type: 'POST', url: 'queryDay.php', data: 'employeeName='+ employeeName +'&hireDate='+ hireDate +'&position='+ position +'&location='+ location +'&interveiwer='+ interveiwer +'&q01='+ q01 +'&q02='+ q02 +'&q03='+ q03 +'&q04='+ q04 +'&q05='+ q05 +'&summary='+ summary, success: function(){ jQuery('#formSubmitted').show(); }, error: function(jqXHR, textStatus, errorThrown){ jQuery('#returnError').html(errorThrown); jQuery('#formError').show(); } }); }; PHP require_once 'config.php'; $employeeName = $_POST['employeeName']; $hireDate = $_POST['hireDate']; $position = $_POST['position']; $location = $_POST['location']; $interviewerID = $_POST['interveiwer']; $q01 = $_POST['q01']; $q02 = $_POST['q02']; $q03 = $_POST['q03']; $q04 = $_POST['q04']; $q05 = $_POST['q05']; $summary = $_POST['summary']; mysql_query("INSERT INTO employee (name, hiredate, position, location) VALUES ('$employeeName', '$hireDate', '$position', '$location')") or die (mysql_error()); $employeeID = mysql_insert_id(); mysql_query("INSERT INTO day (employee, interviewer, datetaken, q01, q02, q03, q04, q05, summary) VALUES ('$employeeID', '$interviewerID', NOW(), '$q01', '$q02', '$q03', '$q04', '$q05', '$summary')") or die (mysql_error());

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  • MySQl - update field by counting data in other table

    - by Qiao
    There are two tables. One is users info "users", one is comments info "comments". I need to create new field "comments" in users table, that contains number of comments of that user. Table "comments" has "user" field with user's id of that comment. What is optimal way to count number of comments by every user? With php you should write script that selects every user and than count number of his comments and then update "comments" field. It is not hard for me, but boring. Is it possible to do it without php, only in MySQL?

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  • Sign Up/Login system for multi-domain/multi-server

    - by David
    I'm thinking about a good solution for implementing a sign up/login system that works across different domains and servers. A working example is Olx (you can register in one domain, and your login will work in the rest of domains). The scenario is that every domain (one per country) has its own database. And there will be 2 servers (for example), each one will have the 50% of the domains (and so the 50% of databases). What would you suggest to start with? Database: MySQL 5.1 Server-side language: PHP 5.3 (I will be using Symfony 1.4, so if someone has some suggestion for this framework, it will be interesting too, although it is optional)

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  • drop down and post data to data base

    - by DAFFODIL
    This is a form which retrieves data from db and displays them in table. At the beginning of each row there will be a check box. If there are 10 rows fetched, I ii check 5 rows and insert them in to diff db but here when, I click drop down box data is getting in to db automatically,bcoz I use onchange event. Any alternative to prevent this to happen. Data should be inserted only when, I click submit button. Any help will be appreciated <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("form1", $con); error_reporting(E_ALL ^ E_NOTICE); $nam=$_REQUEST['select1']; $row=mysql_query("select * from inv where name='$nam'"); while($row1=mysql_fetch_array($row)) { $Name=$row1['Name']; $Address =$row1['Address']; $City=$row1['City']; $Pincode=$row1['Pincode']; $No=$row1['No']; $Date=$row1['Date']; $DCNo=$row1['DCNo']; $DcDate=$row1['DcDate']; $YourOrderNo=$row1['YourOrderNo']; $OrderDate=$row1['OrderDate']; $VendorCode=$row1['VendorCode']; $SNo=$row1['SNo']; $descofgoods=$row1['descofgoods']; $Qty=$row1['Qty']; $Rate=$row1['Rate']; $Amount=$row1['Amount']; } ?> <!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=iso-8859-1" /> <title>Untitled Document</title> <script type="text/javascript"> function ram(id) { var q=document.getElementById('qty_'+id).value; var r=document.getElementById('rate_'+id).value; document.getElementById('amt_'+id).value=q*r; } </script> </head> <body> <form id="form1" name="form1" method="post" action=""> <table width="1315" border="0"> <script type="text/javascript"> function g() { form1.submit(); } </script> <tr> <th>Name</th> <th align="left"><select name="select1" onchange="g();"> <option value="" selected="selected">select</option> <?php $row=mysql_query("select Name from inv "); while($row1=mysql_fetch_array($row)) { ?> <option value="<?php echo $row1['Name'];?>"><?php echo $row1['Name'];?></option> <?php } ?> </select></th> </tr> <tr> <th>Address</th> <th align="left"><textarea name="Address"><?php echo $Address;?></textarea></th> </tr> <tr> <th>City</th> <th align="left"><input type="text" name="City" value='<?php echo $City;?>' /></th> </tr> <tr> <th>Pincode</th> <th align="left"><input type="text" name="Pincode" value='<?php echo $Pincode;?>'></th> </tr> <tr> <th>No</th> <th align="left"><input type="text" name="No2" value='<?php echo $No;?>' readonly="" /></th> </tr> <tr> <th>Date</th> <th align="left"><input type="text" name="Date" value='<?php echo $Date;?>' /></th> </tr> <tr> <th>DCNo</th> <th align="left"><input type="text" name="DCNo" value='<?php echo $DCNo;?>' readonly="" /></th> </tr> <tr> <th>DcDate:</th> <th align="left"><input type="text" name="DcDate" value='<?php echo $DcDate;?>' /></th> </tr> <tr> <th>YourOrderNo</th> <th align="left"><input type="text" name="YourOrderNo" value='<?php echo $YourOrderNo;?>' readonly="" /></th> </tr> <tr> <th>OrderDate</th> <th align="left"><input type="text" name="OrderDate" value='<?php echo $OrderDate;?>' /></th> </tr> <tr> <th width="80">VendorCode</th> <th width="1225" align="left"><input type="text" name="VendorCode" value='<?php echo $VendorCode;?>' readonly="" /></th> </tr> </table> <table width="1313" border="0"> <tr> <td width="44">&nbsp;</td> <td width="71">SNO</td> <td width="527">DESCRIPTION</td> <td width="214">QUANTITY</td> <td width="214">RATE/UNIT</td> <td width="217">AMOUNT</td> </tr> <?php $i=1; $row=mysql_query("select * from inv where Name='$nam'"); while($row1=mysql_fetch_array($row)) { $SNo=$row1['SNo']; $descofgoods=$row1['descofgoods']; $Qty=$row1['Qty']; $Rate=$row1['Rate']; $Amount=$row1['Amount']; ?> <tr> <td><input type="checkbox" name="checkbox" value="checkbox" checked="checked"/></td> <td><input type="text" name="No[<?php echo $i?>]" value='<?php echo $SNo;?>' readonly=""/></td> <td><input type="text" name="descofgoods[<?php echo $i?>]" value='<?php echo $descofgoods;?>' /></td> <td><input type="text" name="qty[<?php echo $i?>]" maxlength="50000000" id="qty_<?PHP echo($i) ?>"/></td> <td><input type="text" name="Rate[<?php echo $i?>]" value='<?php echo $Rate;?>' id="rate_<?PHP echo($i) ?>" onclick="ram('<?PHP echo($i) ?>')";></td> <td><input type="text" name="Amount[<?php echo $i?>]" id="amt_<?PHP echo($i) ?>"/></td> </tr> <?php $i++;} ?> <tr> <td><input type="submit" value="submit" header("location:values to be brought for print page.php");/></td> </tr> </table> <label></label> </form> </body> </html> <?php /*error_reporting(E_ALL ^ E_NOTICE); $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("form1", $con); /*if(checked=checkbox) { mysql_query="INSERT INTO invo (Name, Address, City, Pincode, No, Date, DCNo, DcDate, YourOrderNo, OrderDate, VendorCode, SNo, descofgoods, Qty, Rate, Amount) VALUES ('$_POST[Name]','$_POST[Address]','$_POST[City]','$_POST[Pincode]','$_POST[No]','$_POST[Date]','$_POST[DCNo]','$_POST[DcDate]','$_POST[YourOrderNo]','$_POST[OrderDate]','$_POST[VendorCode]','$_POST[SNo]','$_POST[descofgoods]','$_POST[qty]','$_POST[Rate]','$_POST[Amount]')"; } else { header("location:values to be brought for print page.php"); }*/ header("ins.php"); ?>

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  • Should I be concerned about performance with connections to multiple databases?

    - by Josh Ryan
    I have a PHP app that is running on an Apache server with MySQL databases. Based on the subdomain that users access, I am connecting them to a database (sub1.domain.com connects to database_sub1 and sub2.domain.com connects to database_sub2). Right now there are 10 subdomain-database combos, but that number could potentially grow to well over 100. So, is this a bad thing? Considering my situation, is myslq_pconnect a better way to go? Thanks, and please let me know if more info would be helpful. Josh

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  • magic records being deleted

    - by chris
    i have customised oscommerce to pull in a csv file of products, delete anything thats not with an image/proper description/proper title gets removed. The import runs on a cron job basis pulling information from a supplier, it hasnt run since yesterday but a product has disappeared- Anyone who has used oscommerce will know that, product information is stored over multiple tables. example is- products product_description and so on. the thing that has got me that the information is deleted from the product table but not from the product_description table. The product that is being deleted is a manually input one which carries a special tag/prefix on the model item of the product table. Therefore shouldn't be touched at all. Am clueles as what is going on. Is there mysql integrity checks deleting records? could there be another plugin working on oscommerce?

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  • Order database results by bayesian rating

    - by One Trick Pony
    I'm not sure this is even possible, but I need a confirmation before doing it the "ugly" way :) So, the "results" are posts inside a database which are stored like this: the posts table, which contains all the important stuff, like the ID, the title, the content the post meta table, which contains additional post data, like the rating (this_rating) and the number of votes (this_num_votes). This data is stored in pairs, the table has 3 columns: post ID / key / value. It's basically the WordPress table structure. What I want is to pull out the highest rated posts, sorted based on this formula: br = ( (avg_num_votes * avg_rating) + (this_num_votes * this_rating) ) / (avg_num_votes + this_num_votes) which I stole form here. avg_num_votes and avg_rating are known variables (they get updated on each vote), so they don't need to be calculated. Can this be done with a mysql query? Or do I need to get all the posts and do the sorting with PHP?

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  • search for a winner, returning their rank

    - by incrediman
    Earlier I asked this question, which basically asked how to list 10 winners in a table with many winners, according to their points. This was answered. Now I'm looking to search for a given winner X in the table, and find out what position he is in, when the table is ordered by points. For example, if this is the table: Winners: NAME:____|__POINTS: Winner1 | 1241 Winner2 | 1199 Sally | 1000 Winner4 | 900 Winner5 | 889 Winner6 | 700 Winner7 | 667 Jacob | 623 Winner9 | 622 Winner10 | 605 Winner11 | 600 Winner12 | 586 Thomas | 455 Pamela | 434 Winner15 | 411 Winner16 | 410 These are possible inputs and outputs for what I want to do: Query: "Sally", "Winner12", "Pamela", "Jacob" Output: 3 12 14 623 How can I do this? Is it possible, using only a MySQL statement? Or do I need PHP as well? This is the kind of thing I want: WHEREIS FROM Winners WHERE Name='Sally' LIMIT 1 Ideas?

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  • Need help on how to begin learning website development

    - by Golfy
    OK for the past 10 days I have been trying to figure out where to begin learning website development a dynamic one, So far everyone told me you should start with computer science and I am like What the @#$% how am I suppose to learn computer science without going to school and get a degree but I don't want a degree I just want to learn how to devlope websites. So now I am here and confused about how to put together a website. I get HTML and CSS but still have some problems designing the site now on the other had I am having trouble trying to figure out how php and database mysql is used to put together a website. I have seen videos from lynda.com and still have no Idea after I have watched the video the basic one, one that teach you the Variables, Loops, Strings ext... ok than what happens, how do you build the website with it, that is the question I am real not understand the answer to. Any help will be appreciated.

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  • What is wrong with my SQL syntax here?

    - by CT
    New to SQL. I'm looking to create a IT asset database. Here is one of the tables created with php: mysql_query("CREATE TABLE software( id VARCHAR(30), PRIMARY KEY(id), software VARCHAR(30), key VARCHAR(30))") or die(mysql_error()); echo "Software Table Created.</br />"; This is the output from the browser when I run the script: 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 'VARCHAR(30))' at line 5 I am running a standard LAMP stack on Ubuntu Server 10.04. Thank you.

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  • Get a DB result with a value between two column values

    - by vitto
    Hi, I have a database situation where I'd like to get a user profile row by a user age range. this is my db: table_users username age email url pippo 15 [email protected] http://example.com pluto 33 [email protected] http://example.com mikey 78 [email protected] http://example.com table_profiles p_name start_age_range stop_age_range young 10 29 adult 30 69 old 70 inf I use MySQL and PHP but I don't know if there is some specific tacnique to do this and of course if it's possible. # so something like: SELECT * FROM table_profiles AS profiles INNER JOIN table_users AS users # can I do something like this? ON users.age IS BETWEEN profiles.start_age_range AND profiles.stop_age_range

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  • Improve performance of website

    - by Vinodtiru
    Hi, I have designed a new web site. I have hosted it online. I want it to be of the best performance and load pages faster. This website is designed in php 5.0+ using codeigniter. This is using mysql as DB. I have images on it. I am using Nitobi grid for displaying set of records on page. The rest is everything normal page controls. As i am not so very experienced with website performance factors i would like to get suggestions and details on factors that can improve performance of website. Please let me know how i can improve my performance. Also please let me know if there are any ways to measure the performance of website and also any websites or tools to help test the performance. Any kind of help is appreciated. Thanks in advance. Thanks and Regards Vinod T.

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  • How can I make my search more broad?

    - by user1804952
    I created this search mysql string and it is to literal or maybe un-literal? If I search for Dave for example it will only find items like "Dave something" and not find Dave. $queryArtist = mysql_query("SELECT * FROM artists WHERE artist LIKE '%$ArtistNameSearch%' ORDER BY artist ASC"); I know mysql_query is out dated and will change it to mysqli soon as I get this worked out. Stuck here. An example of it no working is Example of search Could it be becasue the %20 space? I got it figured out, but it still does NOT find one direction or other things even those exist. here is what I have now $queryArtist = mysql_query("SELECT * FROM artists WHERE match(artist) against('$SafeSearchTerm' in boolean mode)");

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  • How to insert multiple check-box values inside database when one or more will be left unchecked?

    - by Sally
    I have a form that contains 5 check boxes. The user may select one or more of these check boxes. The user may select 2 and leave 3 unchecked or select 4 and leave one unchecked and so on, in that case how can I write the php/mysql code that will insert the form data into the database. With just one selection it's easy, I would do: $checkbox_value = $_POST['i_agree']; mysql_query("INSERT INTO terms (user, pass, conditions) VALUES ('$user','$pass','$checkbox_value')"); But how can I write this when there are multiple check box options and only one or more of them will be checked? I want to insert them all in one column called "tags" separated by commas.

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  • What choices to make for an application backend

    - by Saif Bechan
    I am creating an web application and I at the point that i am starting to make backend choices. Now there are a lot of ways to go with this, so I am looking for some good points and back practices. Some of the question i have involve: Should i make a seperate table in the db for admin users Should i extend make some classes to load the admin data and the normal data, or make seperate classes for the admin section Where can i get some information on making different types of users Just some best practices for a backend My application is written in PHP with an MySQL database.

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  • Creating a Variable of Present Date Minus a Past Timestamp

    - by John
    Hello, In the code below, "created" is a field in a MySQL table. This field is of the type "timestamp" and the default is set to "CURRENT_TIMESTAMP" of whenever a given row is created. In the query below, I would like to create a new variable that equals the present date minus the timestamp of "created", rounded off to units of days. I would like the present date to be whenever the query is run. How could I do this? Thanks in advance, John $sqlStr = "SELECT l.loginid, l.username, l.created, ...

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  • using jquery to load data from mysql database

    - by Ieyasu Sawada
    I'm currently using jquery's ajax feature or whatever they call it. To load data from mysql database. Its working fine, but one of the built in features of this one is to load all the data which is on the database when you press on backspace and there's no character left on the text box. Here's my query: SELECT * FROM prod_table WHERE QTYHAND>0 AND PRODUCT LIKE '$prod%' OR P_DESC LIKE '$desc%' OR CATEGORY LIKE '$cat%' As you can see I only want to load the products which has greater than 0 quantity on hand. I'm using this code to communicate to the php file which has the query on it: $('#inp').keyup(function(){ var inpval=$('#inp').val(); $.ajax({ type: 'POST', data: ({p : inpval}), url: 'querys.php', success: function(data) { $('.result').html(data); } }); }); Is it possible to also filter the data that it outputs so that when I press on backspace and there's no character left. The only products that's going to display are those with greater than 0 quantity?

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  • Advanced search functionality

    - by Chris
    I have a website with a jQuery based autocomplete search functionality which works great. Currently though I have just one search box for all categories, what I want is for someone to be able to type in, say for example, dorian gray dvd (in any order) which will search for dorian gray within the dvd category. What this will require then is a bit of magic on the server side to figure out if any of the words are category keywords, and then limit the search by that. What is the best (and quickest) way to do this in PHP / MySQL? I currently have a few trains of thought Search the category table for matches and perhaps order the results by that. Or split up the search terms into an array and separately search the categories for that for a match. Another thought I just had is to concat the category title to the dvd title in the database and match against that, or something similar... but this sounds computationally expensive? Any advice?

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  • How to order results based on number of search term matches?

    - by Travis
    I am using the following tables in mysql to describe records that can have multiple searchtags associated with them: TABLE records ID title desc TABLE searchTags ID name TABLE recordSearchTags recordID searchTagID To SELECT records based on arbitrary search input, I have a statement that looks sort of like this: SELECT recordID FROM recordSearchTags LEFT JOIN searchTags ON recordSearchTags.searchTagID = searchTags.ID WHERE searchTags.name LIKE CONCAT('%','$search1','%') OR searchTags.name LIKE CONCAT('%','$search2','%') OR searchTags.name LIKE CONCAT('%','$search3','%') OR searchTags.name LIKE CONCAT('%','$search4','%'); I'd like to ORDER this resultset, so that rows that match with more search terms are displayed in front of rows that match with fewer search terms. For example, if a row matches all 4 search terms, it will be top of the list. A row that matches only 2 search terms will be somewhere in the middle. And a row that matches just one search term will be at the end. Any suggestions on what is the best way to do this? Thanks!

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  • Is there a way to query if array field contains a certain value in Doctrine2?

    - by dpimka
    Starting out with Symfony2 + Doctrine. I have a table with User objects (fos_user), for which my schema contains a roles column of an 'array' type. Doctrine saves fields of this type by serializing them from php 'array' to 'longtext' (in mysql's case). So let's say I have the following users saved into DB: **User1**: array(ROLE_ADMIN, ROLE_CUSTOM1) **User2**: array(ROLE_ADMIN, ROLE_CUSTOM2) **User3**: array(ROLE_CUSTOM2) Now in my controller I want to select all users with ROLE_ADMIN set. Is there a way to write a DQL query which would directly return me User1 and User2? Or do I need to fetch all users to have Doctrine to unserialize roles column and then for each of them do in_array('ROLE_ADMIN', $user-getRoles())? I have searched the DQL part of the manual, but so far did not find anything similar to my needs...

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  • How to use avg function?

    - by Marcelo
    I'm new at php and mysql stuff and i'm trying to use an avg function but i don't know how to. I'm trying to do something like this: mysql_connect(localhost,$username,$password); @mysql_select_db($database) or die ("Did not connect to $database"); mysql_query("AVG(column1) FROM table1 ") or die(mysql_error()); mysql_close(); echo AVG(column1); (Q1)I'd like to see the value printed in the screen, but i'm getting nothing but an error message. How could I print this average on the screen ? (Q2)If I had a column month in my table1, how could I print the averages by the months ? Sorry for any bad English, and thanks for the attention.

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  • How to make notification to group users with read/unread flag?

    - by user2335065
    I am making a notification system so that when users in a group perform an action, it will notify all the other users in the group. I want the notification to be marked "read" or "unread" for each user of the group. With that, I can easily retreive any unread notification for a user and display it. I am think about creating a notification table that have the following fields. +----------------------+ | notification | +----------------------+ | id | | userid | | content | | status (read/unread) | | time | +----------------------+ My question is: Whether it is the correct way of making the system? As it means that when there is 1,000 users in a group, then I have to insert 1,000 rows to the table. If not, what is the better way of doing this? If it is the way to do this, how can I write the php/mysql codes to do the looping of inserting the rows? Thanks!

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  • SQL select all items of an owner from an item-to-owner table

    - by kdobrev
    I have a table bike_to_owner. I would like to select current items owned by a specific user. Table structure is CREATE TABLE IF NOT EXISTS `bike_to_owner` ( `bike_id` int(10) unsigned NOT NULL, `user_id` int(10) unsigned NOT NULL, `last_change_date` date NOT NULL, PRIMARY KEY (`bike_id`,`user_id`,`last_change_date`) ) ENGINE=MyISAM DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci; In the profile page of the user I would like to display all his/her current possessions. I wrote this statement: SELECT `bike_id`,`user_id`,max(last_change_date) FROM `bike_to_owner` WHERE `user_id` = 3 group by `last_change_date` but i'm not quite sure it works correctly in all cases. Can you please verify this is correct and if not suggest me something better. Using php/mysql. Thanks in advance!

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  • Set up lnux box for hosting a-z [apache mysql php ssl]

    - by microchasm
    I am in the process of reinstalling the OS on a machine that will be used to host a couple of apps for our business. The apps will be local only; access from external clients will be via vpn only. The prior setup used a hosting control panel (Plesk) for most of the admin, and I was looking at using another similar piece of software for the reinstall - but I figured I should finally learn how it all works. I can do most of the things the software would do for me, but am unclear on the symbiosis of it all. This is all an attempt to further distance myself from the land of Configuration Programmer/Programmer, if at all possible. I can't find a full walkthrough anywhere for what I'm looking for, so I thought I'd put up this question, and if people can help me on the way I will edit this with the answers, and document my progress/pitfalls. Hopefully someday this will help someone down the line. The details: CentOS 5.5 x86_64 httpd: Apache/2.2.3 mysql: 5.0.77 (to be upgraded) php: 5.1 (to be upgraded) The requirements: SECURITY!! Secure file transfer Secure client access (SSL Certs and CA) Secure data storage Virtualhosts/multiple subdomains Local email would be nice, but not critical The Steps: Download latest CentOS DVD-iso (torrent worked great for me). Install CentOS: While going through the install, I checked the Server Components option thinking I was going to be using another Plesk-like admin. In hindsight, considering I've decided to try to go my own way, this probably wasn't the best idea. Basic config: Setup users, networking/ip address etc. Yum update/upgrade. Upgrade PHP: To upgrade PHP to the latest version, I had to look to another repo outside CentOS. IUS looks great and I'm happy I found it! cd /tmp #wget http://dl.iuscommunity.org/pub/ius/stable/Redhat/5/x86_64/epel-release-1-1.ius.el5.noarch.rpm #rpm -Uvh epel-release-1-1.ius.el5.noarch.rpm #wget http://dl.iuscommunity.org/pub/ius/stable/Redhat/5/x86_64/ius-release-1-4.ius.el5.noarch.rpm #rpm -Uvh ius-release-1-4.ius.el5.noarch.rpm yum list | grep -w \.ius\. [will list all packages available in the IUS repo] rpm -qa | grep php [will list installed packages needed to be removed. the installed packages need to be removed before you can install the IUS packages otherwise there will be conflicts] #yum shell >remove php-gd php-cli php-odbc php-mbstring php-pdo php php-xml php-common php-ldap php-mysql php-imap Setting up Remove Process >install php53 php53-mcrypt php53-mysql php53-cli php53-common php53-ldap php53-imap php53-devel >transaction solve >transaction run Leaving Shell #php -v PHP 5.3.2 (cli) (built: Apr 6 2010 18:13:45) This process removes the old version of PHP and installs the latest. To upgrade mysql: Pretty much the same process as above with PHP #/etc/init.d/mysqld stop [OK] rpm -qa | grep mysql [installed mysql packages] #yum shell >remove mysql mysql-server Setting up Remove Process >install mysql51 mysql51-server mysql51-devel >transaction solve >transaction run Leaving Shell #service mysqld start [OK] #mysql -v Server version: 5.1.42-ius Distributed by The IUS Community Project And this is where I'm at. I will keep editing this as I make progress. Any tips on how to Configure Virtualhosts for SSL, setting up a CA, setting up SFTP with openSSH, or anything else would be appreciated.

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  • share image with url from database

    - by LauroSkr
    In my PHP web site will have a text converted to jpeg file,so it is a image. I want that i give users option to share the image. every image will have unique url so they can share image on facebook,twitter , do i need to put images with their url in mysql database or you do it in cloud? user will write text and then script will convert it to image and show it as image. then i want to provide user that he/she can share their created image. It would be great if you could provide me with a link or tutorial for my problem. Dont be hard on beginners,you were all in the same boat.. Thanks, LauroSkr

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