Shanti C

correct this line to (==) [icode] if ($username == 'letmein') { [/icode] = means assignment operator. == means comparision operator.

if you are listing all clients names from members table then pass member table id as parameter in java script on click function, which is not difficult. like [CODE]<a href="javascript:showuser('<?php=$clicnt_id;?>');"><?php=$client_name;?></a>[/CODE] The [ICODE]$client_id[/ICODE] and [ICODE]$client_name[/ICODE] are the variables contains proper values fetched from database table. If this is not your requirement …

read this article: [url]http://molendijk.110mb.com/include_menu_tutorial_passing_data/[/url] [url]http://www.cryer.co.uk/resources/javascript/script8.htm?Something=3.%20This%20works%20too[/url]

If you want to update a row in your table , then come up with that table's primary key. based on that id you can update your table and refresh that page which contain table data. Means: add this line after your hearder in line number 25 : [ICODE]<td bgColor="#008080"><font …

Two points have to be noted here: 1. if you use else if conditional loop then it will go for only on condition for your input,then other matched inputs will not return strings. 2.decide your input manner, whether it is in comma separated or space separated or some thing else. …

by url rewriting (help of .htaccess file). read articles on this topic.

if [ICODE]if($critiria=="epitheto_iatrou")[/ICODE] this is true, then how would the following become true.. [ICODE]$sql="SELECT * FROM doctors WHERE epitheto='$critiria'";[/ICODE] if [ICODE]$critiria="epitheto_iatrou"[/ICODE] then [ICODE]epitheto==epitheto_iatrou[/ICODE], this never become true.. change your code into accordingly..

:: is for referencing static properties or methods of a class. -> is for referencing instance properties and methods. check line 4 for static is missing in first example.

use ajax function on drop down onchange.. read some [URL="http://www.skillfusion.com/articles/ajaxDropdown.php"]ajax tutorials[/URL]..

[QUOTE]What to do? When you see a post containing this Russian spam (or any spam for that matter), click the "flag bad post" button next to it and type something creative. [/QUOTE] ok, Sure..

[QUOTE]no body replies what happened? [/QUOTE] post your check login query here...

for error 1 : you need to close while loop brace }. for error 2 : switch case ends in line : 304. But you have written default in line no : 345. So remove unexpected default there. or re arrange your switch cases.

hi, i checked your code, your facility field contains 17,16,14,15. Means these values are not in sorted order. that is the problem.. sort the array while inserting the facility column. why because mysql IN() function works based on binary search. binary search needs to be searched array in sorted manner.

you can use some editors like : [url]http://www.openwebware.com/[/url]

>> check $row['al_id'] getting value or not. If it is same field id in both tables then you have to get it by alias name. and try to echo $row['al_id']; >> try with a static query string : [code] <a class="pics" href="javascript<b></b>:ajaxpage('jobs/list-jobs-detail.php&album=2', 'mainwindow');"> [/code] >> and when you use [B]GET[/B] …

yes, you can do this by php,ajax,mysql combination. check this link: [url]http://www.w3schools.com/ajax/ajax_aspphp.asp[/url]

try this: [code] <?php if(isset($_SESSION['userlevel']) || !empty($_SESSION['userlevel'])) { if ($_SESSION['userlevel'] == 2) { echo "ADMIN"; } else { header('Location:index.php?query=notadmin'); } } else { echo "session not exists.."; } ?> [/code]

check this link: [url]http://stackoverflow.com/questions/520237/how-do-i-expire-a-php-session-after-30-minutes[/url]

like this: [code] <select name="con_st" onChange="getData('current_page.php',this.value);" > [/code] in javascript [code] function getData(page,x) { window.location=page+"?dropdownid="+x; } [/code] in current_page.php write your code and with the query string $_GET['dropdownid'] like this: [code] if(isset($_GET['dropdownid'])) { $q="select * from your_tbl where your_field="+$_GET['dropdownid']; // your code here } [/code]

just intialize a session with username and logincounter like $_SESSION['usernmae'] and $_SESSION['logincounter'] for every try increment the logincounter session for the same username.. if its greater than 3 then dont' allow for login.. or post your code, we will check it out..

by this: [code] foreach($_SESSION as $directory){ for($i=0;$i<count($directory);$i++) { $r_values=$directory[$i]; foreach($r_values as $value) { echo $value;echo "<br>"; } } } [/code]

or create alias names for common fields like : [code] SELECT p.thumb as photo_thumb,....... FROM photo_album as p, video as v ORDER BY photo_album.p DESC LIMIT 10 [/code]

yes, you can do this by using sessions... try this: [code] session_start(); if(!isset($_SESSION['words'])) { session_register(); $_SESSION['words']=array(); } $iterator=count($_SESSION['words']); $_SESSION['words'][$iterator][] = wordCheck($lname, "lastname"); $_SESSION['words'][$iterator][] = wordCheck($addr, "address"); . . . . print_r($_SESSION['words']); [/code]

[url]http://www.daniweb.com/forums/thread252011.html[/url]

[url]http://www.wampserver.com/en/[/url]

Decide how many levels of categories and sub categories you need for your products. Search(Advanced search) a product is very important , end user to see the product easily. Need payment gateway to purchase online. Showing your products images , information user friendly . Reduce number of navigation pages while …

try this: [code] RewriteRule ^album/(.*)(/?)$ /viewAlbum.php?aid=$1 [/code]

yes, check your inserted id once.. [code] // Make sure query inserted user successfully echo mysql_insert_id(); if ( mysql_insert_id()!=0) { die("Error: User not added to database."); } else { // Redirect to thank you page. Header("Location: register.php?op=thanks"); [/code]

check this [URL="http://www.daniweb.com/forums/thread141776.html"]tutorial[/URL].. and also [URL="http://www.roseindia.net/jdbc/display-data-database.shtml"]this[/URL].

put this in process.php [code] <?php if($_SERVER['REQUEST_METHOD']== "POST") { $a = $_POST['a']; // Get the value submitted for "a" $b = $_POST['b']; // Get the value submitted for "b" if ($a > $b) { echo "a is greater than b"; } else { echo "a is greater than b"; } …

Hello all, I am working on a project where i save client machine path into my database. And if end user clicks download files button, then the files from ftp server have to be download in client machine directly. My application resides in web server. Please give me some suggestions …

[QUOTE]i does't work with return it still print Y_constructor_finished [/QUOTE] if you want to stop the execution at return statement, place break statement there.. [code] <?php class X { function X(){ if($_POST['break']=='yes') { //return; break; } } } class Y extends X{ function Y(){ parent::__construct(); print("Y_constructor_finished"); } } $_POST['break']=='yes'; $new_Y=new …

check this [URL="http://www.theserverside.com/discussions/thread.tss?thread_id=5444"]link[/URL].

[QUOTE]<frameset rows="50%,50%"> <frame noresize="noresize" src="limits.php" /> <frame noresize="noresize" src="limits.php" /> </frameset>[/QUOTE] this code is perfectly fine... check your limits.php whether there is any error... execute this page individually in your browser. thanks

you can get array of form field like this: [code] $uimages = $_POST['establishment_image']; foreach( $uimages as $v ) { // single image print $v; } [/code]

pass table name, fields as parameters to function declaration, and write db query in your function definition which is in class. try coding ...

use [URL="http://php.net/manual/en/function.mkdir.php"]mkdir[/URL] to make a directory. Give user login name for this directory. and you can store all uploaded images by this user in this directory by giving this directory path like: assume login name is stored in session $dirname=$_SESSION['login_name']; after creating directory.. save the uploaded images in this folder.. …

what is $result contains?

check this [URL="http://www.roseindia.net/answers/viewqa/Java-Beginners/9058-file-upload-in-jsp.html"]link[/URL] .

What data type you have specified for your points field in your database.. post your table structure and your query...

check this [URL="http://www.daniweb.com/forums/thread51208.html"]thread[/URL]

Hello All, I am new to JSP. Currently i am working on jsp project with mysql. I got stucked with jsp date and time which supports mysql datetime format. [I need to do more time calculations in my project.] Please come up with your suggestions. Thanks in Advance. SHANTI.

$data is multi-dimensional array.. try to print like this: [code] print_r($data[0]); print_r($data[7]); [/code]

use update query to replace that string. Assume photos is your table and imageurl is your field.. Take backup of your table before update all these.. try this: [code] update photos set imageurl = replace(imageurl,'http://domain.com/','') [/code] If you dont' want to update in your database.. then try some php string …

use explode function.. Assume $_POST['textareaname'] is your form textarea name. see this: [code] $names = explode(",", $_POST['textareaname']); print_r($names ); foreach($names as $key=>$value) { echo $value; } [/code]

Try this: [code] RewriteEngine On RewriteRule ^(.*)$ /index.php?page=$1 [L] [/code] The link in the url will not be changed its self, you should have to give [url]http://example.com/images[/url] instead of [url]http://example.com/index.php?page=images[/url] all in your application pages... So that end user can access with your desired link.

[QUOTE]For some reason, it acts like the session variable was not created. My current output: (domain/index.php?page=portfolio&view=#) Quote ... ERROR: NULL VARIABLE My expected output: (domain/index.php?page=portfolio&view=#) Quote ... TEST [/QUOTE] First test your session is created or not.. like this: [code] session_start(); if($_GET['view'] == 0){ // URL would be domain/index.php?page=portfolio&view=0 $_SESSION['catImages'] …

First , this thread is not at all related to PHP, and many codes available in google for this.. Just try with some code and come here for suggestions.. the below is the code for disabling a button: [code] document.myform.mybutton.disabled = true; [/code] Write onclick function for your form checkbox …

remove slash after file name which comes from $post–>post_name;

[QUOTE]if($count==1) { session_register("txtusername"); session_register("txtpassword"); header("location:Main_Dashboard.php"); }[/QUOTE] You have written session_register("txtusername"); But no where assigning value in it like $_SESSION['txtusername']="name"; in logout page: [QUOTE]unset($_SESSION['username']);[/QUOTE] where do you defined $_SESSION['username'] in your code.. If you got any errors, plz post here...