karthik_ppts

mark this thread as solved if your problem solved

Go through this link [url]http://php.net/manual/en/function.odbc-connect.php[/url]

Posting this thread under web design->css/html will give more result.

posting your code with code tags will be helpful for us.

[CODE] select u.name, u.subscription_package, i.image_url from users u,userimages i where u.userid =i.userid; [/CODE]

assign stud_no value to the check box as follows echo "<input type='checkbox' name='check[]' value='".$row['stud_no']."'>"; and store this value in another table while submit. Use **[code]** tags to post your codes

try this [url]http://roshanbh.com.np/2008/01/populate-triple-drop-down-list-change-options-value-from-database-using-ajax-and-php.html[/url]

Replace the line 40 [CODE=php] echo "<tr><td bgcolor=\"white\">".$img."></td>";[/CODE] with [CODE=php]echo "<tr><td bgcolor=\"white\"><img src='foldername/$img'></td>";[/CODE] and no need of this line [CODE=php]echo "<br";[/CODE] if you are using <table> if you need change it as [CODE=php]echo "<br>";[/CODE]

Now what is the output of the above code?

where is the `from` clause in your select query?

Why you use comparison operator(==) on line 3? [CODE=php]$_SESSION['authenticated']==true; [/CODE] It should be [CODE=php]$_SESSION['authenticated']=true; [/CODE]

You can't use the style in js file. js file only accept javascript code. so cut off your style from menu.js

Mark this thread as solved if your problem solved. Dont use this quotes for field_name like 'status'. if you need, you can use like this `status`

Use ajax to call the xyz.php

Why can't you try this with your php file [CODE=php] ini_set("upload_max_filesize","50Mb"); ini_set("post_max_size","50Mb"); ini_set("max_execution_time","500"); ini_set("max_input_time","500"); [/CODE]

Replace this [CODE=php]<option value="<?php echo $row['sid'] ?>"><?php echo $row['name'] ?></option>[/CODE] with [CODE=php]<option value="<?php echo $row['sid'] ?>" <?php if($_POST["u_state"]==$row['sid']) echo "selected"; ?>><?php echo $row['name'] ?></option>[/CODE] for city [CODE]<option value="<?php echo $row['cityid'] ?>" <?php if($_POST["s_city"]==$row['cityid']) echo "selected"; ?>><?php echo $row['cityname'] ?></option>[/CODE]

Mark this thread as solved if your problem solved.

Replace your line 17 [CODE=php]$row = mysql_fetch_array($result) or die(mysql_error());[/CODE] with [CODE=php]$row = mysql_num_rows($result);[/CODE]

emclondon is correct. Chage your code as per the his suggestion. the mail will bw send only once.

In line 148 remove the single quotes from Thank's.. Please use code-tags to post your codes

Mark this thread as solved if your problem solved

In this line [CODE=php]<a href="franchiseDetails.php?city=<?php echo $row3['city'];?>">[/CODE] pass your record id instead of city as follows [CODE=php]<a href="franchiseDetails.php?id=<?php echo $row3['id'];?>">[/CODE]

try this [url]http://roshanbh.com.np/2008/01/populate-triple-drop-down-list-change-options-value-from-database-using-ajax-and-php.html[/url]

Post full code or a block of your code that contains the above line.

put [CODE=php]echo $file['name'];[/CODE] before that if statement [CODE=php]if (!in_array(end(explode(".", strtolower($file['name']))), $allowedExtensions))[/CODE] and see the result.

Mark this thread as solved if your problem solved

Which echo is not displaying? "helo" or "something?

Did the code return any error message?

Refer this and change your code [url]http://www.plus2net.com/php_tutorial/php_change_password.php[/url]

Post your validation code...