kishan112

There is a problem with your query, check your database.. Maybe the field-names are incorrect or your table name [CODE=php]$query="SELECT e_mail,user_name FROM register WHERE register.e_mail = '$email_address'"[/CODE]

Hey guys, I've a problem. I want a layout that looks like the layout.jpg (left) attachment in this post. I use this css code: [CODE=css] .contleft {background:#f2f2e6; clear:left; float:left; padding:10px; border:1px solid #fff; font:14px/17px Arial, Helvetica, sans-serif; border-color:#fff #666661 #666661 #fff; text-shadow:none; margin-top:-25px; margin-bottom:36px; width:500px;} .contright {float: right; clear:right; background:#f2f2e6; …

You can put [CODE]php_flag display_errors off[/CODE] in your .htaccess file

Hi all, I have a script to built thumbnails for some images. It's working great, but when the file name for the original picture contains a single quote, it skips that picture. I use this: imagejpeg($result, "thumb/".$title.".jpg", 100) or die("Cant save image"); $title is the picture name, received direct from …

I have the following script to resize and crop images. I can use jpg and png files, but can't use tif files. Can you please help? [CODE=php] <?php function Image($image, $crop = null, $size = null) { $image = ImageCreateFromString(file_get_contents($image)); if (is_resource($image) === true) { $x = 0; $y = …

You must send the mail as HTML. Here is the code. Try it [CODE=php] <?php $to="o.oloyede@feeslimited.net"; $message="<table width='500' border='1' cellpadding='0' cellspacing='0' bordercolor='#CCCCCC'> <tr> <td><table width='500' border='0' cellspacing='0' cellpadding='8'> <tr> <td colspan='2' bgcolor='#CDD9F5'><strong>APPLICATION FOR $course ?></strong></td> </tr> <tr> <td width='168' bgcolor='#FFFFEC'><strong>PROGRAM:</strong></td> <td width='290' bgcolor='#FFFFEC'><? $course ?></td> </tr> <tr> <td bgcolor='#FFFFDD'><strong>PERIOD:</strong></td> <td …

Hey guys, I have the following table: persons - name - street - number - regioncode - state - city I want to count the amount of people per region per street. I can do it with the following query: [CODE=mysql] SELECT state, city, regioncode, street, COUNT(*) AS personsStreetRegion FROM …

I think you don't really understand how ajax works. Maybe you can take some tuts at [url]http://w3schools.com/[/url] I'm sure it'll solve your problem.

Something like this? [CODE=mysql]SELECT * FROM info WHERE address = ''[/CODE]

You can use php to replace the "," into a "." [CODE=php] $new_number = str_replace(",", $form_number, "."); [/CODE]

You can use the jQuery plugin lightbox2 [url]http://www.huddletogether.com/projects/lightbox2/[/url] You can set up an link to the image and when a user clicks on it, it'll then load and display the picture.. So this won't take to much bandwidth as it loads the pics only after clicking the links

You can try this: [CODE=php] <?php $sql = mysql_query("SELECT * FROM `candidate`"); while ($res = mysql_fetch_array($sql)) { ?> <?php echo $res['name']; ?> <input type='radio' name='president' value='<?php echo $res['can_id']; ?>'>[/CODE]

Hi, I'm working on a simple dating website with member profiles. I need to do something really simple, but I'm too tired to do this.. There is a profile table, login table and a profile_view table: profiles: profile_id, name, picture, etc.. login: session_id, username, time, etc.. profile_view view_id, profile_id, session_id …

It's something like this: [CODE=javascript]document.location('approved.php');[/CODE]

Maybe you can use this: [CODE=MYSQL] (SELECT ammount FROM track1) UNION (SELECT ammount FROM track2) UNION (SELECT ammount FROM track3) [/CODE]

You have to post your code first..

try this: [CODE=PHP] <?php mysql_connect("localhost","dbusername","dbpass"); mysql_select_db("databasename"); $sql = mysql_query("SELECT * FROM `tablename`"); $results = mysql_fetch_array($sql); print_r($results); ?> [/CODE] Is this what you want?

I don't know what you want exactly, but your html should look like this: [CODE=html] <select id="itemsTbx" name="originSel" onchange="checkOriginDest()"> <option value="B">Bacolod</option> <option value="C">Cebu</option> <option value="M">Manila</option> <option value="Z">Zamboanga</option> </select> [/CODE] If you want to get code in a variable you can use this: [CODE=javascript]var cod = document.getElementById('itemsTbx').value();[/CODE]

Can you give details of your database tables?

This is how it's done [CODE=php]if ($cvalue == "Germany" || $cvalue == "England" || $cvalue == "latvia" || $cvalue == "Africa") { echo "Do something" ; }[/CODE] Or you can use an array: [CODE=php] $countries = array("Germany","England","Latvia","Africa"); if (in_array($cvalue,$countries)) { echo "Do something" ; }[/CODE] Remember that in_array() is case-sensitive

You've to make another table ProductCategories containing: CategoryID, ProductID And table product should contain: ProductID, Product_Name, Product_Description table categories is the same: Category_ID, Category In your version of the products table there is functional dependency, so you have to normalise it.. Hope this will help you

You've to try something like this: [CODE=php] if ($row['DayRemind'] == '1') { $sel = "Selected"; } echo "<input type='checkbox' name='DayRemind' value='1' $sel />&nbsp;Day Reminder<br />";[/CODE]

In Line 6 of the query, I think you've to use IN, not ON

Hi, can anyone please rewrite the following sql-query? It's working, but it takes some time to execute. [CODE=sql] SELECT `cd_e_calls`.`bedrijfs_id` AS gegevens_id, `gebeld` , `status` FROM `cd_e_calls` , `cd_e_gegevens` , `cd_e_negatief` WHERE `cd_e_gegevens`.`RvV` != '1' AND `bedrijfs_id` = ( SELECT `cd_e_gegevens`.`id` AS gegevens_id FROM `cd_e_gegevens` WHERE ( SELECT COUNT( `cd_e_calls`.`bedrijfs_id` …

Here is my introduction: Name: Kishan Hair: Black Eyes: Brown Location: Suriname ;) Age: 23 Hobbies: Experimenting on my pc, music, going out with friends and family, sleeping, dancing Relationship Status: uhm.. Education: I'm on the university - Software Engineer (1st year) Work: System & Application administrator.. PHP programmer Languages …