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Member Avatar for phpDave

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]

Member Avatar for srilatha12
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222
Member Avatar for kishan112

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; …

Member Avatar for ko ko
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70
Member Avatar for BzzBee

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

Member Avatar for kishan112
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112
Member Avatar for kishan112

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 …

Member Avatar for kishan112
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96
Member Avatar for kishan112

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 = …

Member Avatar for kishan112
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115
Member Avatar for kayblack

You must send the mail as HTML. Here is the code. Try it [CODE=php] <?php $to="[email protected]"; $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 …

Member Avatar for cjohnweb
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6K
Member Avatar for kishan112

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 …

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Member Avatar for Mouse1989

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.

Member Avatar for kishan112
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116
Member Avatar for whiteyoh

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

Member Avatar for whiteyoh
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83
Member Avatar for jvestby

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

Member Avatar for kishan112
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67
Member Avatar for realnsleo

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

Member Avatar for realnsleo
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148
Member Avatar for takeshi

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]

Member Avatar for cwarn23
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135
Member Avatar for kishan112

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 …

Member Avatar for ppetree
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119
Member Avatar for shiv0013

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

Member Avatar for shiv0013
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140
Member Avatar for ktsangop

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

Member Avatar for ktsangop
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195
Member Avatar for aditi_19
Member Avatar for takeshi

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?

Member Avatar for harry_watson
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158
Member Avatar for cuty_clemz

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]

Member Avatar for BzzBee
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90
Member Avatar for DJDan93
Member Avatar for SgtMash

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

Member Avatar for kireol
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122
Member Avatar for peck3277

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

Member Avatar for peck3277
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Member Avatar for Member #671080

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]

Member Avatar for zeroge
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Member Avatar for gaspan
Member Avatar for Member #119018
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Member Avatar for kishan112

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` …

Member Avatar for kishan112
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Member Avatar for kishan112

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 …

Member Avatar for cortaza
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