Hi!
It seems I'm having a really weird problem with SQL SELECT command....I have table into a mySQL DB wich as 4 entry in it. I created the select command to retrieve the data and only 3 of the 4 entry is displaying. If I enter a 5th entry. Only 4 of the 5 will display....?
Here's the code :
<?php
mysql_connect("server", "user", "pass") or die(mysql_error());
mysql_select_db("db") or die(mysql_error());
$data = mysql_query("SELECT * FROM artists")
or die(mysql_error());
$info = mysql_fetch_array( $data );
while($info = mysql_fetch_array( $data ))
{
echo "
<center>
<table border="1" cellpadding="0" cellspacing="0" width="400">
<tr>
<td width="150"><img src="admin/artists photo/".$info['photo']."" /></td>
<td width="250"><h1>".$info['artist']."</h1></td>
</tr>
</table><br />
</center>
";
}
?>
If anyone sees what I did wrong it would be really appreciated!
Thanks
mysql query with single quotes in a variable
$sitedetails = "INSERT INTO vars (address, sitename, description, ownername, theme) VALUES ('$url', '$sitename', '$description', '$ownername', '$theme') ";mysql_query($sitedetails) or
Object Interfaces
EDIT: Never mind, I just updated to php 5.Hey all,I'm currently experimenting with php object interfaces. However, whenever I try to implement one, I get a php error.interface iTemplate{ public
Need help making a script that moves data into acrhive table
Hi,I'm kinda new to php/mysqlAnd i wrote a simple script to store sales lead for a buisness and they are posted out into a table after they are submitted in a form. The page where you can see all your
IP question
ive got 2 ip addresses both global from same user how would i detect if they are local to each other
Chat Box in PHP
I was thinking in doing a Chat Box in PHP. For that I would use a form with two fields, Nick and Message, then I would store the data in a DB and show them.I have already made the code for this. In
Need help: how to catch acess of undefined class properties
Hello. I am learning OO with PHP and have hit a problem.Some code runs as perfectly valid code, where i would like PHP to issue a warning / error.I guess this is because of the loose typing of PHP,
$variable $variables type question
I need to be able to designate an array element dynamically, so I thought to use a variable variable, but it doesn't work:Code: $test = array(1,2);$num = "[0]";echo $test{"$num"};
Help With editting and deleting form
Hallo !!So look at this image :http://img194.imageshack.us/img194/8272/snapshot5f.png This table prints the titles of entries from a table in a database.. The code that i use for this table is this
help with this code please?
Hello,I am trying to build a remote upload script for my image hosting site.I am using $_GET for testing purposes.this would be the url you would visit:Code:
Mail functionality from localhost to server
Hi I am facing problem of mail functionality.When i tested mail functionality in my localhost it works fine but when i tried it on server it didn't work and also no error it displays.So please give me