help with variable
Posted on
16th Feb 2014 07:03 pm by
admin
I've got a problem, i want to echo some images depending on the $id, however for me to get that id i have to extract it from a query. So i've done that, but i want to extract all id's (not just 1), so i can echo images for all them ids (not just echo the images based on 1 id), so I did a while() loop, however doesnt seem to work
My code:
Code: <?php
include "connect.php";
$user = "Peter";
$sql = mysql_query("select * from game_main where user = '$user'");
while($row = mysql_fetch_array($sql)){
$id = $row['id'];
}
//do i have to do foreach $id as something??
$views = mysql_num_rows(mysql_query("SELECT userip FROM game_views WHERE sub_id = '$id'"));
$sql3 = mysql_query("select monthstamp from game_monthlyfeatured where first = '$id'");
$num3 = mysql_num_rows($sql3);
$sql4 = mysql_query("select weekstamp from game_weeklyfeatured where first = '$id'");
$num4 = mysql_num_rows($sql4);
$sql5 = mysql_query("select daystamp from game_dailyfeatured where first = '$id'");
$num5 = mysql_num_rows($sql5);
$sql6 = mysql_query("select date from site_featured where id = '$id'");
$num6 = mysql_num_rows($sql6);
if($num3 == 1) {
?>
<img src="../images2/game/game_awards_1monthly.gif" width="20" height="20" />
<?php
}
if($num4 == 1) {
?>
<img src="../images2/game/game_awards_topweekly.gif" width="20" height="20" />
<?php
}
if($num5 == 1) {
?>
<img src="../images2/game/game_awards_topdaily.gif" width="20" height="20" />
<?php
}
if ($views > 1000) {
?>
<img src="../images2/game/game_awards_1000views.gif" width="20" height="20" />
<?php
}
if ($views > 25000) {
?>
<img src="../images2/game/game_awards_25000views.gif" width="20" height="20" />
<?php
}
if ($num6 > 0) {
?>
<img src="../images2/game/game_awards_modpick.png" width="16" height="16" />
<?php
}
?>
Hope someone can help, thanks
No comments posted yet
Your Answer:
Login to answer
140
26
Other forums
Sum of Values in an Array
This is probably really simple... but it's been years since I've written anything, so bare with me!<
Attempt to assign property of non-object in...
I'm having issues with the following function in PHP 5...
function getTreeWithChildre
opening a window with after form submission
I know this this forum has nothing to do with JS, but i'm trying to use it with my php script.
<
Option box to change variable
Hello, i need help by making a script!
I need to write a file with option box, so a dropdown
Keeping data in form
How can I keep whatever I write in the form?
paginate search result
Hi, I have a paginations script to display data from my database but i would like to paginate someon
login from external site
Hi my new experience begins, Now what i am trying to do is i make three pages, login.php logout.php
big pagination problem in php
<?php
$connect = mysql_connect("localhost", "root", "")
whats wrong with my code please help!!!
this is the error
Warning: mysql_close(): supplied argument is not a valid MySQL-Link res
MS are the best!!!
Visit http://www.microsoft.com/australia/windows/default.aspx?h=watch-a-demo and click the massive '