Forum Moderators: coopster
texttest3.php
<?php
ini_set("display_errors", "On");
echo("this is text before the picture <br>");
?>
<IMG SRC="imgtest3.php?type=1">
<IMG SRC="imgtest3.php?type=2">
<br>
<?php
echo("this is text after the picture");?>
i can not tell you why it's not working, but i think too that it should. maybe the file displayed then, does not work. simply put the url of the image which makes problems directly into your browsers adress textfield and check it out by hand.
-hakre
localhost/imgtest3.php?type=1 or localhost/imgtest3.php?type=2
both of these work exactly as they should.
The display I get sometimes displays part of the images, usually just a bit at the top then some sort of static looking stuff then it goes all black. Sometimes I get an apache server error as well.
If it is not creating an image, but rather outputing HTML to display an image, then you cannot put the url in the src part of an img tag.
<?php
ini_set("display_errors", "On");
echo("this is text before the picture <br>");
?>
<IMG SRC="imgtest3.php?type=1">
<IMG SRC="imgtest3.php?type=2">
<br>
<?php
echo("this is text after the picture");
?>
imgtest3.php
<?php
//start of main
header("Content-type: image/gif");
ini_set("display_errors", "On");
$link = @mysql_connect('localhost', 'user', 'mypass')
or die("Could not connect to MySQL");
$db = @mysql_select_db('temp2',$link)
or die("Could not select database");
$type=$_GET['type'];
$image=drawimage($link,$type);
imagegif($image);
//end of main
function drawimage($link,$type)
{
if($type==1)$image = imagecreatefromgif("map.gif");
if($type==2)$image = imagecreatefromgif("map2.gif");
$query = "SELECT * FROM table2ll WHERE status=1"; //select green sites
$result = @mysql_query($query,$link) or die("Could not submit query1");
makedots($image, $result, 1);
$query = "SELECT * FROM table2ll WHERE status=2";//select red sites
$result = @mysql_query($query,$link) or die("Could not submit query");
makedots($image, $result,2);
$query = "SELECT * FROM table2ll WHERE status=3";//select clear sites
$result = @mysql_query($query,$link) or die("Could not submit query");
makedots($image, $result,3);
return($image);
}
<?php
//start of main
header("Content-type: image/gif");
ini_set("display_errors", "On");
$type=$_GET['type'];
$image=drawimage($link,$type);
imagegif($image);
//end of main
function drawimage($link,$type)
{
if($type==1)$image = imagecreatefromgif("map.gif");
if($type==2)$image = imagecreatefromgif("map2.gif");
return($image);
}
?>
Try using this and see if it works... If it does, it might be something in your SQL causing it.
anyway, the only difference i can find is within these lines:
[tt]if($type==1)$image = imagecreatefromgif("map.gif");
if($type==2)$image = imagecreatefromgif("map2.gif");[/tt]
i guess that map2.gif does not exists. it's only a guess but i think the error can be found here or the image is to small to be drawn on etc. pp. my shoot is the difference between these to .gif files.
--hakre
Right now the code of imgtest3.php is
<?php
//start of main
header("Content-type: image/gif");
ini_set("display_errors", "On");
$type=$_GET['type'];
$image=drawimage(1,$type);
imagegif($image);
//end of main
function drawimage($link,$type)
{
$image = imagecreatefromgif("map.gif");
return($image);
}
?>
The page that I'm trying to view is:
<IMG SRC="ita.php?type=1">
<IMG SRC="ita.php?type=2">
and ita.php is
<?php
header("Content-type: image/gif");
$image=imagecreatefromgif("map.gif");
imagegif($image);
?>
This works if whatever comes after IMG SRC=" is exactly the same in each case, but if I pass a get parameter it crashes, and if I save one copy of ita as ita2 it crashes. But if I set one to map.gif and the other to map2.gif it works.
also, dig up memory php is allowed to use etc (check phpinfo();), maybe we can dig something up here of your problem.
It would be listed under "Server API" in phpinfo().
