Thinker2501
January 26th, 2005, 04:42 PM
I am using getimagesize and retrieving the file name from an array. In the MySQL database the column the file names are stored in is set to TinyText. I'm using the following code to create the HTML tag for the images:
<?php
if ($result_ar['picfile'] == "") {
echo "<img src=\"images/listings/noimage.jpg\" width=\"100\" height=\"100\">";
} else {
$imagesize = getimagesize("images/listings/"+$result_ar['picfile']); echo "<img src=\"images/listings/"; echo $result_ar['picfile']; echo "\""; echo $imagesize[3]; echo ">";
}
?>
When I go to the page the images are all rendered correctly, but the following error code appears above each one:
Warning: getimagesize(88): failed to open stream: No such file or directory in /home/grubbsf/public_html/index.php on line 122
The "88" is the only the first part of the file name stored in the array. In this case the full name is "88kearny.jpg". Why is it giving that error and still rendering the images correctly? How do I fix it? Thanks in advance.
<?php
if ($result_ar['picfile'] == "") {
echo "<img src=\"images/listings/noimage.jpg\" width=\"100\" height=\"100\">";
} else {
$imagesize = getimagesize("images/listings/"+$result_ar['picfile']); echo "<img src=\"images/listings/"; echo $result_ar['picfile']; echo "\""; echo $imagesize[3]; echo ">";
}
?>
When I go to the page the images are all rendered correctly, but the following error code appears above each one:
Warning: getimagesize(88): failed to open stream: No such file or directory in /home/grubbsf/public_html/index.php on line 122
The "88" is the only the first part of the file name stored in the array. In this case the full name is "88kearny.jpg". Why is it giving that error and still rendering the images correctly? How do I fix it? Thanks in advance.