RE: [PHP] Unable to display images on browser
| From: | joakim dot andersson at cybercom dot se | Date: | Thu, 28 Feb 2002 09:12:27 +0000 |
| Subject: | RE: [PHP] Unable to display images on browser | ||
| Groups: | php.general | ||
| Request: | Send a blank email to php-general+get-86556@lists.php.net to get a copy of this message | ||
You need to insert the following lines after this line:
Header("Content-type: image/gif");
$im = ImageCreateFromString ($fileContent);
ImageGif ($im);
and then remove this line: echo $fileContent;
That should do it.
/Joakim
-----Original Message-----
From: Narvaez, Teresa [mailto:Teresa.Narvaez@DynCorp.com]
Sent: Wednesday, February 27, 2002 8:32 PM
To: 'joakim.andersson@cybercom.se'; php-general@lists.php.net
Subject: RE: [PHP] Unable to display images on browser
Hello, Thanks for your help. This is what I have for file1.php and
ddownloadfile.php. What I want is to click on "Donwnload now" link and be
able to get the file out of the database and display it on the browser.
Thank you in adavance, -Teresa
file1.php
-----------
<?php
while ($row = mysql_fetch_array($result))
{
?>
SOME_HTLM_CODE_TO_DISPLAY_DB_FIELDS_GOES_HERE;
// <img src=\"ddownloadfile.php?fileId=<?php echo $row["PicNum"];
?>\"
>
<a href="ddownloadfile.php?fileId=<?php echo $row["PicNum"]; ?>"
>
Download Now
</a></font>
</td>
</tr>
<?php
ddownloadfile.php
------------------
<?
$dbQuery = "Select PicNum, size, type, description, Image";
$dbQuery .= " FROM Images WHERE PicNum = $fileId";
$result = mysql_query($dbQuery)
or die ("Could not get file list: " . mysql_error() );
echo "Sent Query successfully<br>";
if ( mysql_num_rows($result) == 1)
{
$fileType = @mysql_result($result,0, "type");
$fileContent = @mysql_result($result, 0, "Image");
$filedesc = @mysql_result($result,0, "description");
$filenum = @mysql_result($result,0, "PicNum");
Header("Content-type: image/gif");
echo $fileContent;
}
else
{
echo "Record does not exist";
} // else
?>