Hello,
Thanks for your responses; however, I do not think I have the gd
libraries installed because ImageCreateFromString() was not found. I will
install it.
I have a question: I can display the PNG or GIF image using the
browser. So Why do I need the GD library? Since I store the mime type in
the database I think that a call to header() to tell the browser what type
of mime-type is coming from the database should sufice. Also, I want to
store any type of binary data into my database (PNG, GIF, word, JPG, etc) Am
I missing something?
Thanks,
-Teresa
-----Original Message-----
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
Sent: Thursday, February 28, 2002 4:12 AM
To: [EMAIL PROTECTED]
Subject: RE: [PHP] Unable to display images on browser
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 $fileContent;
That should do it.
/Joakim
-----Original Message-----
From: Narvaez, Teresa [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, February 27, 2002 8:32 PM
To: '[EMAIL PROTECTED]'; [EMAIL PROTECTED]
Subject: RE: [PHP] Unable to display images on browser
Hello, Thanks for your help. This is what I have for file1.php and
ddownloadfile.php. What I want is to click on "Donwnload now" link and be
able to get the file out of the database and display it on the browser.
Thank you in adavance, -Teresa
file1.php
-----------
<?php
while ($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
?>
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