ok i've tried everthing i can think of and it still won't work. below is the code and a link to the page.
http://www.steppinupwebdesign.com/cms/imgupload.php
the main part of this script is the imgprint() function it is near the bottom
[PHP]
if ($_REQUEST[completed] == 1) {
// Need to add - check for large upload. Otherwise the code
// will just duplicate old file ;-)
// ALSO - note that latest.img must be public write and in a
// live appliaction should be in another (safe!) directory.
move_uploaded_file($_FILES['imagefile']['tmp_name'],"latest.img");
$image = addslashes(fread(fopen ($_FILES['imagefile']['tmp_name'],'r'),filesize($_FILES['imagefile']['tmp_name'])));
if (strlen($instr) < 149000) {
mysql_query ("insert into pix (title, imgdata, location) values (\"".
$_REQUEST[whatsit].
"\", \"".
$image.
"\")");
} else {
$errmsg = "Too large!";
}
}
imgprint();//calls the image printer after images are uploaded
// If this is the image request, send out the image
/*
if ($_REQUEST[gim] == 1) {
header("Content-type: image/jpeg");
print $bytes;
exit ();
}
*/
?>
<html><head>
<title>Upload image</title>
<body bgcolor=white><h2>latest picture</h2>
<font color=red><?= $errmsg ?></font>
<center><img src=?gim=1 width=144><br>
<table>
<tr>
<td>
<?
function imgprint()
{
$result = mysql_query("SELECT * FROM pix");//this table has an four coloms; title(unique), imgdata, id, location
$rowcount = mysql_num_rows($result);
$i = 1;
while ($i < $rowcount)
{
$row = mysql_fetch_row($result);
$title = $row[title];
echo '<img src="imagedisp.php?title=$title" />';
$i++;
}
}
?>
</table>
<img src="imagedisp.php?title=asdf">
<b><?= $title ?></center>
<form enctype=multipart/form-data method=post>
<input type=hidden name=MAX_FILE_SIZE value=150000>
<input type=hidden name=completed value=1>
browse: <input type=file name=imagefile><br>
comments: <input name=whatsit><br>
then: <input type=submit></form><br>
</body>
</html>
[/PHP]
and here is imgdisp.php
[PHP]
$title = $_GET[title];
$result = "SELECT imgdata, type FROM pix WHERE title='$title'";
$Show = mysql ( $hDB, $hSQL );
$rows = mysql_num_rows($hShow);
if($rows<1){
// no image matches this query
}
else{
// at least one image has this title
$getPhoto = mysql_fetch_object($Show);
// we need to determine the mime type
$Type = $getPhoto->type;
// and send the correct header to the browser
Header("Content-type: $Type");
// now send the image
$Body = $getPhoto->body;
echo $Body;
flush();
}
[/PHP]
thanks eric
316
