Hi all
I am using a PHP script from Codewalkers.com to work with a
database
of images (or image filename). The code is as follows:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0
Transitional//EN" "http://
www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<?php
// index.php - by Hermawan Haryanto <hermawan codewalkers.com>
// Example PHP Script, demonstrating Storing Image in
Database
// Detailed Information can be found at http://www.codewalkers.com
// database connection $conn =
mysql_connect("localhost", "username",
"password") OR DIE (mysql_error());
mysql_select_db ("testimages", $conn) OR
DIE (mysql_error());
// Do this process if user has browse the
// file and click the submit button
if ($_FILES) {
$image_types = Array ("image/bmp",
"image/jpeg",
"image/pjpeg",
"image/gif",
"image/x-png");
if (is_uploaded_file
($_FILES["userfile"]["tmp_name"])) {
$userfile = addslashes (fread (fopen
($_FILES["userfile"]["tmp_name"],
"r"), filesize
($_FILES["userfile"]["tmp_name"])));
$file_name =
$_FILES["userfile"]["name"];
$file_size =
$_FILES["userfile"]["size"];
$file_type =
$_FILES["userfile"]["type"];
if (in_array (strtolower ($file_type), $image_types)) {
$sql = "INSERT INTO image " .
"(image_type, image,
image_size, image_name, image_date) ";
$sql.= "VALUES (";
$sql.= "'{$file_type}', '{$userfile}',
'{$file_size}',
" . "'{$file_name}', NOW())";
mysql_query ($sql, $conn);
Header("Location:".$_SERVER["PHP_SELF"])
;
exit();
}
}
}
// Do this process if user has clicked
// a file name to view or remove
if ($_GET) {
$iid = $_GET["iid"];
$act = $_GET["act"];
switch ($act) {
case rem:
$sql = "DELETE FROM image WHERE image_id=$iid";
mysql_query ($sql, $conn);
Header("Location:./index.php");
exit();
break;
default:
print "<img src= "image.php?iid=$iid
">";
break; } }
?>
<html>
<head>
<title>Storing Images in DB</title>
</head>
<body>
<form method="post"
enctype="multipart/form-data">Select Image
File:
<input type="file" name="userfile"
size="40" />
<input type="submit"
value="submit">
</form>
<?php
$sql = "SELECT * FROM image ORDER BY image_date
DESC";
$result = mysql_query ($sql, $conn);
if (mysql_num_rows($result)>0) {
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$i++;
$str .= $i.". ";
$str .= "<a
href='index.php?iid=".$row["image_id"]."
'>"
. $row["image_name"]."</a> ";
$str .= "[".$row["image_date"]."]
";
$str .= "[".$row["image_size"]."]
";
$str .= "[<a
href='index.php?act=rem&iid=".
$row["image_id"]
. "'>Remove</a>]<br>";
}
print $str;
}
?>
</body>
</html>
When running this script I get the following error message:
Parse error: parse error, unexpected ';' in
C:htdocsPhotosindex.php
on line 63
FYI in DW line 63 is: if
(mysql_num_rows($result)>0) {
I do not find the reason for this error on lines 62 or 63.
Can anyone
help me out?
Thanx in advance.
Plato
--~--~---------~--~----~------------~-------~--~----~
You received this message because you are subscribed to the
Google Groups "PHP & MySQL" group.
To post to this group, send email to phpmysqlgooglegroups.com
To unsubscribe from this group, send email to
phpmysql-unsubscribegooglegroups.com
For more options, visit this group at http://
groups.google.com/group/phpmysql?hl=en
-~----------~----~----~----~------~----~------~--~---
|
