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by hugon » Thu Nov 14, 2002 5:03 am
I keep getting parse error for this line
- Code: Select all
if ($userfile_size > 524288000) { echo ("This file to big for saving at our server. Sorry!"); }
I've taken this from this forum. don't remember where.
form.html
- Code: Select all
<form enctype=multipart/form-data action=add_pic.php method=post>
Which file??? <input name=picture type=file>
Your name is .... <input type=text name=username>
<input type=submit value="OK!">
</form>
add_pic.php
- Code: Select all
<?
// so if have a file we will save him, but we must test is this file a picture or not
$userfile_name=$HTTP_POST_FILES['picture']['name'];
$userfile_size=$HTTP_POST_FILES['picture']['size'];
$userfile_type=$HTTP_POST_FILES['picture']['type']
if ($userfile_size > 524288000) { echo ("This file to big for saving at our server. Sorry!"); }
if ($userfile_type == "image/gif" or $userfile_type == "image/jpeg" or $userfile_type == "image/png")
{
// In this case file is image and we are go to the dir where are pictures situated
// create a dir
$destination="upload/$username/"
mkdir ($destination, 0700); // I hope you want save pic into the $user - directory ...
$fd =$destination;
$fd .=$$userfile_name;
if (file_exists($fd))
{
move_uploaded_file ($userfile_name, $destination);
copy($userfile_name);
echo("<a href=http://www.fruir.org/fileform/simple_hotmail/showpic.php?link=$fd>link</a>");
}
}
?>
showpic.php
- Code: Select all
<?
echo("<img src=$link>");
?>
-
hugon
- New php-forum User

-
- Posts: 2
- Joined: Thu Nov 14, 2002 4:10 am
by WiZARD » Thu Nov 14, 2002 5:46 am
Hi!
As I see you so new in PHP.
You have few sintax error in your code
- Code: Select all
<?
// so if have a file we will save him, but we must test is this file a picture or not
$userfile_name=$HTTP_POST_FILES['picture']['name'];
$userfile_size=$HTTP_POST_FILES['picture']['size'];
$userfile_type=$HTTP_POST_FILES['picture']['type'];
if ($userfile_size > 524288000) { echo ("This file to big for saving at our server. Sorry!"); }
if ($userfile_type == "image/gif" or $userfile_type == "image/jpeg" or $userfile_type == "image/png")
{
// In this case file is image and we are go to the dir where are pictures situated
// create a dir
$destination="upload/$username/";
mkdir ($destination, 0700); // I hope you want save pic into the $user - directory ...
$fd =$destination;
$fd .=$$userfile_name;
if (file_exists($fd))
{
move_uploaded_file ($userfile_name, $destination);
copy($userfile_name);
echo("<a href=http://www.fruir.org/fileform/simple_hotmail/showpic.php?link=$fd>link</a>");
}
}
In next time see carefully :wink:
-

WiZARD
- Moderator

-
- Posts: 1257
- Joined: Thu Jun 20, 2002 10:14 pm
- Location: Ukraine, Crimea, Simferopol
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