Resizing image & then inserting into db

[jasonmartin]

I have the following code that takes an image from a form & copies it to a directory on the server. It then resizes the image using ImageMagick's mogrify. I then want to take that image & insert it into the database, but I can't seem to find a way to get it to work.

I know that the copying of the image to the server & the resizing of the image work, but I can't find a way to insert it into the database properly so that I can display the image later.

Any help is greatly appreciated.

Thanks,

Jason Martin

<?
$dbuser = '****';
$dbhost = '*';
$dbpass = '*';
$dbname = '****';

MYSQL_CONNECT("localhost","nobody","");
mysql_select_db("opequon_motors");

?>
<?

if(isset($userfile)) {

srand(time());

$pid = (rand(111111,999999));

if(copy("$userfile", "/tmp/$pid.jpg"))
{
	//print("file copied to /tmp<br>");
}
else
{
	//print("file not copied to  /tmp<br>");
}

exec("mogrify -density 72x72 -geometry 200x200! /tmp/$pid.jpg");

$data = fread(fopen("/tmp/$pid.jpg", "rw"), filesize($pid.jpg));



$result=MYSQL_QUERY("INSERT INTO images (bin_data,filename,filesize,filetype) "."VALUES ('$data','$userfile_name','$userfile_size','$userfile_type')");

$id= mysql_insert_id();

MYSQL_CLOSE();

} else {
print "<FORM ACTION=add_employee2.php enctype= 'multipart/form-data' METHOD=POST>
<INPUT TYPE=hidden NAME=MAX_FILE_SIZE VALUE=10000000>
<INPUT TYPE=FILE NAME=userfile><INPUT TYPE=SUBMIT VALUE=Mogrify></FORM>";
}

?>

[Anon]

What goes wrong? Check your MySQL query for errors and error messages.

What's the output of this PHP expression?:

<?php
print 'MQ_GPC statusis:'.get_magic_quotes_gpc();
?>

[jasonmartin]

I get MQ_GPC statusis:1 when I place it in my code.

I forgot to place the following code that calls the results. I just use the code_name.php?id=a_number to call the picture up, but nothing shows. That is because in the database nothing is getting stored in the bin_data column.

So, I'm pretty sure $data variable where I am trying to place the resized image into, is the problem. I just don't know what is the correct command to place that image into that variable.

Thanks,

Jason Martin

<?
if($id) {
@MYSQL_CONNECT("localhost","nobody","");
@mysql_select_db("opequon_motors");
$query = "select bin_data,filetype from images where id=$id";
$result = @($query);
$data = @MYSQL_RESULT($result,0,"bin_data");
$type = @MYSQL_RESULT($result,0,"filetype");
Header( "Content-type: $type");
echo $data;
};
?>

[Anon]

You simply need more error checking in your code, in this phase of development:
1. After each mysql function call,
check if it went OK. If not
call the MySQL-function which
will tell you why something went
wrong:
http://www.php.net/manual/en/function.mysql-error.php

2. Don't use @ in front of function calls
   if you experience errors. That way
   you are hiding potential explanations
   for yourself.

[Anon]

?> print "Hey"