Pulling hair out

[Anon]

please have a look at this where am I going wrong ?? Its is not updating the Mysql file and not updloading the file ??

UserID would = any name

<form enctype="multipart/form-data" method="post" action="<?php echo $PHP_SELF ?>">
picture name (please put a name in for your file but leave your clan name at
the begining)<br>
<input type="Text" name="member_pic" size="25" value="<? echo $UserID ?>">
<br>
<br>
picture<br>
<input type="File" name="picture" size="25">
<br><br>
<input type="submit" name="submit" value="Upload">
</form>
<?php

if ($submit) {

$db = mysql_connect("localhost","username","pass"); 

mysql_select_db("mydb",$db); 

$sql = "UPDATE members ";
		"SET member_pic = '$member_pic'";
		"WHERE UserID = '$UserID'";

exec("cp $picture /home/localhost/public_html/images/$member_pic"); 

echo "member_pic: $member_pic<br>\n"; 
echo "temp file: $picture<br>\n"; 
echo "file name: $picture_name<br>\n"; 
echo "file size: $picture_size<br>\n"; 
echo "file type: $picture_type<br>\n"; 
echo "<br>\n"; 
echo "<img src=images/$member_pic><br>\n"; 

}

?>

[Anon]

I have the file uploading now. I forgot to set permissions on the server. But its still not updating the Mysql database

[Anon]

Check you $sql variable:

$sql = "UPDATE members ";
"SET member_pic = '$member_pic'";
"WHERE UserID = '$UserID'";

If this is not a typo $sql only contains:
UPDATE members

try this:
$sql = "UPDATE members SET member_pic = '$member_pic' WHERE UserID = '$UserID'";

Troy

[Anon]

That's what I was going to say. Put your SQL query all on one line, because I think the semi-colons are messing it up. Echo out sql and I bet you'll only get "UPDATE members"

Also, PHP has a copy function. I'd recommend using that instead of the exec() command.

---John Holmes...

[Anon]

I have already tried it as a one liner But still does not work. ?? I cant see why. I can put the same sql command into phpMyAdmin and works fine .??

Something is a miss.

How about INPUT INTO but Im not sure if you could use WHERE UserID = '$UserID'

Any further ideas ??

Thanks

[Anon]

Umm...where are you actually executing your sql statement? I don't see any mysql_query() command??

mysql_query($sql,$db);

Are you still having trouble uploading the image, too?

You have to be careful with file uploads. Someone could type in '/etc/passwd/' into your file box, then you are copying the password file to someplace under your webroot. then someone just types in the url and views your passwd file. The php page has some scripts posted for checking the file and making sure it's coming from the right place...it should always come from the /tmp directory, or whatever you have setup in your php.ini file.

---John Holmes...

[Anon]

OK here ir is again and I have added the Mysql_querry to it and it comes up with no errors.

<form enctype="multipart/form-data" method="post" action="<?php echo $PHP_SELF ?>">
Image name (please put a name for your file but leave your clan name at
the begining)<br>
<input type="Text" name="member_pic" size="25" value="<?php echo $UserID ?>">
<br>
<br>
picture<br>
<input type="File" name="picture" size="25">
<br><br>
<input type="submit" name="submit" value="Upload">
</form>
<?php

if ($submit) {

$db = mysql_connect("localhost","username","password"); 

mysql_select_db("mydb",$db); 

$sql = "UPDATE members SET member_pic = '$member_pic' WHERE UserID = '$UserID'"; 

$result = mysql_query($sql, $db) or die ("Database error: ".mysql_error($db));

exec("cp $picture /home/public_html/images/$member_pic"); 

echo "member_pic: $member_pic<br>\n"; 
echo "temp file: $picture<br>\n"; 
echo "file name: $picture_name<br>\n"; 
echo "file size: $picture_size<br>\n"; 
echo "file type: $picture_type<br>\n"; 
echo "<br>\n"; 
echo "<img src=images/$member_pic><br>\n"; 

}

?>

This updloading the file to the site no prob. But is not adding the pic name to the mysql database ?? I just cannot see why.

Lee Smith (Thanks for all the help)

[Anon]

I think I've got it. You're not passing along UserID with your form, so that after you submit the form, UserID is not set. Therefore the query doesn't give any errors, but it doesn't insert because it's looking for UserID = ''

You'll have to pass along UserID as a hidden value within the form.

<input type=hidden name=UserID value="<?php echo $UserID; ?>">

That should solve your problems...hopefully while you've still got hair.

---John Holmes...

[Anon]

Although not the problem here, error checking is your friend :

$db = mysql_connect("localhost","username","password")
or die("Could not connect");

mysql_select_db("mydb","$db")
or die("Could not select db");

[Anon]

No that has not worked as well. When putting it as hidden it just shows nothing ?? here is what I put

<input type="hidden" name="member_pic" size-"25" value="<?php echo $UserID?>">

This did not work. I am really getting pissed off now. I am not getting any errors on the page the file is uploading lovely. I do believe its a problem with the form not passing on the info. But I cannot see why. ??? Any more help would be nice

Lee Smitn (Virtually no hair now) !!

[Anon]

You have to use the following:

<input type=hidden name="UserID" value="<?php echo $UserID; ?>">

When you first load the page, UserID is set, and it echos into the text box to create the image name. However, when you press submit, UserID is not passed along anywhere in the form, therefore it doesn't exist anymore. This whole thing is a two step process. UserID exists in the first step, but not in teh second. You try to use UserID in your WHERE statement. UserID doesn't exist, so it'll equal out to an empty string. Passing along UserID as a hidden form element preserves this value. The name of the hidden element and the variable must match to pass it.

Hopefully that explains it.

---JH

[Anon]

Sorry John

This is not working at all. I have put in what you have said and this does not work at all. When I put the value echo UserID into the value first it puts it within the form. So in return when you submite the form it does become text so it cant be hidden. ??? When I use you code it does not even give me a text box.

Lee Smith

[Anon]

You're form should look like this:

<form enctype="multipart/form-data" method="post" action="<?php echo $PHP_SELF ?>"> 
picture name (please put a name in for your file but leave your clan name at 
the begining)<br> 
<input type="Text" name="member_pic" size="25" value="<? echo $UserID; ?>"> 
<input type="hidden" name="UserID" value="<?php echo $UserID; ?>">
<br> 
<br> 
picture<br> 
<input type="File" name="picture" size="25"> 
<br><br> 
<input type="submit" name="submit" value="Upload"> 
</form>

[Anon]

John

A big big thankyou this has worked. But i cant seem to understand why you have 2 inputs ?? Could you explain to me so that I fully understand. e-mail [e-mail] if ya dont mind.

And again thanks to all who helped

Lee Smith