Help Default Image

[weber2nd11]

Hello everyone, I have an easy (or atleast I think it should be extremely easy) issue, that I have become stuck on. I am just starting to learn PHP. I am developing a site for one of my classes where users can log in, and access an edit profile page. This page displays various personal information, and an image (if the user has uploaded one).

I am using Florian Dittmer's solution to this problem (http://www.phpbuilder.com/columns/florian19991014.php3?aid=38&print_mode=1). The code works fine, I just want to add a bit of code to display the image "noimage.gif" which is located in the same directory as the php files, if no image has been uploaded.

I set the default value for the field "filename" to "default image".

This is what I want, if "filename" is "default image", echo "noimage.gif", else echo "whatever image has been uploaded".

Here's the code I have:

<?php

// getdata.php3 - by Florian Dittmer <[e-mail]>
// Example php script to demonstrate the direct passing of binary data
// to the user. More infos at http://www.phpbuilder.com
// Syntax: getdata.php3?id=<id>

if($id) {

@MYSQL_CONNECT("host","user","password");

@mysql_select_db("hbgalumn_authenticate");

$query = "select bin_data,filetype from binary_data where id=$id";
$result = @MYSQL_QUERY($query);
$data = @MYSQL_RESULT($result,0,"bin_data");
$type = @MYSQL_RESULT($result,0,"filetype");

Header( "Content-type: $type");

while ($row = mysql_fetch_array($result)) {

    if ($row["filename"] == "default image") {

    echo "noimage.gif";

    }

    else {


    echo $data;

    }

}

}

?>

HELP PLZ!?!?! 😃

[cahva]

There was some errors in your script. You used mysql_fetch_array with $row['filename'] but you didnt have it (filename field) in your query.
Try with this:

<?php

// getdata.php3 - by Florian Dittmer <[e-mail]>
// Example php script to demonstrate the direct passing of binary data
// to the user. More infos at http://www.phpbuilder.com
// Syntax: getdata.php3?id=<id>

if($id) {
@MYSQL_CONNECT("host","user","password");
@mysql_select_db("hbgalumn_authenticate");

$query = "select bin_data,filetype,filename from binary_data where id=$id";
$result = @mysql_query$query);

if (mysql_num_rows($result)>0) {
	list ($data,$type,$filename)=mysql_fetch_row($result);
	if ($filename=='default image') {
		header( "Content-type: img/gif");
		$fp = fopen('noimage.gif', 'rb');
		fpassthru($fp);
		exit;
	} else {
		header( "Content-type: $type");
		echo $data;
		exit;
	}
}
?>

[weber2nd11]

Hello again. Thanks for you response Cahva. I tried using your idea, however still no luck :-/ I did notice a couple things. A missing "(" in the $result = @$query); definition. Also, $data and $type were not defined before the if statement. I fixed (or atleast I like to think I did 🙂 ) this, and this is the code I have now. However, still no luck..

<?php

// getdata.php3 - by Florian Dittmer <[e-mail]>
// Example php script to demonstrate the direct passing of binary data
// to the user. More infos at http://www.phpbuilder.com
// Syntax: getdata.php3?id=<id>

if($id) {
@MYSQL_CONNECT("host","user","pass");
@mysql_select_db("hbgalumn_authenticate");

$query = "select bin_data,filetype,filename from binary_data where id=$id";
$result = @($query);
$data = @MYSQL_RESULT($result,0,"bin_data");
$type = @MYSQL_RESULT($result,0,"filetype");

if (mysql_num_rows($result)>0) {
list ($data,$type,$filename)=mysql_fetch_row($result);
if ($filename=='default image') {
header( "Content-type: img/gif");
$fp = fopen('noimage.gif', 'rb');
fpassthru($fp);
exit;
} else {
header( "Content-type: $type");
echo $data;
exit;
}
}
?>

Any ideas? Thanks again!!!! 😃

[cahva]

It was defined in this:

list ($data,$type,$filename)=mysql_fetch_row($result);
// eg. same as:
$data=mysql_result($result,0,'bin_data');
$type=mysql_result($result,0,'filetype');
$filename=mysql_result($result,0,'filename');

You dont have to fetch the information before mysql_num_rows().
What does it say? Does it fetch the information okay?

Add die(mysql_error()); after $result row. It should tell you if you have something wrong in your query.

[weber2nd11]

The code works correctly if there is an image uploaded in the bin_data field in the database (but doesn't show noimage.gif if no image uploaded in bin_data). I dropped the only record I have in the database at this point, and inserted a record for my uname w/ null value for the bin_data field, and "default image" for the filename field.

However, it displays the last image that was contained in bin_data for the last instance of the record (displays image that was in bin_data in the record that I dropped)... If I upload a new image into bin_data, it displays that image.. I tried viewing the page on my roommate's computer, and it does the same thing.. displays the last image for my username, although the bin_data field is null..

I have to apologize for my lack of experience and knowledge of the language and it's use.

<?php

if($id) {
@MYSQL_CONNECT("host","user","password");
@mysql_select_db("hbgalumn_authenticate");

$query = "select bin_data,filetype,filename from binary_data where id=$id";
$result = @($query);

if (mysql_num_rows($result)>0) {
list ($data,$type,$filename)=mysql_fetch_row($result);

if ($filename=='default image') { 
    header( "Content-type: img/gif"); 
    $fp = fopen('noimage.gif', 'rb'); 
    fpassthru($fp); 
    exit; 
} else { 
    header( "Content-type: $type"); 
    echo $data; 
    exit; 
} 

}
die(mysql_error());
}
?>

Did I place the error command correctly?
If I place it anywhere else, it results in a broken image..

Thanks again for your help on this matter.

[weber2nd11]

WHOHOO! I figured it out. The problem was in the way I have the values initially set up in the database. Whenever the user clicks the confirmation link in an e-mail, it automatically adds an instance in the binary_data table w/ their the user's uname, auto-increments id, and assigns the value "" for all the remaining fields. This was the problem, since in my add function I specified the "" value to be added to the database for the filename field, it did not insert the "default image" value I had intended it to. I simply changed the

if ($filename=='default image') {

to

if ($filename=='') {

and the code works like a charm now!!! 😃

If anybody else was interested in doing something similar to this, the code is now:

<?php

// getdata.php3 - by Florian Dittmer <[e-mail]>
// Example php script to demonstrate the direct passing of binary data
// to the user. More infos at http://www.phpbuilder.com
// Syntax: getdata.php3?id=<id>

if($id) {
@MYSQL_CONNECT("host","user","password");
@mysql_select_db("databasename");

$query = "select bin_data,filetype,filename from binary_data where id=$id";
$result = @($query);

if (mysql_num_rows($result)>0) {
list ($data,$type,$filename)=mysql_fetch_row($result);

if ($filename=='') { 
    header( "Content-type: img/gif"); 
    $fp = fopen('noimage.gif', 'rb'); 
    fpassthru($fp); 
    exit; 
} else { 
    header( "Content-type: $type"); 
    echo $data; 
    exit; 
} 

}

}
?>

to call the image from an html or php document located in the same directory as the "getdata.php3" file, simply use the same img tag and url that was specified by Florian Dittmer's article located (http://www.phpbuilder.com/columns/f...mp;print_mode=1):

<img src="getdata.php3?id=<id>">

Thank you very much cahva!!!! Couldn't have done it without you. Well maybe, but I surely would have vowed to never program again, probably been bald, and very bitter at age 23. J/k

Thanks again, greatly appreciated. Hope this helps somebody else.