just quickly....simple

Anon

what the is up with the error..

Warning: Undefined variable: submit....

isnt it just saying if the submit button is pressed then do this...

<?php
if ($submit) {
$db = mysql_connect("localhost","root");
mysql_select_db("antiques",$db);
mysql_query($sql) or die(mysql_error());
$sql = "INSERT INTO general (name,description,price,image_name) VALUES ('$name','$description','$price','$image_name')";
}
?>

van__

This question was already answered in your original thread:

http://www.phpbuilder.com/forum/read.php3?num=2&id=181442&thread=181417

Anon

thanks sorry but in this suituation why does it not show a error?....

<body>

<?php

if ($submit) {

// process form

$db = mysql_connect("localhost", "root");

mysql_select_db("antiques",$db);

$sql = "INSERT INTO general (name,description,price,image_name) VALUES ('$name','$description','$price','$image_name')";

$result = mysql_query($sql);

echo "Thank you! Information entered.\n";

} else{

// display form

name:<input type="Text" name="name"><br>

description:<input type="Text" name="description"><br>

price:<input type="Text" name="price"><br>

image name:<input type="Text" name="image_name"><br>

</form>
<?php

} // end if

?>
</body>
</html

Anon

Because the first time it runs (i.e. when the form hasn't been submitted yet), there's no $submit variable. That's why you need to use isset().

Diego

Anon

Thanks for your reply... could you please look at this code...it DOES cause an error the PREVIOUS one that u just replyed to does NOT. why?

<b>Name</b><br>
<input type="Text" name="name" size="25">
<br>
<b>Description</b><br>
<input type="Text" name="description" size="25">
<br>
<b>Price</b><br>
<input type="Text" name="price" size="25">
<br>
<td><b>Image upload</b><br></td>
</tr>
<tr>
<td><input type="File" name="userfile" size="30" maxlength="255"><br></td>
</tr>
<tr>
<td><b>target filename: </b><br></td>
<td><input type="Text" name="image_name" size="30" maxlength="^{255"><br></td>}
</tr>
<tr>
<tr>
<td><b>upload diretory: </b><br></td>
<td>c:\program files\apache group\realdocs\files\antiquesbasic\images</td>
</tr>
<td colspan="2" align="CENTER">
<INPUT TYPE="submit" name="submit" VALUE="upload">
</td>
</tr>
</FORM>
<?php
if ($submit) {
$db = mysql_connect("localhost","root");
mysql_select_db("antiques",$db);
mysql_query($sql) or die(mysql_error());
$sql = "INSERT INTO general (name,description,price,image_name) VALUES ('$name','$description','$price','$image_name')";
}
?>
</body>

Anon

What kind of error? I copy and pasted your script, ran it and it ran without any error...

Diego

Anon

Sorry, I misread your post.

Both of your scripts DO produce an error (well, not an error, but a warning) also.

Diego

Anon

problem is that the information is not actually getting into the db as far as i can see, when i query the db to find out whats in it, it returns nothing that wasnt already in there

Anon

would it work better , or at all, if i just made the submit through a function in the up_go.php and passed it the variables like this, hasn't worked as far as i can tell....

<?php
function do_upload($filename,$image_name) {
$file = basename($filename);
$tmp_upload_path = "c:\windows\";
$new_file_name = "c:\program files\apache group\realdocs\files\antiquesbasic\images\".$image_name;
if (!copy($tmp_upload_path.$file, $new_file_name)) echo "failed to copy file<br>\n";
return;}

function do_update($name,$description,$price,$image_name){
$db = mysql_connect("localhost", "root");
mysql_select_db("antiques",$db);
$sql = "INSERT INTO general (name,description,price,image_name) VALUES ('$name','$description','$price','$image_name')";
$result = mysql_query($sql);
}

Anon

You're never calling mysql_query(), that's why you are not getting anything in the db, because you didn't send to it!

Diego