$_POST probme

[pser]

Hi, i have some problem i hope you could help me with.

I don't get this to work:

$qallitem = "SELECT concat(left(Type,3),ItemTypeid) as Item, Type as ItemType From ItemType;";

if(!$query_allitem = mysql_query($qallitem, $connection))
handle_mysql_error($connection);

while ($row_allitems = mysql_fetch_object($query_allitem))
{
print "$var1= $_POST['$row_allitems->Item']";
}

The problem is the print line in while loop.

I have a page where a user can fill in number in a certain amount of input boxes I don't know which one. So i've created a query which list all available item in that table. Then i want to check the $_POST variable if it contains a value, if it does i'm using that variable further.

Regards
Paal

[pser]

Solved.

while ($row_allitems = mysql_fetch_object($query_allitem))
{
$tmp = $row_allitems->Item;
$tmp . $_POST[$tmp] . "<p>";

..
..
if check bla bla set then do something
}

[tsinka]

Hi,

I hope I got what you want to do 🙂

Try this inside the while loop:

echo "$var1= ".$_POST[$row_allitems->Item]."<br>\n"; 

This will use $row_allitems->Item as key to read out the $_POST value stored with that key or do this inside the while loop:

if (isset($_POST[$row_allitems->Item]) {
    echo "$var1= ".$_POST[$row_allitems->Item]."<br>\n"; 
} else {
    echo "No value submitted for ".$row_allitems->Item."<br>\n";
}

[tsinka]

Ok,

seems like I was too slow 🙂

The main problem was that you used $_POST inside a string, you can do that but you then have to put curly braces around it or just do it the way like in my post.

[pser]

Thanks anyway. Sometimes you just get blind into code 🙂