Geting a row out of a db

Anon

im trying to let the user of my website be able to amend information that is in a table. Im listing the rows with a radio button next to each row, this bit works, the bit i cant get to work is geting wot is already in that row our of the database so the user knows wot they are chaging in the next page,
ive use

mysql_query (" select field1, field2 table where id = $id ");
echo $id;
echo $field1;

$id is showing the data that it is supposed to, this is the value for the radio button so i know that bit is working
$field1 is staying blank so it is not geting the data out from the database

What am i doing wrong, does anybody know?

geeeek

Darkness,

You need to put single quotes around around the $ID in your query. So, it should look like this:
mysql_query (" select field1, field2 table where id = '$id' ");

When you are having this type of problem, a good way to research the problem would be to have your page echo the query string. Try rewriting it like this:

$query = "select field1, field2 table where id = '$id'";
$result = mysql_query($query) or die(mysql_error());

echo $query;
echo $id;
echo $field1;

This way, you'll know the server thinks you are asking for.

Best of luck!

Anon

Well ive tried it like that and im still having the same problem. I did have an error at first but that was because i had missed out a 'from' between the fields and table, i put that in and still the same problem.
Does anybody else have any ideas?

geeeek

Ok, try this:

$query = "SELECT field1, field2 FROM table WHERE id = '$id'";
$result = mysql_query($query) or die(mysql_error());

while($data = mysql_fetch_array($result))
{
list($field1, $field2) = $data;
print("field1 = $field1 | field2 = $field2\n");
}

Anon

That still just left it all blank, i tried putting mysql_fetch_row in as well but that done exactly the same thing.