[Resolved] create an "ALL" option for each menu--HELP!

baitubai

I have multiple drop down menus, and each of them needs an "all" option. what shall I do to create it and how mysql will look like to get the data when selecting this option?

So far I have something like this:

echo '<b>Host </b>';
echo '<select name="hostfield" >';
echo '<option value=*> ALL </option>';
...
echo '</select>';

echo '<b> Program </b>';
echo '<select name="programfield">';
echo '<option value=*> ALL </option>';
...
echo '</select>';
...
...
$hostId=$_POST['hostfield'];
$programId=$_POST['programfield'];
$q="SELECT date,time,host,msg FROM table WHERE host='$hostId' AND program='$programId' AND ... ;";
....

But I am geting no data.
Thanks.

Trs2988

I believe that the wildcard for MySQL is %. So change the values to '%', and it should work.

Trs2988

Actually, I'm wrong on that. I tested it, and it didn't work :rolleyes:

I don't know how to do it, then.

baitubai

I changed my codes to this:

$q="SELECT field1 field2 f3 f4 FROM table WHERE ";
$where = "";

if($field1!="ALL") {
$where.=" AND field1=."$field1;
}
if($field2!="ALL") {
$where.=" AND field2=".$field2;
}
if($f3..
......
}

$sql=$q.$where;
print $sql;

It works when choose "ALL" option for all of the menus. But doesn't work otherwise. The sql looks like right: SELECT .... WHERE f1="t1" AND f2="p1"...
But it told me :Unknown column 't1' in 'where clause'.

Any ideas?

baitubai

never mind.
just a syntax error.