SQL = "WHERE figureName like '$number.%' " ?

Cherphas

Hey,

I created an SQL statement:

$sql = "SELECT image FROM table01 WHERE figureName like '$number.%' ORDER BY image";

an I think I'm having problems with the "like" of the WHERE clause.

The Number before the ".%", depending on what the user clicked before, could be 1 or 2. So $number.% could reprensent 1.2, 1.3, 1.4, 2.2, 2.3 etc.
For example, if Iclicked on Number 2, I want everything from the database that starts with 2, like 2.1, 2.2, 2.3 etc. But the it won't work. The statement seems logic to me, so does anybody see anything wrong in there?

Thanks

stolzyboy

add an or die after your mysql_query statement to see where the problem lies

$result = mysql_query($sql) or die(mysql_error());

Cherphas

I do have that, it doesn't give me an error, probably because there's no syntax error. The problem is that the statement I wrote doesn't give me what I want. But I can't see any other way of writing it.

stolzyboy

try adding a slash () before the ., since a . is the concatenation operator in php, it is probably screwing it up

Cherphas

doesn't work. 🙁

stolzyboy

you could do if this to get all the numbers

let say you clicked to and then $number would equal 2

$highnumber = number + 1;

$sql = "select * from tablename where number between $number and $highnumber;";

you may have to play with addition or something to get this to work correctly, but i think you will get the idea