Why does this query only show one result?

Question

The query below will be used a search script. For some reason it won\'t return all results where either condition is true. What am i doing wrong?

$sql = \"SELE         


        

Answer 1

Your loop overwrites the value of $name and $code with each loop, so all you will eventually see is the value of the last cycle.

while( $row = mysql_fetch_array( $query ) ) {
  $name = htmlspecialchars($row['name']);
  $code = htmlspecialchars($row['id_code']);
}

You can either echo those values from within your loop, or push them onto a collection someplace:

while ( $row = mysql_fetch_array( $query ) ) {
  $names[] = htmlspecialchars( $row["name"] );
  $codes[] = htmlspecialchars( $row["id_code"] );
}

Or you could put both values into a single array:

$set = array();
while ( $row = mysql_fetch_array( $query ) ) {
  $set[] = array( 
    "Name" => htmlspecialchars( $row["name"] ),
    "Code" => htmlspecialchars( $row["id_code"] )
  );
}

At this point, you have loaded all of the names and codes into arrays (or an array) that can be manipulated after your loop has run its course.

print_r( $names ); // or $set

Redundant Actions

Additionally, you have some redundant code:

$result = mysql_query($sql);
$query  = mysql_query($sql) or die ("Error: ".mysql_error());

This runs your query twice - no need for that.

$num_rows1 = mysql_num_rows($result);
$rows = mysql_num_rows($result);

This is counting the number of rows returned, twice. Again, no need for that.