SQL PROBLEMS

[Anon]

$numrows is saying that i am getting 20 results, but print_r ($TYPE); only returns 1 result. Please help

---

$col=mysql_field_name($column,$j );
$SQL=mysql_field_name($column,($j+1));
$SQL2= "Select $SQL FROM $dbase WHERE ident LIKE \"a\" GROUP By id ";
echo $SQL2;
$TYPE= mysql_db_query($database_name, $SQL2, $cid);
$numrows = mysql_num_rows($TYPE);
echo $numrows."<br>";
$TYPE2= mysql_fetch_array($TYPE);
print_r ($TYPE);
echo $TYPE2[0];echo $TYPE2[1];echo $TYPE2[2]." ---<br>";

[crusty_php]

$TYPE is only the result identifier.

[Anon]

My mistake. it only returns one response with:

print_r ($TYPE2);

[Anon]

Results: $SQL is good. it returns what i want in an array in mysql
$numrows =20
print_r ($TYPE2)=Array ( [0] => textfield [type1] => textfield )

shouldn't there be a lot more data results in the array?

---

$j=6;
$column = mysql_list_fields($database_name,$dbase,$cid);
$col_type = mysql_field_name($column,($j+1));
$SQL= "Select $col_type FROM $dbase WHERE ident LIKE \"a\" GROUP By id";
$TYPE= mysql_db_query($database_name, $SQL, $cid);
$TYPE2= mysql_fetch_array($TYPE);
$numrows = mysql_num_rows($TYPE);

echo $SQL;
echo " ---<br>";
echo $numrows;
echo " ---<br>";
print_r ($TYPE2);
echo " ---<br>";