Query Error PHP select from where clause

Question

Im try to get datatable from query, which select from where clause, but not working :

<?php
  include "koneksi.php";
    $query=mysql_query("SELECT @rownum := @rownum + 1 AS urutan,t.      id_data, nama_aut, judul_abs, file_abs, status_lolos, status_bayar
  FROM datauser   WHERE status_lolos = "Sudah" t,
  (SELECT @rownum := 0) r");
   $data = array();
   while($r = mysql_fetch_assoc($query)) {
   $data[] = $r;
}

 $i=0;
   foreach ($data as $key) {
    // add new button
   .....
$i++;
 }
   $datax = array('data' => $data);
echo json_encode($datax);
?>

error

Parse error: syntax error, unexpected 'Sudah' (T_STRING)

But if not use where clause , thats working fine :

  <?php
  include "koneksi.php";
    $query=mysql_query("SELECT @rownum := @rownum + 1 AS urutan,t.       id_data, nama_aut, judul_abs, file_abs, status_lolos, status_bayar
  FROM datauser   t, 
  (SELECT @rownum := 0) r");
   $data = array();
   while($r = mysql_fetch_assoc($query)) {
   $data[] = $r;
}
 ...

Anyone can help, im newbie..

Change `WHERE status_lolos = "Sudah"` to `WHERE status_lolos = 'Sudah'` or `WHERE status_lolos = \"Sudah\"`. — Styphon, Mar 10 '17 at 11:49
Look at the color highlighting. Also update from `mysql_` to `mysqli` or `pdo`. — chris85, Mar 10 '17 at 11:51
when i use 'Sudah' or \"Sudah\" that giving error ''Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given... — hahaha, Mar 10 '17 at 11:54
That means the query failed, use error reporting. http://php.net/manual/en/function.mysql-error.php (also note the big warning on this page) — chris85, Mar 10 '17 at 12:03
REPLACE WHERE status_lolos = "Sudah" t WITH WHERE status_lolos = 'Sudah' — user1544541, Mar 10 '17 at 12:06
Because you're not checking to see if your query is successful. You should do `if ( ! $query = mysql_query(...) ) { echo mysql_error(); }` and it will give you the error. — Styphon, Mar 10 '17 at 12:07
when i try WHERE status_lolos = "Sudah" to WHERE status_lolos = 'Sudah', that not working.. — hahaha, Mar 10 '17 at 12:11

score 0 · Answer 1 · answered Mar 10 '17 at 11:49

0

no need to use double qoute for Sudah:

 FROM datauser   WHERE status_lolos = "Sudah" t,

answered Mar 10 '17 at 11:49

Therichpost

1,759
2
14
19

score 0 · Answer 2 · answered Mar 10 '17 at 12:15

0

$query=mysql_query("SELECT @rownum := @rownum + 1 AS urutan,t.id_data, nama_aut, judul_abs, file_abs, status_lolos, status_bayar FROM datauser t WHERE status_lolos = 'Sudah', (SELECT @rownum := 0) r");

answered Mar 10 '17 at 12:15

user1544541

193
3

Query Error PHP select from where clause

2 Answers2