jQuery模式执行正常,但是当我想在
PHP成功执行表单时显示警告消息时,我不知道我错过了什么?
$.post( 'name.PHP',{ ime: ime,location: location },function(data){ if (data.success) { alert("form posted!"); } else { $("#dialog-form").dialog("open"); } },"json" );
========================== PHP ============
if ($result == false) { echo json_encode(array('success' => false,'result' => 0)); exit; } echo json_encode(array('success' => true,'result' => $result)); $sql2 = "DELETE FROM $names WHERE name='$ime'"; $result2 = MysqL_query($sql2);