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深入extjs与php参数交互的详解_php技巧

WBOYOriginal: 2016-05-17 08:58:50869browse

  HelloWorld

 $userName = $_POST["userName"];
 $password = $_POST["password"];
 $msg = "";
 if($userName=="lowkey" && $password=="zq19890319") {
  $msg = "登陆成功";
 } else {
 $msg = "登陆失败";
 }
 echo($msg);
?>

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深入extjs与php参数交互的详解_php技巧

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