Question

我正在使用的代码工作正常，但我有点卡住尝试从成功函数加载新页面。最简单的方法是，如何过滤来自php后端的数据来做出决策？例如，我在后端testajax.php页面和ajax成功函数测试中为某些条件设置了一个条件。在php中，它将使用if语句。我对jquery很新，所以我会很感激如何实现这一点。感谢

$.ajax({
  type: "POST",
  url: "testajax.php",
  data: data,
  success: function (data) {

   $('#login_message').html(data);
   //$(ok).val('Logged In');
   //$("#login").get(0).reset();
   //$("#form").dialog('close');
   $.mobile.changePage( "admin/index.php/", { transition: "slideup"} );
  },
   error:function (xhr, ajaxOptions, thrownError){
   alert('There was an exception thrown somewhere');
   alert(xhr.status);
   alert(thrownError);
  }
});

testajax.php

<?php
  // test wether the user session is already set
  session_start();
  $userpost = mysql_real_escape_string($_POST['user']);

  if (!isset($_SESSION['user'])) {
    $_SESSION['user'] = $userpost;
   if($_SESSION['user']=='admin') {
     echo 'Welcome Admin';
     // for example, test data in ajax for 
     // this flag and if equal 1 then load admin/index.php
     $flag = 'admin';
     // end of flag
     echo $flag;
  }
  else
     echo 'Unknown User';
  }
?>

Answer 1

没有什么可以阻止你在jquery代码中使用if语句，所以你可以使用ajax回调：

if (data == 'Unknown User') {
  alert('Unknown User');
} else {
  //Do some other stuff here
}

Answer 2

<?php
  // test wether the user session is already set
  session_start();
  $userpost = mysql_real_escape_string($_POST['user']);

  if (!isset($_SESSION['user'])) {
    $_SESSION['user'] = $userpost;
   if($_SESSION['user']=='admin') {        
     $flag = 'admin';        
     echo $flag;
  }
  else
     echo 'Unknown User';
  }
?>

ajax part

$.ajax({
  type: "POST",
  url: "testajax.php",
  data: data,
  success: function (data) {

   $('#login_message').html(data);
   if(data==='admin')
   { 
    //redirect
    $.mobile.changePage( "admin/index.php/", { transition: "slideup"} );
   }

  },
   error:function (xhr, ajaxOptions, thrownError){
   alert('There was an exception thrown somewhere');
   alert(xhr.status);
   alert(thrownError);
  }
});

Answer 3

如果请求是ajax请求，您可以检查该请求，并仅输出ajax部分所需的内容。

    if(strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) == 'xmlhttprequest')
    {
      //request by ajax
    }
    else
    {
      //regular request.
    }

jQuery ajax从成功加载页面

3 个答案: