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此代码假设接受用户值，然后给出相应的警报框，但它不起作用。我可以知道这段代码有什么问题吗？我做了很多谷歌搜索（很多指向这个网站），但我似乎找不到解决方案？请提前帮助和感谢。

索引.html

<html>
<head>
<title>the title</title>
   <script type="text/javascript" src="http://code.jquery.com/jquery-1.9.0.min.js"></script>

<script type="text/javascript">

 function check() {
      var    one = $('#first').val();
       var   two = $('#second').val();
       var  three = $('#third').val();
       var  four = $('#fourth').val();  

    $.post("result.php", { 
    first: one,
    second: two,
    third: three,
    fourth: four
},function(data)
        {

         if (data=="yay") //for no input
         { alert("yay");
         }
         else 
         {
         alert("nay");
         }

         }


        } 

</script>

</head>
<body>

        <form name="form">
    <input type="text" id="first">
    <input type="text" id="second">
     <input type="text" id="third">
    <input type="text" id="fourth">
    <input type="button" value="get" onclick="check();">
    </form>


</body>
</html>

结果.php

<?php

    $one = $_POST["first"];
        $two = $_POST["second"];
         $three =$_POST["third"];
         $four = $_POST["fourth"];
         if($one > 5)        
         {

         echo "yay";
         }
     elseif($two > 10 )        {

         echo "yay";  }

             elseif($three > 15 )        
             {

         echo "yay";  
             }

             elseif($four > 20 )        
             {

         echo "yay";  
             }

           else{
               echo "nay";
           }

?>
4

1 回答 1

1

json 数据类型

    ///<script ....
     $.post("result.php", { 
    first: one,
    second: two,
    third: three,
    fourth: four
},function(data){
    if (data.response=="yay") { alert("yay"); }
    else { alert("nay"); }
},"json");

//</script>

<?php

    $one = $_POST["first"];
    $two = $_POST["second"];
    $three =$_POST["third"];
    $four = $_POST["fourth"];

   if($one > 5)        
         {

         $response = "yay";
         }
     elseif($two > 10 )        {

         $response = "yay";  }

             elseif($three > 15 )        
             {

         $response = "yay";  
             }

             elseif($four > 20 )        
             {

         $response = "yay";  
             }

           else{
               $response = "nay";
           }

           echo json_encode(array("response" => $response));

?>
于 2013-05-27T17:35:49.890 回答