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php+jquery+ajax+json的一个最简单实例

来源:互联网 收集:自由互联 发布时间:2021-06-30
html页面: html head meta http-equiv="content-type" content="text/html;charset=utf-8" / script type="text/javascript" src="jquery-1.8.2.min.js"/script script type="text/javascript" $(function(){ $("#send").click(function(){ var cont = $(

html页面:

<html>
<head>
<meta http-equiv="content-type" content="text/html;charset=utf-8" />
<script type="text/javascript" src="jquery-1.8.2.min.js"></script>
<script type="text/javascript">
 $(function(){
      $("#send").click(function(){
       var cont = $("input").serialize();
       $.ajax({
            url:'ab.php',
            type:'post',
            dataType:'json',
            data:cont,
            success:function(data){
             var str = data.username + data.age + data.job;
             $("#result").html(str);
        }
   });
  });  
 });
</script>
</head>
<body>
<div id="result">一会看显示结果</div>
<form id="my" action="" method="post">
           <p><span>姓名:</span> <input type="text" name="username" /></p>
          <p><span>年龄:</span><input type="text" name="age" /></p>
          <p><span>工作:</span><input type="text" name="job" /></p>
</form>
<button id="send">提交</button>
</body>
</html>

/*******************************************/

php页面:

<?php
header("Content-type:text/html;charset=utf-8");
        $username = $_POST['username'];
        $age = $_POST['age'];
        $job = $_POST['job'];
        $json_arr = array("username"=>$username,"age"=>$age,"job"=>$job);
        $json_obj = json_encode($json_arr);
        echo $json_obj;
?>

使用post方式

<script type="text/javascript">  $(function(){   $("#send").click(function(){       var cont = {username:$("input")[0].value,age:$("input")[1].value,job:$("input")[2].value};       var url = 'ab.php';       $.post(url,cont,function(data){        var res = eval("(" + data + ")");//转为Object对象       var str = res.username + res.age + res.job;     $("#result").html(str);    });   });    }); </script>

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