zoukankan      html  css  js  c++  java
  • ajax在php中应用实例

    1,ajax分为$.ajax(),$.get(),$.post(),$.getJSON() 几种形式,实例如下:

    <html>
    <meta http-equiv="Content-Type" content="text/html;charset=utf-8">
    <script type="application/javascript" src="../js/jquery-1.7.2.js"></script>
    <script type="application/javascript">
    $(document).ready(function(){
        $("#ajaxBut").click(function(){
            $.ajax({
                'type':'get',
                'url':'test4.php',
                'dateType':'json',
                'data':$("input").serialize(),
                'success':function(ret){
                    alert(ret);
                }
            });
    
        });
        $("#getBut").click(function(){
            $.get("test4.php",$("input").serialize(),function(ret){
                alert(ret);
            });
        });
        $("#postBut").click(function(){
            $.post("test5.php",$("input").serialize(),function(ret){
                alert(ret);
            });
        });
        $("#jsonBut").click(function(){
            $.getJSON("test4.php",$("input").serialize(),function(ret){
                alert(ret);
            });
        });
    
    });
    </script>
    <body>
    <form>
        <h1>user Login</h1>
        username:<input type="text" name="user" id="user" /><br/>
        password:<input type="password" name="password" id="password"/><br/>
        <input type="button" name="but" id = "ajaxBut" value="ajaxLogin" />
        <input type="button" name="but" id = "postBut" value="postLogin" />
        <input type="button" name="but" id = "getBut" value="getLogin" />
        <input type="button" name="but" id = "jsonBut" value="jsonLogin" />
    </form>
    </body>
    </html>
    

    test4.php

    <?php
    $username = $_GET['user'];
    $password = $_GET['password'];
    $ret = "fail";
    if($username == 'zhangsan' && $password == '123'){
        $ret = "success";
    }
    echo json_encode($ret);

    test5.php

    <?php
    $username = $_POST['user'];
    $password = $_POST['password'];
    $ret = "fail";
    if($username == 'zhangsan' && $password == '123'){
        $ret = "success";
    }
    echo json_encode($ret);
    

    2,ajax跨域获取数据,使用到jsonp,实例如下:

     $.getJSON("http://www.ganji.com/test6.php?callback=?", $("input").serialize() , function(data){
        if(data){
            console.log(data);
        }
     });
    

    test6.php

    $str = 'OK';
    $callback = $_GET('callback');
        if (!empty($callback)) {
            header("content-type: application/x-javascript; charset=UTF-8");
            echo $callback . '(' . $str . ')';
        } else {
            echo $str;
        }
    }      
    

      

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  • 原文地址:https://www.cnblogs.com/usa007lhy/p/5905763.html
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