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  • jq之ajax以及json数据传递

    <html>
     <head>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />     
    <title>闭包演示</title>
    <script src="js/jquery-2.1.1.min.js"></script>    
    <script type="text/javascript">   

        function transfer(){
    var dat = $( "#sel" ).val();
    //console.log(dat)
       $.ajax({

       type:"POST",                                       //post 同 get

      url:"ts1.php",

      data:{data:dat},               
    //console.log(data);
      success:function(data){
        var data = eval('(' + data + ')');
        console.log(data);
        console.log(data.authors[0]);return;
        //alert(data);
       $( '#sub' ).val( data );

      }

    })

    }
     </script>
     </head>
     
     <body>
      <select id = "sel" onchange = "transfer()">
       <option value = "welcome">欢迎</option>
       <option value = "hello">你好</option>
       <option value = "pity">可惜</option>
       <option value = "ok">好的</option>
      </select>
      <input type = "text" id = "sub" value = "name" />
     </body>
    </html>


    <?php
    //$data[] = 'hello';
    //
    //$data = $_POST['data'];
    //echo $data;
    $array = array('a'=>'a1','b'=>'b1');

    //print_r($array['a']) ;
    $data='{
    "title": "narto",
    "authors": [
    "Nicholas C. Zakas","a2"
    ],
    "edition": 3,
    "year": 2011
    }';
    $data=json_decode($data, true);
    $data=json_encode($data);
    print_r($data);



    $json = '{"a":1,"b":2,"c":3,"d":4,"e":5}';
    $json=json_decode($json, true);
    //print_r($json);

    // 演示可能出现的数据库部分 注:未连接数据库, 该部分只是演示说明
    // $sql = 'SELECT * FROM `TABLE` WHERE ID = '.$data.';';
    // $query = mysql_query( $sql );
    // $result = mysql_fetch_row( $query );
    //输出响应

    ?>

    打印的

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  • 原文地址:https://www.cnblogs.com/lsc-boke/p/6136271.html
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