如何在jquery中解码/分离php编码的json? [英] How to decode / separate php encoded json in jquery.?

问题描述

我想将编码的json解码为jquery，并将所有分离的代码设置为不同的表单元素.在这里，我发布了所有代码.

I want to decode my encoded json to jquery and set all separated code to different form element. Here I post my all code.

php代码.

<?php
    if(isset($_POST['passid']) === true && empty($_POST['passid'])=== false)
{
    //$q = $_POST['increament'];[![enter image description here][1]][1]
    mysql_connect('localhost','root', '');
    mysql_select_db('kmk_inst');
    $query = mysql_query("select Q_Id, QP_Name, Question from question_paper where Q_Id = '".mysql_real_escape_string(trim($_POST['passid']))."'")
    or die("Wrong Query".mysql_error());
    //echo (mysql_num_rows($query) !== 0) ? mysql_result($query, 0 , 'QP_Name'): 'passid not found';
    //$row = mysql_fetch_array($query);

    if ( mysql_num_rows($query) !== 0 ) {
        $data = json_encode(mysql_fetch_assoc($query));
        echo $data;
    }
?>

HTML代码

<a class="btn btn-primary col-xs-12 col-lg-2" style="float: right; margin:10px;" id="savenext" name="savenext" onclick="return dataPass()">Save & Next</a></span>
<input type="text" style="width: 410px;text-align: center; margin: 0px;" class="onlyNumber form-control pull-left" id="ques" value="1" name="ques" />
<input type="text" style="width: 410px;text-align: center; margin: 0px;" class="onlyNumber form-control pull-left" id="QPt" value="1" name="QPt" />

** jQuery **

** jquery **

$('a#savenext').on('click',function(){
var passid = $('input#ques').val();
//var name = $('input#QPt').val();
if($.trim(passid) != '')
{
        $.post('retrivedata.php', { passid: passid}, function(data){

            //$.each(data, function(i, name) {
            //      alert(name.Q_Id);
            //});
            $('div#datatable').text(data);
            //$('input#QPt').text(QPt);
        });
    }
});

我是jquery Ajax和JavaScript的新手，所以我无法处理它.任何帮助表示赞赏.

I am new to jquery Ajax and JavaScript so I can`t handle it. any help appreciate.

目前我正在得到这样的结果.

Currently I am getting result like this.

推荐答案

在php中设置适当的标头

Set the proper header in the php

header('Content-Type: application/json');
$data = json_encode(mysql_fetch_assoc($query));

jquery将负责解析

jquery will take care of the parsing

使用附录将数据添加到页面中

to add the data to your page use append

      $('body').append('<form><p>'+data.Question+'<input value="'+data.QP_Name+'"></from>');

var data = {
  "Q_Id": "1",
  "QP_Name": "test1",
  "Question": "Which is indian capital.?"
};
$('#ques').val(data.Question);
$('#QPt').val(data.QP_Name);
$('body').append('<form><p>' + data.Question + '<input value="' + data.QP_Name + '"></from>');

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<a class="btn btn-primary col-xs-12 col-lg-2" style="float: right; margin:10px;" id="savenext" name="savenext" onclick="return dataPass()">Save & Next</a></span>
<input type="text" style="width: 410px;text-align: center; margin: 0px;" class="onlyNumber form-control pull-left" id="ques" value="1" name="ques" />
<input type="text" style="width: 410px;text-align: center; margin: 0px;" class="onlyNumber form-control pull-left" id="QPt" value="1" name="QPt" />

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