2

我有一个 json 文件，我想构建一个允许我在文件中添加/编辑元素的表单。是否有一个 jQuery 函数/方法可以让我能够在外部 json 文件中发布和附加元素？

不确定这是否会有所帮助，但当前的 json 结构如下：

 [ { "cast" : "",
    "director" : "",
    "genre" : "",
    "id" : false,
    "nrVotes" : 0,
    "plot" : "",
    "rating" : 0,
    "runtime" : 0,
    "title" : "",
    "year" : false
  },
  { "cast" : "Tim Robbins, Morgan Freeman, Bob Gunton, ",
    "director" : "Frank Darabont",
    "genre" : "Crime Drama ",
    "id" : "0111161",
    "nrVotes" : 968510,
    "plot" : "Two imprisoned men bond over a number of years, finding solace and eventual redemption through acts of common decency.",
    "rating" : 9.3000000000000007,
    "runtime" : 142,
    "title" : "The Shawshank Redemption",
    "year" : "1994"
  }]

非常感谢您提前！！（：

4

2 回答 2

5

如果您使用 jQuery 的getJSON或parseJSON()，则您有一个可以操作的 javascript 对象。例如：

$.getJSON( "/test.json", function( data ) {
  // now data is JSON converted to an object / array for you to use.
  alert( data[1].cast ) // Tim Robbins, Morgan Freeman, Bob Gunton

  var newMovie = {cast:'Jack Nicholson', director:...} // a new movie object

  // add a new movie to the set
  data.push(newMovie);      
});

您现在要做的就是保存文件。您可以使用 jQuery.post() 将文件发送回服务器以便为您保存。

更新：发布示例

//inside getJSON()

var newData = JSON.stringify(data);
jQuery.post('http://example.com/saveJson.php', {
    newData: newData
}, function(response){
    // response could contain the url of the newly saved file
})

在您的服务器上，使用 PHP 的示例

$updatedData = $_POST['newData'];
// please validate the data you are expecting for security
file_put_contents('path/to/thefile.json', $updatedData);
//return the url to the saved file
于 2013-10-20T04:27:18.440 回答
0 
<html>
<head>
    <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css" integrity="sha384-BVYiiSIFeK1dGmJRAkycuHAHRg32OmUcww7on3RYdg4Va+PmSTsz/K68vbdEjh4u" crossorigin="anonymous">
    <script type="text/javascript" src="http://code.jquery.com/jquery-1.4.3.min.js" ></script>
</head>
<body>
    <?php
        $str = file_get_contents('data.json');//get contents of your json file and store it in a string
        $arr = json_decode($str, true);//decode it
         $arrne['name'] = "sadaadad";
         $arrne['password'] = "sadaadad";
         $arrne['nickname'] = "sadaadad";
         array_push( $arr['employees'], $arrne);//push contents to ur decoded array i.e $arr
         $str = json_encode($arr);
        //now send evrything to ur data.json file using folowing code
         if (json_decode($str) != null)
           {
             $file = fopen('data.json','w');
             fwrite($file, $str);
             fclose($file);
           }
           else
           {
             //  invalid JSON, handle the error 
           }

        ?>
    <form method=>
</body>

数据.json

{  
  "employees":[  
  {  
     "email":"11BD1A05G9",
     "password":"INTRODUCTION TO ANALYTICS",
     "nickname":4
  },
  {  
     "email":"Betty",
     "password":"Layers",
     "nickname":4
  },
  {  
     "email":"Carl",
     "password":"Louis",
     "nickname":4
  },
  {  
     "name":"sadaadad",
     "password":"sadaadad",
     "nickname":"sadaadad"
  },
  {  
     "name":"sadaadad",
     "password":"sadaadad",
     "nickname":"sadaadad"
  },
  {  
     "name":"sadaadad",
     "password":"sadaadad",
     "nickname":"sadaadad"
  }
   ]
}
于 2016-08-23T02:05:29.537 回答