3

文件

索引.php

<form name="postform" action="post.php" method="post" enctype="multipart/form-data">
<table class="postarea" id="postarea">
    <tbody>
<tr>    <td class="postblock">Date:</td><td><input type="text" name="startDate"></td></tr>
<tr>    <td class="postblock">Title:</td><td><input type="text" name="headline"></td></tr>
<tr>    <td class="postblock">Article:</td><td><textarea id="text" rows="5" cols="30" type="text" name="text"></textarea> </td> </tr>
<tr>    <td class="assetblock">Image address:</td><td><input type="text" name="media"></td></tr>
<tr>    <td class="assetblock">Image caption:</td><td><input type="text" name="caption"></td></tr>

<tr>    <td class="postblock"></td><td> <input type="submit" value="Submit Entry"> </td>    </tr>
</tbody>
</table>
</form>
</form>

post.php

<?php
// check if a form was submitted
if( !empty( $_POST ) ){
// convert form data to json format
$json = json_encode( $_POST );
// make sure there were no problems
//if( json_last_error() != JSON_ERROR_NONE ){
    //exit;  // do your error handling here instead of exiting
// }
$file = 'entries.json';
// write to file
//   note: _server_ path, NOT "web address (url)"!
file_put_contents( $file, $json, FILE_APPEND);
}

问题

我正在尝试创建一个允许我的用户向站点添加条目的表单。条目存储在 JSON 文件中。我创建了一个提交到 JSON 表单的表单，其格式如下：

{"startDate":"example",
"headline":"example",
"text":"example",
"media":"example",
"caption":"example"}

媒体和标题与图像、视频或其他多媒体相关，我需要将它们存储为单独的对象，如下例所示。

 {"startDate":"example",
 "headline":"example",
 "text":"example",
 "asset" :{"media":"example",
          "caption":"example"}
 }
4

2 回答 2

9 
<?php
// check if a form was submitted
if( !empty( $_POST ) ){

// convert form data to json format
    $postArray = array(
      "startDate" => $_POST['startDate'],
      "headline" => $_POST['headline'],
      "text" => $_POST['text'],
      "asset" => array(
         "media" => $_POST["media"],
         "caption" => $_POST['caption']
        )
    ); //you might need to process any other post fields you have..

$json = json_encode( $postArray );
// make sure there were no problems
//if( json_last_error() != JSON_ERROR_NONE ){
    //exit;  // do your error handling here instead of exiting
// }
$file = 'entries.json';
// write to file
//   note: _server_ path, NOT "web address (url)"!
file_put_contents( $file, $json, FILE_APPEND);
}
于 2012-09-28T15:54:36.923 回答
1

你知道json_encode吗？只需使用给定结构将辅助数组传递给此函数

创建一个新数组 ie 并将其传递给 json_encode

$array = array(
    "startDate" => $_POST['startDate'],
    //..
    "assets" => array(
            "media" => $_POST["media"]
            //..
    )
);
$json = json_encode($array);
于 2012-09-28T15:45:05.987 回答