<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>
<?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);
}
我正在尝试创建一个允许我的用户向网站添加条目的表单。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"}
}
答案 0 :(得分: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);
}
答案 1 :(得分:1)
你知道json_encode吗?只需使用给定结构将关联数组传递给此函数
创建一个新数组,然后将其传递给json_encode
$array = array(
"startDate" => $_POST['startDate'],
//..
"assets" => array(
"media" => $_POST["media"]
//..
)
);
$json = json_encode($array);