Question

我有json文件，它具有以下json对象结构

{
  "Project_3": {
      "link": "",
      "title": "",
      "desc": ""
  },
  "Project_2": {
      "link": "",
      "title": "",
      "desc": ""
  },
  "Project_1": {
      "link": "",
      "title": "",
      "desc": ""
  }
}

现在我想要的是像这样更新这个json文件

Project_3 = Project_2

Project_2 = Project_1

我的意思是Projects_3的内容应该与Projects_2的内容一样，Project_2的内容应该与Project_1的内容一样，Project_1的内容应该是表单数据的下一个输入

Uptil现在这是我在php中尝试过的。

$projects = json_decode(file_get_contents('../json/recent_projects.json','w'));

$projects->Project_3 = $projects->Project_2;
$projects->Project_2 = $projects->Project_1;

$projects->Project_1->link  = htmlspecialchars($_POST['project_link']);
$projects->Project_1->title = htmlspecialchars($_POST['project_name']);
$projects->Project_1->desc  = htmlspecialchars($_POST['project_desc']);

$fh = fopen("../json/recent_projects.json", 'w') or die('File cannot be opened');
fwrite($fh, json_encode($projects,JSON_UNESCAPED_UNICODE));
fclose($fh);

但是，Project_3和Project_2的内容始终保持不变，只有Project_1更新。我不明白为什么会发生这种情况。

修改

我得到了我的问题的答案，但为什么在这个过程中没有使用。我没有得到！热烈欢迎任何帮助或建议！

Answer 1

测试了您的代码，但Project_3和Project_2确实发生了变化。我得到的错误是Project_2和Project_1都被$_POST变量更新了。所以我只是unset() Project_1。现在它的工作就像你想要的那样。

<?php 

$json = '{
  "Project_3": {
      "link": "L3",
      "title": "T3",
      "desc": "D3"
  },
  "Project_2": {
      "link": "L2",
      "title": "T2",
      "desc": "D2"
  },
  "Project_1": {
      "link": "L1",
      "title": "T1",
      "desc": "D1"
  }
}';


$projects = json_decode($json);

echo "<pre>";
print_r($projects);

$projects->Project_3 = $projects->Project_2;
$projects->Project_2 = $projects->Project_1;

unset($projects->Project_1);


$projects->Project_1->link  = htmlspecialchars($_POST['project_link']);
$projects->Project_1->title = htmlspecialchars($_POST['project_name']);
$projects->Project_1->desc  = htmlspecialchars($_POST['project_desc']);

echo "<pre>";
print_r($projects);

$fh = fopen("../json/recent_projects.json", 'w') or die('File cannot be opened');
fwrite($fh, json_encode($projects,JSON_UNESCAPED_UNICODE));
fclose($fh);

?>

输出：

stdClass Object
(
    [Project_3] => stdClass Object
        (
            [link] => L3
            [title] => T3
            [desc] => D3
        )

    [Project_2] => stdClass Object
        (
            [link] => L2
            [title] => T2
            [desc] => D2
        )

    [Project_1] => stdClass Object
        (
            [link] => L1
            [title] => T1
            [desc] => D1
        )

)

stdClass Object
(
    [Project_3] => stdClass Object
        (
            [link] => L2
            [title] => T2
            [desc] => D2
        )

    [Project_2] => stdClass Object
        (
            [link] => L1
            [title] => T1
            [desc] => D1
        )

    [Project_1] => stdClass Object
        (
            [link] => L5
            [title] => T5
            [desc] => D5
        )

)

无法在PHP中更新json文件

1 个答案: