Question

从我的对象中提取数据没有问题。我的问题是编辑字符串中的数据并重新编码。每次我尝试编辑对象时，它都会删除对象中的所有数据，只保存我编辑的内容。

我会认为这有效，但事实并非如此。有什么建议？ （以下显示在对象模式下，我也尝试将其作为关联数组并获得相同的结果）

    $jsonString = '[{ "stuff" : [{"name" : "name", "description" : "description", "id" : "id",}], "morestuff" : []}]';
    $name = 'new name';
    $description = 'new description';
    $obj_json = json_decode($jsonString);
    $obj_json->stuff->name = $name;
    $obj_json->stuff->description = $description;
    $newJsonString = json_encode($obj_json);

以下是打印的内容：

{ "stuff" : {"name" : "new name", "description" : "new description"}}

Answer 1

你的代码似乎是正确的，但试试这个（也许有修改对象的东西..）：

$obj_json = json_decode($jsonString, true); //as associative array
$obj_json['stuff']['name'] = $name;
$obj_json['stuff']['description'] = $description;
$newJsonString = json_encode($obj_json);

使用json作为关联数组

Answer 2

做你问的问题没问题：

<?php

$jsonString = '{
    "stuff": {
        "name": "Original name",
        "description": "Original description",
        "foo": "Another field"
    }
}';
$name = "New name";
$description = "New description";

$obj_json = json_decode($jsonString);
$obj_json->stuff->name = $name;
$obj_json->stuff->description = $description;
$newJsonString = json_encode($obj_json);

echo $newJsonString . PHP_EOL;

...打印：

{"stuff":{"name":"New name","description":"New description","foo":"Another field"}}

你可能正在阅读或写错了属性。

修改

仔细观察，您的数据被包装在一个数组中，而stuff本身也是一个数组：

$jsonString = '[{ "stuff" : [{"name" : "name", "description" : "description", "id" : "id",}], "morestuff" : []}]'; ^ ^ ^ ^ | \______________________________________________________________/ | \_______________________________________________________________________________________________/

修改＃2 ：事实上，您的数据为not valid JSON而json_decode()返回null：

$jsonString = '[{ "stuff" : [{"name" : "name", "description" : "description", "id" : "id",}], "morestuff" : []}]'; $obj_json = json_decode($jsonString); var_dump($obj_json, json_last_error());

NULL int(4)

错误＃4是JSON_ERROR_SYNTAX：语法错误，格式错误的JSON

如何编辑使用json_decode（）创建的PHP对象？

2 个答案: