在PHP中组合2个或更多JSON

我当前的JSON值如下所示：

{"vegetable_names":["vegetables 1","vegetables 2","vegetables 3"]}{"grade":["XXL","A","B","S"]}{"packages":["Carton Boxes","Baskets"]}

{"vegetable_names":["vegetables 1","vegetables 2","vegetables 3"],"grade":["XXL","A","B","S"],"packages":["Carton Boxes","Baskets"]}

但我希望输出如下：

{"vegetable_names":["vegetables 1","vegetables 2","vegetables 3"]}{"grade":["XXL","A","B","S"]}{"packages":["Carton Boxes","Baskets"]}

{"vegetable_names":["vegetables 1","vegetables 2","vegetables 3"],"grade":["XXL","A","B","S"],"packages":["Carton Boxes","Baskets"]}

---

我猜你得到了你描述的Json。如果不是，我建议你遵循保罗·克罗维拉的解决方案，从一开始就做到正确

假设您没有选择，那么将Json解码为一个对象或数组，合并这两个或更多个对象，然后将它们再次编码为Json如何。 丑陋，但是如果Json是由独立于您的源代码创建的，您可以这样做。

只是一些数据结构的文本表示。从JSON字符串中恢复数据结构，使用PHP对其进行修改（或创建新的数据结构，如果更合适），再次将更新/新的数据结构编码为

JSON

您作为输入发布的“JSON”似乎根本不是

JSON

。它是三个

JSON

字符串的串联。这样不行

// Input JSON strings
$json1 = '{"vegetable_names":["vegetables 1","vegetables 2","vegetables 3"]}';
$json2 = '{"grade":["XXL","A","B","S"]}';
$json3 = '{"packages":["Carton Boxes","Baskets"]}';

// Restore the data structures as arrays
$data1 = json_decode($json1, TRUE);
$data2 = json_decode($json2, TRUE);
$data3 = json_decode($json3, TRUE);

// Combine them; it seems all you need is a simple merging
$data = array_merge($data1, $data2, $data3);

// Encode the combined arrays as JSON again
$output = json_encode($data);

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echo{“蔬菜名称”：[“蔬菜1”、“蔬菜2”、“蔬菜3”]，“等级”：[“XXL”、“A”、“B”、“S”]，“包装”：[“纸箱”、“篮子”]不过，说真的，我们需要查看您的代码以提供任何有用的帮助。如果您不澄清数据是如何生成的，您的问题可能会被解决。