JSON比较元素或元素的存在

Question

如何在json中查找和比较值？

{"data 1": {
    "product": "54",
    "info": {
        "browser": "Safari",
        "isp": "isp sample",
        "width": 500,
        "ip": 109.24.55.121,
        "uid": 124,
        "location": 10.23,123.90
    }
}
{"data 2": {
    "campaign": "2",
    "info": {
        "browser": "Mozilla",
        "isp": "isp sample",
        "width": 500,
        "ip": 109.24.55.121,
        "uid": 231,
        "location": 10.23,123.90
    }
}
{"data 3": {
    "campaign": "2",
    "info": {
        "browser": "Mozilla",
        "isp": "isp sample",
        "width": 500,
        "ip": 109.24.55.121,
        "uid": 227,
        "location": 10.23,123.90
    }
}
{"same data": {
    "campaign": "2",
    "info": {
        "browser": "Mozilla",
        "isp": "isp sample",
        "width": 500,
        "ip": 109.24.55.121,
        "uid": 227,
        "location": 10.23,123.90
    }
}
{"same data": {
    "campaign": "2",
    "info": {
        "browser": "Mozilla",
        "isp": "isp sample",
        "width": 500,
        "ip": 109.24.55.121,
        "uid": 227,
        "location": 10.23,123.90
    }
}

我想输出what are there，similarities.on，what string，what values。

Answer 1

$data = json_decode($your_json_data);

foreach($data as $datum) {
    .... do some processing ...
}

json_encode/json_decode将在原生PHP数组和原生Javascript数组/对象之间进行转换。您没有说明希望使用哪种语言进行处理，但由于JSON是javascript，因此在javascript端不需要进行解码。只需加载数据即可开始。在PHP上，您需要首先将其解码为原生PHP数组。