PHP JSON数据选择

Question

我如何从以下JSON中选择每个标题的数据？我有JSON解码,但我不知道如何选择我想要的部分.

{
    "responseData": {
        "results": [
            {
                "title": "Justin Giesbrecht 749",
                "titleNoFormatting": "Justin Giesbrecht 749",
            },
            {
                "title": "Gopher dunes 09",
                "titleNoFormatting": "Gopher dunes 09",
            },
            {
                "title": "dirtbike Justin",
                "titleNoFormatting": "dirtbike Justin",
            },
            {
                "title": "A Warming",
                "titleNoFormatting": "A Warming",
            }
        ],
        "cursor": {
            "pages": [
                {
                    "start": "0",
                    "label": 1
                },
                {
                    "start": "4",
                    "label": 2
                }
            ],
            "estimatedResultCount": "6",
            "currentPageIndex": 0,
        }
    },
    "responseDetails": null,
    "responseStatus": 200
}

我以为会是这样的,但我什么都没得到:

echo "Response ". $jsonS->responseData->results[1]->title;

Answer 1

实际上你已经正确阅读了标题部分,这是JSON无效的.

将JSON复制到JSON验证器/ lint(例如http://www.jsonlint.com/)将显示您在几个地方的最后一个对象属性之后有其他(逗号)(确切地说,在每个'titleFormatting'属性之后5个位置)之后'currentPageIndex').

如果你修复这些错误并使用json_decode例如解析它:

$jsonS = json_decode($json_text);

然后你自己的代码:

echo "Response " . $jsonS->responseData->results[1]->title;

将输出第二个(索引1是第二个索引)结果标题

响应Gopher dunes 09