如何从json中提取这些值

考虑到这一点：

{
    "type": "https://tools.ietf.org/html/rfc7231#section-6.5.1",
    "title": "One or more validation errors occurred.",
    "status": 400,
    "traceId": "|a134a743-4f46942d175af9d6.",
    "errors": {
        "$.payment[0].invoiceDate": [
            "The JSON value could not be converted to System.Nullable`1[System.Int32]. Path: $.payment[0].invoiceDate | LineNumber: 0 | BytePositionInLine: 533."
        ]
    }
}

如何提取$.payment[0].invoiceDate和JSON值。。。变成变量

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使用SQL Server 2016。

这里有三种不同的方法

declare @json           nvarchar(max)=N'{
    "type": "https://tools.ietf.org/html/rfc7231#section-6.5.1",
    "title": "One or more validation errors occurred.",
    "status": 400,
    "traceId": "|a134a743-4f46942d175af9d6.",
    "errors": {
        "$.payment[0].invoiceDate": [
            "The JSON value could not be converted to System.Nullable`1[System.Int32]. Path: $.payment[0].invoiceDate | LineNumber: 0 | BytePositionInLine: 533."
        ]
    }
}';

/* #1 without OPENJSON */
select json_value(@json, '$.errors."$.payment[0].invoiceDate"[0]') newCol;

/* #2 with OPENJSON and JSON_VALUE */
select json_value(value, '$[0]') newCol
from openjson(@json,'$.errors');

/* #3 with OPENJSON */
select value newCol
from openjson(@json,'$.errors."$.payment[0].invoiceDate"');

每个输出都是相同的

newCol
The JSON value could not be converted to System.Nullable`1[System.Int32]. Path: $.payment[0].invoiceDate | LineNumber: 0 | BytePositionInLine: 533.

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我的尝试在第一步就失败了。只需从中选择JSON_VALUEvalue“$.errors”作为例外IDOPENJSON@json，以“$.errors”为例；给出消息13607，级别16，状态4，第18行JSON路径未正确格式化。在位置2处发现意外字符。选择*，json_valuevalue，$[0]”来自openjson@json，“$错误”