Question

使用MS SQL Server 2017，我有2个表：

贷款：列包括贷款，OrganizationID
组织：列包括OrganizationID

我的目标是生成JSON，其中包含一个包含组织的贷款编号的字符串数组。

我正在按照以下方式使用SQL：

SELECT
    OrganizationID,
    (   SELECT '[' + STRING_AGG('''' + Loan + '''', ',') + ']' 
        FROM Loan 
        WHERE Loan.OrganizationID = Organization.OrganizationID
    ) AS [Loans]
FROM Organization 
WHERE OrganizationID = 1
FOR JSON PATH, WITHOUT_ARRAY_WRAPPER;

我想得到：

{
    "OrganizationID": 1,
    "Loans": ['Test 001','Test Loan 123','Test Loan 234']
}

但是，我实际得到的是用引号将“贷款”值包裹起来的：

{
    "OrganizationID": 1,
    "Loans": "['Test 001','Test Loan 123','Test Loan 234']"
}

我知道我可以做到：

SELECT
    OrganizationID,
    (   SELECT Loan 
        FROM Loan 
        WHERE Loan.OrganizationID = Organization.OrganizationID
        FOR JSON PATH
    ) AS [Loans]
FROM Organization 
WHERE OrganizationID = 1
FOR JSON PATH, WITHOUT_ARRAY_WRAPPER;

产生这个：

{
    "OrganizationID": 1,
    "Loans": [{
        "Loan": "Test 001"
    }, {
        "Loan": "Test Loan 123"
    }, {
        "Loan": "Test Loan 234"
    }]
}

但是，我的要求是生成一个简单的字符串数组，而不是一个Loan对象的数组。

关于如何塑造JSON的建议将不胜感激！

Answer 1

JSON字符串使用"作为分隔符，而不是'。另外，我们应注意转义（以防万一）。考虑到这些事项，JSON_QUERY将帮助我们：

SELECT
    OrganizationID,
    JSON_QUERY((   
        SELECT '[' + STRING_AGG('"' + STRING_ESCAPE(Loan, 'json') + '"', ',') + ']' 
        FROM Loan 
        WHERE Loan.OrganizationID = Organization.OrganizationID
    )) AS [Loans]
FROM Organization 
WHERE OrganizationID = 1
FOR JSON PATH, WITHOUT_ARRAY_WRAPPER;

SQL FOR JSON发出简单数组

1 个答案: