SQL结果添加

Question

我有带有表的数据库：

• 贷款
• 文档
• DocumentTemplate

文档具有loanId和documentTeplateId作为外键。

每个贷款都有多个文档。

我想选择没有文档loanId的每笔贷款（或teplateId 100）。

目前，我对以下SQL感到困惑：

SELECT l.id as loanId, d.id as documentId, d.document_templateid as documentTeplateId 
FROM loan as l
LEFT JOIN document as d ON (d.loanid = l.id)
WHERE d.document_templateid != 100
ORDER BY loanId DESC

很明显，它会给我类似的信息。

但这不是我想要的。

有什么建议吗？

Answer 1

SELECT l.id as loanId
FROM loan as l
LEFT JOIN document as d
ON (d.loanid = l.id)
WHERE d.document_templateid != 100
ORDER BY loanId  DESC
GROUP BY loanId

“ GROUP BY loanId”会将具有相同loanId的行分组为一行，从而删除重复项。您只能选择loan.id使其正常工作，因为您似乎表明这是您唯一需要的值，它非常适合您的情况。

Answer 2

您只需要不同的贷款ID：

SELECT distinct l.id as loanid
FROM loan as l
LEFT JOIN document as d
ON (d.loanid = l.id)
WHERE d.document_templateid != 100
ORDER BY loanId desc

Answer 3

我认为您想要聚合和一个having`子句：

SELECT l.id as loanid
FROM loan l LEFT JOIN
     document d
     ON d.loanid = l.id
GROPU BY l.id
HAVING SUM( d.document_templateid = 100 ) = 0;

如果您只想考虑有单据的贷款，则不需要JOIN：

SELECT d.loanid
FROM document d
GROPU BY d.loanid
HAVING SUM( d.document_templateid = 100 ) = 0;