我有以下查询: $stmt = $conn->query("SELECT * FROM research as r LEFT JOIN research_participants as rp ON rp.research_no = r.id LEFT JOIN researcher as rc ON rc.id = rp.researcher_id where r.id = $researchid");这个查询工作得很好,但是,我希望它只选择状态=“完成”的研究,所以我将其更改为以下内容:$stmt = $conn->query("SELECT * FROM research WHERE status= 'done' as r LEFT JOIN researcher as rp ON r.researcher_id = rp.id");不幸的是,它不起作用。我还尝试了以下方法: $stmt = $conn->query("SELECT * FROM research as r LEFT JOIN researcher as rp ON r.researcher_id = rp.id WHERE status= 'done'");但即使这样也行不通。研究表如下所示: 在此处输入图像描述
- 2 回答
- 0 关注
- 124 浏览
添加回答
举报
0/150
提交
取消