cre*_*eam 1 mysql sql select left-join
当应该有一些结果时它返回0行.
这是我第一次尝试使用JOIN,但从我所看到的这看起来非常正确,对吧?
"SELECT pending.paymentid, pending.date, pending.ip, pending.payer, pending.type, pending.amount, pending.method, pending.institution, payment.number, _uploads_log.log_filename
FROM pending
LEFT JOIN _uploads_log
ON pending.paymentid='".$_GET['vnum']."'
AND _uploads_log.linkid = pending.paymentid"
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我需要返回两个pending.paymentid和_uploads_log.log_filename等于的每个表中的指定值$_GET['vnum]
这样做的正确方法是什么?为什么我没有得到任何结果?
如果比我更有经验的人能指出我正确的方向,我会非常感激.
编辑
对于pending主键paymentid,_uploads_log主要是被调用的col log_id并且log_filename被列为索引.
试试这个
SELECT pending.paymentid,
pending.date,
pending.ip,
pending.payer,
pending.type,
pending.amount,
pending.method,
pending.institution,
payment.number,
_uploads_log.log_filename
FROM pending
LEFT JOIN _uploads_log
ON _uploads_log.linkid = pending.paymentid
WHERE _uploads_log.log_filename = '" . $_GET['vnum'] . "'
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