这感觉应该很容易，但我就是做不到，我已经尝试了所有的方法。
我如何才能仅仅获得此查询返回的行数：
(SELECT Alias1 . *,
Alias2 . *
FROM Table1 AS Alias1
LEFT JOIN Table2 AS Alias2 USING ( Col1 )
WHERE Alias1.Col1 != 'blah1'
AND Alias1.Col2 LIKE 'blah2'
AND ( Alias1.Col3 LIKE 'blah3'
OR Alias1.Col3 LIKE 'blah4' )
AND Alias1.Col4 = 'blah5'
AND Alias2.Col7 LIKE 'blah6'
ORDER BY Alias1.Col6 DESC
LIMIT 50)
UNION
(SELECT Alias1 . *,
Alias2 . *
FROM Table1 AS Alias1
LEFT JOIN Table2 AS Alias2 USING ( Col1 )
WHERE Alias1.Col1 != 'blah1'
AND Alias1.Col2 LIKE 'blah2'
AND ( Alias1.Col3 LIKE 'blah3'
OR Alias1.Col3 LIKE 'blah4' )
AND Alias1.Col4 = 'blah5'
AND Alias2.Col5 LIKE 'blah6'
ORDER BY Alias1.Col6 DESC
LIMIT 50)
UNION
(SELECT Alias1 . *,
Alias2 . *
FROM Table1 AS Alias1
LEFT JOIN Table2 AS Alias2 USING ( Col1 )
WHERE Alias1.Col1 != 'blah1'
AND Alias1.Col2 LIKE 'blah2'
AND ( Alias1.Col3 LIKE 'blah3'
OR Alias1.Col3 LIKE 'blah4' )
AND Alias1.Col4 = 'blah5'
ORDER BY Alias1.Col6 DESC
LIMIT 50)
我想我可以只使用PHP来执行查询，然后对结果执行mysql_num_rows，但我想直接使用SQL来执行，因为我听说这样做更快，因为它节省了一步。
谢谢！
编辑：
以下错误(#1060-列名"Col1"重复(：
SELECT COUNT(*) FROM (
(SELECT Alias1 . *,
Alias2 . *
FROM Table1 AS Alias1
LEFT JOIN Table2 AS Alias2 USING ( Col1 )
WHERE Alias1.Col1 != 'blah1'
AND Alias1.Col2 LIKE 'blah2'
AND ( Alias1.Col3 LIKE 'blah3'
OR Alias1.Col3 LIKE 'blah4' )
AND Alias1.Col4 = 'blah5'
AND Alias2.Col7 LIKE 'blah6'
ORDER BY Alias1.Col6 DESC
LIMIT 50)
UNION
(SELECT Alias1 . *,
Alias2 . *
FROM Table1 AS Alias1
LEFT JOIN Table2 AS Alias2 USING ( Col1 )
WHERE Alias1.Col1 != 'blah1'
AND Alias1.Col2 LIKE 'blah2'
AND ( Alias1.Col3 LIKE 'blah3'
OR Alias1.Col3 LIKE 'blah4' )
AND Alias1.Col4 = 'blah5'
AND Alias2.Col5 LIKE 'blah6'
ORDER BY Alias1.Col6 DESC
LIMIT 50)
UNION
(SELECT Alias1 . *,
Alias2 . *
FROM Table1 AS Alias1
LEFT JOIN Table2 AS Alias2 USING ( Col1 )
WHERE Alias1.Col1 != 'blah1'
AND Alias1.Col2 LIKE 'blah2'
AND ( Alias1.Col3 LIKE 'blah3'
OR Alias1.Col3 LIKE 'blah4' )
AND Alias1.Col4 = 'blah5'
ORDER BY Alias1.Col6 DESC
LIMIT 50)
) a
以下是我的错误(#1064-您的SQL语法有一个错误；(：
SELECT COUNT(*) FROM (
(SELECT Alias1 . *,
Alias2 . *
FROM Table1 AS Alias1
LEFT JOIN Table2 AS Alias2 USING ( Col1 )
WHERE Alias1.Col1 != 'blah1'
AND Alias1.Col2 LIKE 'blah2'
AND ( Alias1.Col3 LIKE 'blah3'
OR Alias1.Col3 LIKE 'blah4' )
AND Alias1.Col4 = 'blah5'
AND Alias2.Col7 LIKE 'blah6'
ORDER BY Alias1.Col6 DESC
LIMIT 50) a
UNION
(SELECT Alias1 . *,
Alias2 . *
FROM Table1 AS Alias1
LEFT JOIN Table2 AS Alias2 USING ( Col1 )
WHERE Alias1.Col1 != 'blah1'
AND Alias1.Col2 LIKE 'blah2'
AND ( Alias1.Col3 LIKE 'blah3'
OR Alias1.Col3 LIKE 'blah4' )
AND Alias1.Col4 = 'blah5'
AND Alias2.Col5 LIKE 'blah6'
ORDER BY Alias1.Col6 DESC
LIMIT 50) b
UNION
(SELECT Alias1 . *,
Alias2 . *
FROM Table1 AS Alias1
LEFT JOIN Table2 AS Alias2 USING ( Col1 )
WHERE Alias1.Col1 != 'blah1'
AND Alias1.Col2 LIKE 'blah2'
AND ( Alias1.Col3 LIKE 'blah3'
OR Alias1.Col3 LIKE 'blah4' )
AND Alias1.Col4 = 'blah5'
ORDER BY Alias1.Col6 DESC
LIMIT 50) c
) z
将整个查询用括号括起来(使其成为子查询(，并从中进行选择计数(*(：
select count(*) from (YOUR HUGE QUERY UNION YOUR HUGE QUERY) a
您需要为列Col1指定一个唯一的别名，因为它用于需要唯一列相互匹配的join中。如果遇到重复的列，它将抛出错误1060。检查此项：SQL查询中的列名重复
在您的案例中，MySQL与表1和表2中的Col1混淆，因此出现错误。一旦你使用了它，你就可以SELECT COUNT(*) from YourQuery
将整个查询包装为
SELECT COUNT(*) FROM (<Your query>)
真正使用mysql_num_rows。对于联合，数据库无论如何都需要构建一个临时表，这是一件代价高昂的事情。即使可以直接在SQL中执行，也不会因为要求MySQL提供该数字而节省太多。
当您只使用mysql_num_rows函数时，数据不会发送到PHP。
您需要为列Col1指定一个唯一的别名，因为它用于需要唯一列相互匹配的join中。如果遇到重复的列，它将抛出错误1060。检查此项：SQL查询中的列名重复
在您的案例中，MySQL与表1和表2中的Col1混淆，因此出现错误。一旦您使用了它，您就可以使用SELECT COUNT(*) from YourQuery