MySQL：如何计算复杂查询返回的行数

这感觉应该很容易，但我无法得到它，我试过了一切。MySQL：如何计算复杂查询返回的行数

我怎样才能只是获得了该查询返回的行数：

(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，然后做的结果mysql_num_rows，但我想直接用SQL来完成它，因为我听说它更快，因为它节省了一个步骤。

谢谢！

编辑：

以下是给我的错误（＃1060 - 重复列名 'Col1中'）：

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

以下是给我的错误（＃1064 - 你有一个错误您的SQL语法）：

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

来源

2011-08-22 ProgrammerGirl

为什么包装整个查询放在括号中（使它成为子查询）并对其执行SELECT COUNT（*）失败？它是否返回错误信息？ – FrustratedWithFormsDesigner

当您将它包装在Select Count（*）FROM（您的长查询）AS CountThis中时会发生什么。你有没有得到一个特定的错误，或者你得到错误的计数？ – Kibbee

你不提供一个不使用'mysql_num_rows（）'的好理由。这是有原因的...使用它。你为什么期望它很慢？ – Brad

你需要为你的列Col1指定一个唯一的别名，因为它在join中使用，它需要唯一的列来相互匹配。如果遇到重复列，它将引发错误1060。检查此：Duplicate column names in SQL query

在你的情况下，MySQL与表1和表2中的Col1混淆，因此错误。一旦你已经使用过，你可以SELECT COUNT(*) from YourQuery

来源

2011-08-22 16:12:10 reggie

工作正常！另外，只需删除'，Alias2。 *'来自所有3个SELECTS。谢谢reggie！ – ProgrammerGirl

环绕你整个查询括号（使它成为一个子查询），并从中做一个SELECT COUNT（*）：

select count(*) from (YOUR HUGE QUERY UNION YOUR HUGE QUERY) a

来源

2011-08-22 14:08:14

这给了我一个错误。 – ProgrammerGirl

实际错误是：＃1248 - 每个派生表都必须有自己的别名 – ProgrammerGirl

尝试在每个UNIONED查询后放置一个别名。即从（（选择...）中选择计数（*）联合（选择...）b联合（选择...）c）a –

整个包住查询作为

SELECT COUNT(*) FROM (<Your query>)

来源

2011-08-22 14:08:46 Stefan

这给了我以下错误：＃1248 - 每个派生表必须具有自己的别名 – ProgrammerGirl

真正使用mysql_num_rows。对于联盟来说，数据库无论如何都需要建立一个临时表，这是很昂贵的事情。即使可以直接在SQL中执行操作，也不会通过让MySQL给你那个数字来节省很多。

无论如何，当您只使用mysql_num_rows函数时，数据不会发送到PHP。

来源

2011-08-22 16:02:54 akirk

您需要为您的列Col1指定一个唯一的别名，因为它用于需要唯一列彼此匹配的join。如果遇到重复列，它将引发错误1060。检查此：Duplicate column names in SQL query

在你的情况下，MySQL与表1和表2中的Col1混淆，因此错误。一旦你使用了，你可以SELECT COUNT(*) from YourQuery

来源

2011-08-22 16:10:45 reggie

工作正常！另外，只需删除'，Alias2。 *'来自所有3个SELECTS。谢谢reggie！ – ProgrammerGirl

MySQL：如何计算复杂查询返回的行数

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