blocks|key|3738390|text|使用UNION+ALL组合所有表上的SELECT查询，以获得所有表中的ID总数。改为使用UNION从所有表中获取distinct+ID的总数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3738391|$variable+=+mysql_num_rows(mysql_query("
++++"SELECT+id+FROM+table1+"+.
++++"UNION+ALL+"+.
++++"SELECT+id+FROM+table2+"+.
++++"UNION+ALL+"+.
++++"SELECT+id+FROM+table3+"+.
++++"UNION+ALL+"+.
++++"SELECT+id+FROM+table4+"+.
++++"UNION+ALL+"+.
++++"SELECT+id+FROM+table5"));|code-block|syntax|javascript|3738392|entityMap^0|2|9|I|6|18|5|1K|8|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|L|$]]

Use <code>UNION ALL</code> to combine the <code>SELECT</code> queries on all the tables to get the total number of IDs from all the tables. Use <code>UNION</code> instead to get total number of <code>distinct</code> IDs from all the tables.

<pre><code>$variable = mysql_num_rows(mysql_query("
 "SELECT id FROM table1 " .
 "UNION ALL " .
 "SELECT id FROM table2 " .
 "UNION ALL " .
 "SELECT id FROM table3 " .
 "UNION ALL " .
 "SELECT id FROM table4 " .
 "UNION ALL " .
 "SELECT id FROM table5"));
</code></pre>

blocks|key|5947031|text|不要从你的数据库中获取所有的行，只获取id的个数。让MySQL代替PHP来计算。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5947032|++++$variable1+=+mysql_query("SELECT+COUNT(id)+FROM+table1");
++++$variable2+=+mysql_query("SELECT+COUNT(id)+FROM+table2");
++++$variable3+=+mysql_query("SELECT+COUNT(id)+FROM+table3");
++++$variable4+=+mysql_query("SELECT+COUNT(id)+FROM+table4");
++++$variable5+=+mysql_query("SELECT+COUNT(id)+FROM+table5");

++++$variable+=+$variable1+%2B+$variable2+%2B+$variable3+%2B+$variable4+%2B$variable5;|code-block|syntax|javascript|5947033|既不要加入联合也不要加入，如果你只是需要计数，这是一项繁重的工作。|5947034|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Don't get all rows from your DB, just get the num of id's. Let MySQL count instead of PHP. 

<pre><code> $variable1 = mysql_query("SELECT COUNT(id) FROM table1");
 $variable2 = mysql_query("SELECT COUNT(id) FROM table2");
 $variable3 = mysql_query("SELECT COUNT(id) FROM table3");
 $variable4 = mysql_query("SELECT COUNT(id) FROM table4");
 $variable5 = mysql_query("SELECT COUNT(id) FROM table5");

 $variable = $variable1 + $variable2 + $variable3 + $variable4 +$variable5;
</code></pre>

Don't make UNION or JOIN neither, it's a heavywight job if you just need the count.

blocks|key|388771|text|SELECT+count(t1.id)%2Bcount(t2.id)%2Bcount(t3.id)%2Bcount(t4.id)%2Bcount(t5.id)+from+table1+as+t1,+table2+as+t2,+table3+as+t3,+table4+as+t4,+table5+as+t5|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|388772|entityMap^0|0|41|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

<code>SELECT count(t1.id)+count(t2.id)+count(t3.id)+count(t4.id)+count(t5.id) from table1 as t1, table2 as t2, table3 as t3, table4 as t4, table5 as t5</code>

blocks|key|4218503|text|SELECT+COUNT(id)+FROM+table2|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4218504|等。|unstyled|4218505|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>SELECT COUNT(id) FROM table2
</code></pre>

etc.

blocks|key|4218569|text|尝试如下的联合语法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4218570|$variable+=+mysql_num_rows(mysql_query("
SELECT+*+FROM(
++SELECT+id+FROM+table1
++UNION
++SELECT+id+FROM+table2
++UNION
++SELECT+id+FROM+table3
++UNION
++SELECT+id+FROM+table4
++UNION
++SELECT+id+FROM+table5
)unionNum
"));|code-block|syntax|javascript|4218571|完成，祝你好运|4218572|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

try union syntax like this:

<pre><code>$variable = mysql_num_rows(mysql_query("
SELECT * FROM(
 SELECT id FROM table1
 UNION
 SELECT id FROM table2
 UNION
 SELECT id FROM table3
 UNION
 SELECT id FROM table4
 UNION
 SELECT id FROM table5
)unionNum
"));
</code></pre>

Finish, good luck

blocks|key|3738444|text|第一种方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3738445|$query+=+mysql_query("
++++select+*+from(
++++++++SELECT+id+FROM+table1
++++++++union
++++++++SELECT+id+FROM+table2
++++++++union
++++++++SELECT+id+FROM+table3
++++++++union
++++++++SELECT+id+FROM+table4
++++)
");
while+($row+=+mysql_fetch_assoc($query))+{
++++//some+operations
}
$variable+=+mysql_num_rows($query);|code-block|syntax|javascript|3738446|第二种方法：|3738447|$query+=+mysql_query("
++++select+count(*)+cnt+from(
++++++++SELECT+id+FROM+table1
++++++++union
++++++++SELECT+id+FROM+table2
++++++++union
++++++++SELECT+id+FROM+table3
++++++++union
++++++++SELECT+id+FROM+table4
++++)
");

while+($row+=+mysql_fetch_assoc($query))+{
++++$variable+=+$query['cnt'];
}|3738448|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

First Method:

<pre><code>$query = mysql_query("
 select * from(
 SELECT id FROM table1
 union
 SELECT id FROM table2
 union
 SELECT id FROM table3
 union
 SELECT id FROM table4
 )
");
while ($row = mysql_fetch_assoc($query)) {
 //some operations
}
$variable = mysql_num_rows($query);
</code></pre>

Second Method:

<pre><code>$query = mysql_query("
 select count(*) cnt from(
 SELECT id FROM table1
 union
 SELECT id FROM table2
 union
 SELECT id FROM table3
 union
 SELECT id FROM table4
 )
");

while ($row = mysql_fetch_assoc($query)) {
 $variable = $query['cnt'];
}
</code></pre>

I'm looking to count the amount of fields in 5 tables and display a result. I currently do this for a single table

<pre><code> $variable = mysql_num_rows(mysql_query("SELECT id FROM table1"));
</code></pre>

What is the most efficient way to include and count all <code>id</code> from <code>table2</code>, <code>table3</code>, <code>table4</code> and <code>table5</code>?

Thanks

mysql_num_rows from multiple tables

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我希望计算5个表中的字段数量，并显示结果。我目前对单个表执行此操作。    $variable = mysql_num_rows(mysql_query("SELECT id FROM table1"));包含和统计来自table2、table3、table4和table5的所有id的最有效方法是什么谢谢

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