mysql - Comparison of two tables in MYSQL

Question

I have two tables. One table (table1) has 28500 rows and the another (table2) has 17450 rows. I would like to compare these tables and find rows that do not exist in table1.

SELECT * FROM table1 WHERE ID NOT IN (SELECT DISTINCT(ID) FROM table2)

Any suggestions?

Answer 1

尝试这个：

SELECT table1.*
FROM table1
LEFT OUTER JOIN table2
ON table1.id = table2.id
WHERE table2.id IS NULL

LEFT OUTER JOIN链接从table1开始的两个表，如果table2没有链接行，table2的所有字段都将为空。所以，如果你把你的WHERE条件 table2.id 设置为空，你只会得到 table1 中不存在于 table2 中的行

Answer 2

您可以通过执行左外连接并检查所有不存在的行来解决此问题。根据您是否要查找表 2 中的表 1 或表 1 中的表 2 中不存在的值，请尝试以下操作。

SELECT *
FROM table1
LEFT OUTER JOIN table2 ON (table1.id = table2.id)
WHERE table2.id IS NULL;


SELECT *
FROM table2
LEFT OUTER JOIN table1 ON (table1.id = table2.id)
WHERE table2.id IS NULL;

SQL 小提琴：http ://sqlfiddle.com/#!2/a9390/8

Answer 3

使用此查询：

SELECT 
    * 
FROM 
    table2 
LEFT JOIN 
    table1
ON 
    table2.primary_key = table1 .primary_key
WHERE 
    table1 .primary_key IS NULL
;

Answer 4

好吧，如果您想要 PHP 中的答案，那么就是：

$sql=mysql_query("SELECT * FROM table1");
while($row=mysql_fetch_array($sql))
{
    $id=$row['id'];
    $sql2=mysql_query("SELECT * FROM table2 WHERE id='$id'");
    $check=mysql_num_rows($sql2);
    if($check==0)
    {
        echo $id." is not in table1<br>";
    }
}

我希望这对你有帮助

Answer 5

如果您想按所有列比较 2 个表（完全比较，而不仅仅是按 ID 等单个特定列），您可以使用这种方法：

SELECT column1, column2, column3
FROM
 (
   SELECT t1.column1, t1.column2, t1.column3
   FROM t1
   UNION ALL
   SELECT t2.column1, t2.column2, t2.column3
   FROM t2
)  t
GROUP BY column1, column2, column3
HAVING COUNT(*) = 1
ORDER BY column3

基于示例：http ://www.mysqltutorial.org/compare-two-tables-to-find-unmatched-records-mysql.aspx

Answer 6

SELECT * FROM table1
where id not in (
    SELECT table2.id
    FROM table2
    LEFT OUTER JOIN table1
    ON (
        table1.id = table2.id 
        and table1.col1 = table2.col1 
        and table1.col2 = table2.col2
        ...
    )
)

那里table1比table2有更多的行