Question

我有2个数据库，DB1中只有1个表，DB2中有2个表。 DB1.table1中的每条记录都被拆分并分别存储在DB1.table1和DB @ .table2中。

For example, DB1 has a table1 which looks like

        Student_Name   Id   Address   Attendance   Marks
        ------------   --   -------   ----------   -----
        John            1   90th st       70         90

The records that are transferred from DB1.table1 are stored in DB2.table and DB2.table2 in the following manner

DB2.table 1: Id   Student_Name   Address   
             --   ------------   -------
             1     John          90th st


DB2.table 2: Id   Attendance   Marks
             --   ----------   -----
             1     70            90

我想编写一个测试用例，以确保将DB1中的所有数据都复制到DB2。我写了一些查询来确定DB1中的记录是否未复制到DB2。除了找出丢失的记录之外，我还想逐列检查每条记录，以确保DB1和DB2中的值相同。

从上面的例子中，我想检查ID = 1，如果DB2.table1 Student_name = DB1.table1 Student_name，DB2.table1 Address = DB1.table1 Address，依此类推..

如果我有1000列怎么办？我应该写一个长脚本来检查每一列吗？不，这是进行此类测试的最佳方式吗？有没有可以使用的工具或者我应该写下脚本？

Answer 1

这会在Id和db1.table1中找到db2.table1中没有匹配项的db2.table2行。

它假定两个表中的列名称相同，并且db2.table1或db2.table2中存在的任何列都应在db1.table1中具有匹配的列名称。因此，如果db2.table2有一个名为Foo的列，db1.table1也必须有一个名为Foo的列。如果db2.table1包含名为Bar的列，则db1.table1也必须包含名为Bar的列。如果该列存在于db2但不存在于db1中，则会出现MySQL错误。

希望这就是你要找的东西！

header("Content-type: text/plain");

// connect with mysqli

// get a list of columns in db2.table1 and db2.table2
$columns = array();
$query = mysqli_query("SELECT table_name, column_name FROM information_schema.columns WHERE table_schema = 'db2' AND table_name IN ('table1', 'table2')");
while ($row = $mysqli_fetch_assoc($query)) {
    $columns[$row["table_name"]][] = "db1.table1.{$row["column_name"]} = db2.{$row["table_name"]}.{$row["column_name"]}";
}

$query = mysqli_query("
    SELECT db1.table1.Id
    FROM
        db1.table1
        LEFT JOIN db2.table1
            ON ". implode(" AND ", $columns["table1"]) ."
        LEFT JOIN db2.table2
            ON ". implode(" AND ", $columns["table2"]) ."
    WHERE
        db2.table1.Id IS NULL
        OR db2.table2.Id IS NULL
");

while (list($id) = mysqli_fetch_row($query)) {
    echo "ID {$id} does not match\n";
}

Answer 2

您可以使用union all和group by：

执行此操作

select Student_Name, Id, Address, Attendance, Marks,
       (case when max(which) = 'db1' then 'Missing in db2'
             when min(which) = 'db2' then 'Missing in db1'
        end) as why
from ((select 'db1' as which, Student_Name, Id, Address, Attendance, Marks
       from db1.table1 t
      ) union all
      (select 'db2' as which, t1.Student_Name, t1.id, t1.Address, t2.Attendance, t2.Marks
       from db2.table1 t1 join
            db2.table2 t2
            on t1.id = t2.id
      )
     ) u
group by Student_Name, Id, Address, Attendance, Marks
having count(distinct which) = 1

Answer 3

此查询的结果将为您提供DB2中与DB1不同的行数，如果行不存在但它也应该计数。如果结果为0，一切正常。

SELECT
    COUNT(*) AS difference
FROM
    DB1.table1 AS d1t1
    LEFT JOIN DB2.table1 AS d2t1
         USING (Id)
    LEFT JOIN DB2.table2 AS d2t2
         USING (Id)
WHERE
    (d1t1.Student_Name <> d2t1.Student_Name)
    OR (d1t1.Address <> d2t1.Address)
    OR (d1t1.Attendace <> d2t2.Attendace)
    OR (d1t1.Marks <> d2t2.Marks)

测试2个mysql数据库之间的内容

3 个答案: