Question

我有两张具有相同结构的表格。（参见下文）

records = id (int, auto increment), name (varchar 250), age (int), address (varchar 250)
backup = id (int), name (varchar 250), age (int), address (varchar 250)

两个表内容3条记录。现在，我想按行的每列比较每行的这两个表，如果有一行与比较表上的相应行不匹配则生成并出错。

表“记录”内容：

row 1 : column id = 1, column name = name1, column age = 12, column address = USA
row 2 : column id = 2, column name = name2, column age = 17, column address = USA
row 3 : column id = 3, column name = name3, column age = 32, column address = USA

表“备份”内容：

row 1 : column id = 1, column name = name1change, column age = 12, column address = USA
row 2 : column id = 2, column name = name2, column age = 17, column address = USA
row 3 : column id = 3, column name = name3, column age = 32, column address = USA

从上面的表格内容可以看出，在表“备份行1，列名”中有一个更改，name1（记录表）更改为name1change（备份表）。因此，当表记录的第1行与表备份的第1行进行比较并找到差异时，它将获得行号和列名，后跟错误文本（参见下文）

echo $errorRow . "with" . $errorColumnOfTheErrorRow . "didn't match, there's is changes";

我希望循环到具有相同过程的记录结束（比较）。怎么做？任何想法，帮助，建议和建议将不胜感激。谢谢！

到目前为止，我尝试的是（参见下文）

$sql = "SELECT DISTINCT * FROM (SELECT a.* FROM records a UNION SELECT b.* FROM backup b ) c";
$result = mysqli_query($this->db,$sql);
$sql = "SELECT COUNT(*), id FROM (SELECT DISTINCT * FROM (SELECT a.* FROM records a UNION SELECT b.* FROM backup b ) c ) d GROUP BY id;";
$result = mysqli_query($this->db,$sql);

echo $result;

如上面的查询，我试图显示不匹配的行，但遗憾的是，没有工作。

Answer 1

你可以用一种很好的方式做到这一点：第一步：将所有数据放在一起并仅保留不同的（=不相同的）行：

SELECT DISTINCT *
FROM (
SELECT a.*
FROM records a
UNION
SELECT b.*
FROM backup b ) c

第二步：计算ID出现次数超过1次的所有记录：

SELECT COUNT(*), id 
FROM (
SELECT DISTINCT *
FROM (
SELECT a.*
FROM records a
UNION
SELECT b.*
FROM backup b ) c ) d
GROUP BY id
HAVING COUNT(*)>1;

这将为您提供行不相同的确切ID。

第三步：获取包含更改数据的行：

SELECT * FROM (
SELECT "current data" AS description,f.* FROM 
records f
INNER JOIN (
SELECT id 
    FROM (
    SELECT DISTINCT *
    FROM (
    SELECT a.*
    FROM records a
    UNION
    SELECT b.*
    FROM backup b ) c ) d
    GROUP BY id
    HAVING COUNT(*)>1 ) e ON f.id=e.id
UNION 
SELECT "backup data" AS description, g.* FROM 
records g
INNER JOIN (
SELECT id 
    FROM (
    SELECT DISTINCT *
    FROM (
    SELECT a.*
    FROM records a
    UNION
    SELECT b.*
    FROM backup b ) c ) d
    GROUP BY id
    HAVING COUNT(*)>1 ) h ON g.id=h.id;
) i ORDER BY id ASC, description ASC;

它不漂亮，但它有效：）

将逐行和每行列表与另一个表进行比较

1 个答案: