Question

我有一个名为table1的表它有两个名为“name1”和“name2”的字段我想将这两列相互比较如果name1有一个name2中不存在的值，我想知道填写一个新的标志列，反之亦然

table1 : 
----
name1  name2
----
jack   wheile
--
james  jack
--

>        
sample output : 
--
name1  name2   flag
--
jack   wheile  jack exist in column2
--
james  jack    jack exist in column1
--

任何帮助？

Answer 1

尝试此查询：

UPDATE TABLE1 SET FLAG = CONCAT('exist in ', IF(NAME1 LIKE '%jack%', column1, column2)) 
WHERE NAME1 LIKE '%jack%' OR NAME2 LIKE '%jack%';

Answer 2

如果从Access本身

中执行，此查询应该达到所需的结果

UPDATE table1 SET
    flag = Switch(DCount("*", "table1", "[name2]=""" & Replace([name1], """", """""") & """")>0, [name1] & " exists in column 2", DCount("*", "table1", "[name1]=""" & Replace([name2], """", """""") & """")>0, [name2] & " exists in column 1")

编辑：

或者，以下方法可以证明效率更高，但需要两次UPDATE查询

UPDATE table1 SET flag = [name1] & " exists in column 2"
WHERE 
    [name1] IN
        (
            SELECT x.name1 
            FROM 
                table1 AS x 
                INNER JOIN 
                table1 AS y 
                    ON x.name1 = y.name2
        )

和

UPDATE table1 SET flag = [name2] & " exists in column 1"
WHERE 
    [name2] IN
        (
            SELECT y.name2 
            FROM 
                table1 AS x 
                INNER JOIN 
                table1 AS y 
                    ON x.name1 = y.name2
        )

比较同一个表中的两列，并使用结果更新新列

2 个答案: