使用字符操作功能显示名称以“E”结尾的所有员工姓名 [英] Using character manipulation function display all the employees name whose name end with 'E'
本文介绍了使用字符操作功能显示名称以“E”结尾的所有员工姓名的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
PLZ找到解决问题的方法
我尝试过:
SELECT ASCII('E')
FROM DUAL;
SELECT LAST_NAME
FROM EMPLOY_INFORMATION
WHERE SUBSTR(LAST_NAME,-1,0)= CHR(69);
解决方案
SELECT Last_Name
FROM Employ_Information
WHERE RIGHT (Last_Name, 1 )= ' E'
或者您可以尝试:
< pre lang =SQL> SELECT Last_Name
FROM Employ_Information
WHERE Last_Name LIKE ' %E'
从employ_information中选择last_name,其中ASCII(RIGHT(last_name,1))= 69
PLZ FIND ME WAYS TO SOLVE ABOVE QUERRY
What I have tried:
SELECT ASCII('E')
FROM DUAL;
SELECT LAST_NAME
FROM EMPLOY_INFORMATION
WHERE SUBSTR(LAST_NAME, -1, 0) = CHR(69);
解决方案
SELECT Last_Name FROM Employ_Information WHERE RIGHT(Last_Name, 1) = 'E'
Or you could try:
SELECT Last_Name FROM Employ_Information WHERE Last_Name LIKE '%E'
select last_name from employ_information where ASCII(RIGHT(last_name,1))=69
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