获取所有没有前导零的行

Question

我需要SELECT column1没有前导0的所有行。 column1的数据类型为String。

column1

0123455677
0987654321
2345567887
0233445566
3422245666

Answer 1

ANSI SQL回答：

select * from tablename
where substring(column1 from 1 for 1) <> '0'

某些dbms产品改为SUBSTR(column1, 1, 1)，或LEFT(column1, 1)。

Answer 2

我认为你在谈论一个字符字段，因为数字字段不会显示前导0，除非你格式化...无论如何

对于你要选择的字符字段......

WHERE column1 NOT LIKE '0%'

如果是数字字段，您可以选择...

WHERE column1 >= 1000000000

Answer 3

有多种方法可以实现这一目标。这里有三个用于oracle dbms的例子（我不知道你的dbms）。

-- using substr
select * from
(select '0123' t from dual union 
 select '123' t from dual union 
 select '5123' t from dual union 
 select '413123' t from dual union 
 select '0454' t from dual) tab
 where substr(tab.t,0,1) != 0;
-- using regex
 select * from
(select '0123' t from dual union 
 select '123' t from dual union 
 select '5123' t from dual union 
 select '413123' t from dual union 
 select '0454' t from dual) tab
 where regexp_like(tab.t,'^[^0]');
-- using the like operator
 select * from
(select '0123' t from dual union 
 select '123' t from dual union 
 select '5123' t from dual union 
 select '413123' t from dual union 
 select '0454' t from dual) tab
 where tab.t not like '0%';

Answer 4

问题已得到解答，但您也可以尝试

SELECT LEFT(column1, 1) AS firstLetter
FROM <table>
WHERE colum1 => 1

它与字符串完美配合。

Answer 5

你需要检查第一个字符是否为零。使用SUBSTR。

SELECT COLUMN FROM TABLE WHERE SUBSTR(COLUMN,1,1) <> '0';