一个利用substr后转换成number出错的问题

此问题总结自itpub，原文链接http://www.itpub.net/873682.html

引入问题
SQL> select cashier_no,sum(substr(lpad(time,4),1,2)*60+substr(lpad(time,4),3,2)) mintus from invoice_time
where cashier 2 _no=112
group 3 by cashier_no;

CASHIER_NO MINTUS
---------- ----------
112 9117

SQL> select cashier_no,sum(substr(lpad(time,4),1,2)*60+substr(lpad(time,4),3,2)) mintus from invoice_time
2 group by cashier_no;
ERROR:
ORA-01722: invalid number

no rows selected
我想问一下，下面这条语句是那里错了？？

为了分析问题，我们测试如下：

SQL> select sum(substr(lpad(5,4),1,2)*60+substr(lpad(5,4),3,2)) from dual;

select sum(substr(lpad(5,4),1,2)*60+substr(lpad(5,4),3,2)) from dual

ORA-01722: 无效数字

SQL> select sum(substr(lpad(55,4),1,2)*60+substr(lpad(55,4),3,2)) from dual;

select sum(substr(lpad(55,4),1,2)*60+substr(lpad(55,4),3,2)) from dual

ORA-01722: 无效数字

SQL> select sum(substr(lpad(555,4),1,2)*60+substr(lpad(555,4),3,2)) from dual;

SUM(SUBSTR(LPAD(555,4),1,2)*60
------------------------------
355

大家看到，在time不足3位长度，就出错了呢，那究竟是哪里的问题？让我接着做个测试。

SQL> select substr(lpad(55,4),1,2) from dual;

SUBSTR(LPAD(55,4),1,2)
----------------------

SQL> select substr(lpad(55,4),1,2)*60 from dual;

select substr(lpad(55,4),1,2)*60 from dual

ORA-01722: 无效数字

大家可以看到，原来是不足两位，在取出前两位是什么呢，肯定不是null了，验证如下

SQL> select 1 from dual where substr(lpad(55,4),1,2) is null ;

1
----------

SQL> select 1 from dual where 1=1;

1
----------
1

大家都知道lpad在默认情况下补空格的

SQL> select lpad(55,4) from dual;

LPAD(55,4)
----------
55

这样找到问题根源就知道了，可以增加处理，让lpad补'0'来处理，避免这个错误。

SQL> select sum(substr(lpad(5,4,'0'),1,2)*60+substr(lpad(5,4,'0'),3,2)) from dual;

SUM(SUBSTR(LPAD(5,4,'0'),1,2)*
------------------------------
5

如果有问题，请大家指正！

[@more@]批

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