Q:How to find name start with '%' in sql |
Q:如何找到的名字开始与“%”在SQL |
I want to search name which start with '%' like '%smit' in oracle sql i have tried with escape '%\%' but not able to find.. |
我想搜索的名字开始与“%”是“他”在Oracle的SQL 我已经尝试逃脱“%”,但无法找到.. |
answer1: | 回答1: |
You need to tell Oracle that you are using an escape character:
Anything in the search pattern after the character specified through the escape option will be treated "as is". So the above searches for any value that starts with the character %. The second % is again treated as a regular wildcard (because it's not prefixed with the escape character) You can also choose a different escape character:
This is nothing Oracle specific, this is how the LIKE operator is defined in the SQL standard |
您需要告诉Oracle您正在使用转义字符:
在搜索模式中的任何字符后,通过转义选项指定将被视为“是”。因此,上面搜索任何以字符%开头的值。第二%再处理为一个一般的通配符(因为它不是在前面加上转义字符) 您也可以选择一个不同的转义字符:
这是什么预言具体,这是怎样的操作是在SQL标准定义 |
oracle oracle11g oracle-sqldeveloper |