mysql使用substr替代like
2018-06-19 本文已影响0人
fangqi179
背景
查找以【,2018】开头的记录;
LIKE
WHERE m.path LIKE ',2018%'
SUBSTR
WHERE SUBSTR(m.path,1,5) = ',2018'
- 列名;
- 起始位置,从1开始;
- 结束位置;
查找以【,2018】开头的记录;
WHERE m.path LIKE ',2018%'
WHERE SUBSTR(m.path,1,5) = ',2018'