部分我有以下URL字符串如何URL字符串分隔成使用正则表达式
/category/category1/2010/12/10/id
我需要串分成部分。我想这个模式
(?:(.*))?(?:(?:\/)?([0-9]{4}))?(?:(?:\/)?([0-9]{1,2}))?(?:(?:\/)?([0-9]{1,2}))?(?:(.*))?
但如果URL是这样/category/category1/
它不能正常工作
,我需要以下瓦尔
path = '/category/category1/';
year = '';
month = '';
day = '';
id ='';
如果URL是这样/category/category1/2010
,我需要下列增值税
path = '/category/category1/2010';
year = '2010';
month = '';
day = '';
id ='';
如果URL是这样的/category/category1/2010/12/
,我需要以下瓦尔
path = '/category/category1/2010/12';
year = '2010';
month = '12';
day = '';
id ='';
如果URL是这样/category/category1/2010/12/10
,我必须满足下列条件瓦尔
path = '/category/category1/2010/12/10';
year = '2010';
month = '12';
day = '10';
id ='';
如果URL是这样/category/category1/2010/12/10/id
,我需要以下瓦尔
path = '/category/category1/2010/12/10/id';
year = '2010';
month = '12';
day = '10';
id ='id';
是否有可能使preg_match
和正则表达式?
如果有人打电话'/分类/组别/ id'或任何其他组合你没有什么名单?为什么不使用普通的旧查询字符串,例如'?类别= 1&d = 2010-12-10&id'? – Gordon 2011-05-14 16:33:35