LUA正则表达式不完全指南

2014-02-23 15:14| 查看: 8424 |作者: 小雪

string.match(s,pattern[,init])
匹配第一个符合条件的项目
s代表目标字符串
pattern代表你要匹配的规则，见上面正则表格
init代表开始匹配的位置，默认为1，可以是负数，可不填

s = "1number123xyz"
x = string.match(s,"%d+",1) --匹配数字，一次或多次匹配，从第一号位开始找
-->x = 1 --返回值是1

x = string.match(s,"%d+",2) --匹配数字，一次或多次匹配，从第二号位开始找
-->x = 123 --返回值是123

x = string.match(s,"%d+",-4) --匹配数字，一次或多次匹配，从右起第四号位开始向右找
-->x = 3

x = string.match(s,"%d+",-3) --匹配数字，一次或多次匹配，从右起第四号位开始向右找
-->x = nil

x = string.match(s,"%d*",-3) --匹配数字，零次或多次匹配，从右起第四号位开始向右找
--> 完成匹配，返回空白

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