[英]cannot parse String with Java Regex
I have a string formatted as below: 我有一个格式如下的字符串:
source1.type1.8371-(12345)->source2.type3.3281-(38270)->source4.type2.903..
It's a path, the number in () is the weight for the edge, I tried to split it using java Pattern as following: 它是一个路径,()中的数字是边缘的权重,我试图使用java Pattern将其拆分如下:
[a-zA-Z.0-9]+-{1}({1}\\d+){1}
[a-zA-Z_]+.[a-zA-Z_]+.(\\d)+-(\\d+)
[a-zA-Z.0-9]+-{1}({1}\\d+){1}-{1}>{1}
hopefully it split the string into fields like 希望它将字符串分成像这样的字段
source1.type1.8371-(12345)
source2.type3.3281-(38270)
..
but none of them work, it always return the whole string as the field. 但它们都不起作用,它总是将整个字符串作为字段返回。
It looks like you just want String.split("->")
( javadoc ). 看起来你只需要
String.split("->")
( javadoc )。 This splits on the symbol ->
and returns an array containing the parts between ->
. 这将拆分符号
->
并返回一个包含->
之间部分的数组。
String str = "source1.type1.8371-(12345)->source2.type3.3281-(38270)->source4.type2.903..";
for(String s : str.split("->")){
System.out.println(s);
}
Output 产量
source1.type1.8371-(12345)
source2.type3.3281-(38270)
source4.type2.903..
It seems to me like you want to split at the ->
's. 在我看来,你想要分开
->
。 So you could use something like str.split("->")
If you were more specific about why you need this maybe we could understand why you were trying to use those complicated regexes 所以你可以使用类似
str.split("->")
如果你更具体地说明你为什么需要这个,也许我们可以理解为什么你试图使用那些复杂的正则表达式
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