Java Regex - Split string on spaces - Ignore space

2019-03-04 07:43发布

This question already has an answer here:

I'm looking for regex to do the following in Java:

String originalString = "";
String splitString[] = originalString.spilt(regex);

Some test cases:

Original1: foo bar "simple"
Spilt1: { "foo", "bar", "\"simple\"" }

Original2: foo bar "harder \"case"
Spilt2: { "foo", "bar", "\"harder \"case\"" }

Original3: foo bar "harder case\\"
Spilt3: { "foo", "bar", "\"harder case\\"" }

Some snippets I have come across:

# Does not react to escaped quotes
 (?=([^\"]*\"[^\"]*\")*[^\"]*$)
# Finds relevant quotes that surround args
(?<!\\)(?:\\{2})*\"

Thanks!

2条回答
啃猪蹄的小仙女
2楼-- · 2019-03-04 07:58

Regex like this will work for simple cases:

("(.+?)(?<![^\\]\\)")|\S+

But I would not suggest to use RegEx for this task, but take a look at CSV parsers instead.

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别忘想泡老子
3楼-- · 2019-03-04 08:20
String stringToSplit = This is the string to split;

String[] split = stringToSplit.split("character to split at");

In this case, split[0] would result as ' This ', split[1] would be ' is ', split[2] would be ' the ', split[3] ' string ' split[4] ' to ' split[5] ' split '.

At this point you can do

var0 = split[0];
var1 = split[1];
var2 = split[2];

Where var0 would equal "This" And so on...

Hope this helps.

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