在空格之前解析文本 Java 正则表达式匹配，在空格之后解析数字以生成 CSV

Question

At the moment I'm using this simple regex:

[^\s]

Which I cobbled together with the help of these docs .

It can grab the following information:

However the full dataset looks like this:

#### LOGS ####
CONSOLE:
makePush            2196
makePush            638
makePush            470
opAdd           8342
opAdd           288
opStop          133
0x
DEBUG:
#### TRACE ####
PUSH32          pc=00000000 gas=10000000000 cost=3

PUSH32          pc=00000033 gas=9999999997 cost=3
Stack:
00000000  0000000000000000000000000000000000000000000000000000000000000005

PUSH32          pc=00000066 gas=9999999994 cost=3
Stack:
00000000  0000000000000000000000000000000000000000000000000000000000000005
00000001  0000000000000000000000000000000000000000000000000000000000000005

ADD             pc=00000099 gas=9999999991 cost=3
Stack:
00000000  0000000000000000000000000000000000000000000000000000000000000005
00000001  0000000000000000000000000000000000000000000000000000000000000005
00000002  0000000000000000000000000000000000000000000000000000000000000005

ADD             pc=00000100 gas=9999999988 cost=3
Stack:
00000000  000000000000000000000000000000000000000000000000000000000000000a
00000001  0000000000000000000000000000000000000000000000000000000000000005

STOP            pc=00000101 gas=9999999985 cost=0
Stack:
00000000  000000000000000000000000000000000000000000000000000000000000000f

Finally I need my result to look like this:

makePush, 2196
makePush, 638
makePush, 470
opAdd, 8342
opAdd, 288
opStop, 133

And the regex I've provided is certainly not robust enough to capture that.

What I'm trying to do is:

• Ignore any string in the input that doesn't have the form makePush 2196

• For lines that are of the form depicted above...

  • Split it into three groups"

    first word , whitespace , second word

• Finally I want to save a csv of the form:

  first word , second word

Answer 1

Try this?

/([a-zA-Z]+)[\t ]+(\d+)/g

where

• ([a-zA-Z]+) matches a single word literals
• [\\t ]+ matches horizontal white spaces
• (\\d+) matches the number literals

Answer 2

Try this (idea from Pshemo but use \\w+)

        Pattern pattern = Pattern.compile("^(\\w+)\\s+(\\d+)$");
        Matcher matcher = pattern.matcher(str);
        while (matcher.find())
        {
            System.out.println(matcher.group(1)+", "+matcher.group(2));
        }