Write a Regex for a string

Question

Hi I have a file with a lot of bad data lines. I've identified the lines with bad data. The file is very big that it cant be done manually. The problem may reoccur in future so I'm writing a small tool in java to remove the bad segments based on a input regex and remove it.

An example of Bad data is

ABC*HIK*UG*XY\17

I'm trying to write a regex for the above string. So far

Only "(^ABC)" works and ABC is removed.

When I use this nothing happens.

"(^ABC*.XY\17$)"

Please give your inputs.

EDITED:

The answer is working perfect but

If my input files contains this

ABC
123
ABC*HIK*UG*XY\17
1025
KHJ*YU*789

I should get output like

ABC
123
1025
KHJ*YU*789

but I'm getting like this

ABC
123

1025
KHJ*YU*789

Do you have access to Perl or Python? I'd use those over Java... — Ryan J, Jan 15 '15 at 17:30
Can you provide a better explanation of 'bad data', and what you're trying to get from it? — DaaaahWhoosh, Jan 15 '15 at 17:30
Your edit is a different question that requires your java code to solve. I suggest that you ask it as a separate question (and remove your edit) because your original is already answered. — Software Engineer, Jan 15 '15 at 18:01

Avinash Raj · Accepted Answer · 2015-01-15T17:56:42.667

1

Change your pattern to,

"^ABC.*XY\\\\17$"

In java, you need to escape the backslash three more times in-order to match a single \ character. And the pattern to match any character zero or more times must be like .* not *. And also you don't need to put your pattern inside a capturing group.

String s = "ABC\n" + 
        "123\n" + 
        "ABC*HIK*UG*XY\\17\n" + 
        "1025\n" + 
        "KHJ*YU*789";
System.out.println(s.replaceAll("(?m)^ABC.*XY\\\\17\n?", ""));

Output:

ABC
123
1025
KHJ*YU*789

Since we are using anchors in our regex, we need to add the modifier. In our case, we need to add multi-line modifier (?m)

edited Jan 15 '15 at 17:56

answered Jan 15 '15 at 17:31

Avinash Raj

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Perfect!! Works like magic. Can you tell me how to remove the empty line?? – Oracle Nerd Jan 15 '15 at 17:34
to match an empty line use this regex `^$` . For the above case, i think you need this `"^ABC.*XY\\\\17\n"`. This matches also the newline character following the matched characters. – Avinash Raj Jan 15 '15 at 17:38
use `\s*` if there contain any spaces. `"^\\s*ABC.*XY\\\\17$"` . Still you have any problem, then please add the exact input to your question along with the expected output. – Avinash Raj Jan 15 '15 at 17:46
@OracleNerd i already told you to add a newline character at the last. – Avinash Raj Jan 15 '15 at 17:52
It isnt working. If i add a new line char, The line isnt replaced at all – Oracle Nerd Jan 15 '15 at 17:53
OMG, `(?m)^ABC.*XY\\\\17\n?` this is the one, You need to add `(?m)` multiline modifier. – Avinash Raj Jan 15 '15 at 17:54
Still.. I m getting the empty line. :( – Oracle Nerd Jan 15 '15 at 17:57
I get it.. Its working for your code. But it is not working for my text file. – Oracle Nerd Jan 15 '15 at 18:05
@OracleNerd it may because of carriage return character. I think you're on windows. So try this `"(?m)^ABC.*XY\\\\17[\r\n]*"` . Still if it's not work then check for the existence of spaces next to the `\17` . In that case try `"(?m)^ABC.*XY\\\\17\\s*"` – Avinash Raj Jan 15 '15 at 18:07
`"(?m)^ABC.*XY\\\\17\\S*\\s*"` – Avinash Raj Jan 15 '15 at 18:17
That wasn't helpfull either. Wrote two lines of java code to solve the problem `outputLine = outputLine.trim(); if (outputLine.isEmpty()) continue;` Thanks all – Oracle Nerd Jan 15 '15 at 19:43

Write a Regex for a string

1 Answers1