How to escape special characters in a regex pattern in java?

Question

I have a string :

Patient: ${ss.patient.howard.firstName} ${ss.patient.howard.lastName} Gender: ${ss.patient.howard.sex} Birthdate: ${ss.patient.howard.dob} ${ss.patient.howard.addressLine1} Phone: (801)546-4765

and I am trying to replace ${..} substrings with other strings so it would look like:

Patient: firstName lastName Gender:sex Birthdate: dob addressLine1 Phone: (801)546-4765

I am trying examleString.replaceFirst("\\${(.*?)\\}", (String) TestcaseContext.getCache().get(matcher.group(1))) and getting, Illegal repetition near index 1 \${(.*?)\} — sophist_pt
– sophist_pt, Commented Mar 30, 2016 at 19:26
Must you use regular expressions? I would consider the Apache Commons Lang StrSubstitutor for this particular application. It does exactly what you want. — KevinO
– KevinO, Commented Mar 30, 2016 at 19:31

anubhava · Accepted Answer · 2016-03-30 19:34:42Z

You can use this regex with a capturing group:

String myString = "Patient:\n${ss.patient.howard.firstName} ${ss.patient.howard.lastName}\nGender: ${ss.patient.howard.sex}\nBirthdate: ${ss.patient.howard.dob}\n${ss.patient.howard.addressLine1}\nPhone: (801)546-4765"; myString = myString.replaceAll("\\$\\{[^}]+?\\.([^.}]+)}", "$1"); System.err.println(myString);

([^.}]+) is the capturing group before } and after the last DOT.

RegEx Demo

Output:

Patient: firstName lastName Gender: sex Birthdate: dob addressLine1 Phone: (801)546-4765

Thanks @anubhava, good to know about the capturing group. It did the trick.

user2705585 · Accepted Answer · 2016-03-30 19:32:25Z

You could try searching for ${string.string.string OR } and replace with empty string or nothing like this.

Regex: \${[a-z]+\.[a-z]+\.[a-z]+\.|}

Replacement to do: Replace with nothing or empty string.

Regex101 Demo

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