CODEWITHSUNDEEP
java
slack king
slack king

Reputation: 33

check for number of breaklines and sometimes too many breaklines and replace it with only a full stop

I would like to replace the string that contains breaklines and replace it with ".". However, in order to replace it, my method is to check how many breaklines it has and then perform replacing. This will make it add too many ".". Example

st = "I have something nice \n\n\n\n\n\n\n\n there"

    String st = st.replace('\r', ' ').replace('\n\n\n', '.').replace('\n', '.').replace('\n\n', '.').replace('\n\n\n\n', '.');

My current way if I keep adding toooooo many replace then it will be: (Is there a smarter way to do so?)

I have something nice ........... there

My expected output:

I have something nice . there

Upvotes: 0

Views: 43

Answers (3)

Mohsen
Mohsen

Reputation: 4633

This will help you:

st.replaceAll("(?:\\s*\n)+", ".");

\\s*\n will escape all endlines which can be followed by spaces.

Also it will eliminate the space between the last char and the "." which is ponctuation rule.

Upvotes: 1

xingbin
xingbin

Reputation: 28269

Use String.replaceAll with regex:

String st = "I have something nice \n\n\n\n\n\n\n\n there";

st = st.replaceAll("\n+", "."); // \n+ matches one or mutiple line breaks
System.out.println(st);//I have something nice . there

Upvotes: 1

Jan Gassen
Jan Gassen

Reputation: 3534

You can use a simple regex to replace all occurrences at one:

String st = "I have something nice \n\n\n\n\n\n\n\n there";
String replaceAll = st.replaceAll("\n+", ".");

Outputs:

I have something nice . there

Upvotes: 3

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