how to perform intelligent string check in android

Question

I have a long string text for example:

String2:

12345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123

String3:

123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890 234567890

I want to show first 144 characters in string and I need to show read more. If the 144th character is not equal to space as shown in string 3 then I need to check for space character after 144th character and then I need to break and show read more.

I am able to check whether 144th character is a space by using following code:

String.charAt(144) == ' ';

but if 144th character is not space then how can I wrap the string to break it when next space character is encountered.

Do you want something like if(String.charAt(144) != ' ') return myString.split(" ")[0];? — logi-kal
– logi-kal, Commented Jul 23, 2016 at 10:35
if 144th character does not equal space then I want to read the string, I want to read the word completely and then break after the word.. — keshav kowshik
– keshav kowshik, Commented Jul 23, 2016 at 10:36

Thomas Ayoub · Accepted Answer · 2016-07-23 10:37:39Z

2

You can use this code:

// String to be scanned to find the pattern.
String line = "123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890 234567890";
String pattern = "(.{144}[^ ]*).*";

// Create a Pattern object
Pattern r = Pattern.compile(pattern);

// Now create matcher object.
Matcher m = r.matcher(line);
if (m.find( )) {
 System.out.println(m.replaceAll("$1..."));
} else {
 System.out.println(line);
}

See demo.

It will split on the space after the 144^th character of the string, no matter how far it is. With your example, the output is:

123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890...

answered Jul 23, 2016 at 10:37

Thomas Ayoub

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