Java-read lines of files and extract only the helpfull info

Question

I have files file1, file2 contains contents such:

[2017-02-01 10:00:00 start running [error:yes] [doing:no] [finish:] [remind:] [alarmno:123456789] [logno:12345678

Answer 1

You may use

^\[?(\d[\d-]+).*?\[alarmno:(\w*)].*?\[logno:(\w*)].*?\[type:\w*:([^\]]*)]

See the regex demo

Details:

• ^ - start of string
• \[? - an optional [
• (\d[\d-]+) - Group 1: a digits and 1 or more digits or -s
• .*? - any 0+ chars other than line break chars as few as possible
• \[alarmno: - a [alarmno: substring
• (\w*) - Group 2: 0+ word chars
• ] - a literal ]
• .*? - any 0+ chars other than line break chars as few as possible
• \[logno: - a literal [logno: substring
• (\w*) - Group 3: 0+ word chars
• ] - a ]
• .*? - any 0+ chars other than line break chars as few as possible
• \[type: - a [type: substring
• \w* - 0+ word chars
• : - a colon
• ([^\]]*) - Group 4: 0+ chars other than ]
• ] - a ]

Java demo:

String s = "[2017-08-17 08:00:00 Comming in [Contact:NO] [REF:] [REF2:] [REF3:] [Name:+AA] [Fam:aa] [TEMP:-2:0:-2:0:-2] [Resident:9:free] [end:0:]";
Pattern pat = Pattern.compile("^\\[*(\\d[\\d: -]+\\d).*?\\[Name:([^]]*)].*?\\[Fam:(\\w*)].*?\\[Resident:\\w*:([^]]*)]");
Matcher matcher = pat.matcher(s);
if (matcher.find()){
    System.out.println("Date: " + matcher.group(1));
    System.out.println("Name: " + matcher.group(2)); 
    System.out.println("Fam: " + matcher.group(3)); 
    System.out.println("Resident: " + matcher.group(4)); 
} 

Output:

Date: 2017-08-17 08:00:00
Name: +AA
Fam: aa
Resident: free