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I have a string that contains the receive part of an email header. It's like from abc.xyz.com (abc.xyz.com. [112.35.4.152]) by xx.yy.com with ESMTPS.

How can I extract 112.35.4.152 from this string?

technophyle
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  • well the simple case would be to use indexOf and substring for `[` and `]` – Scary Wombat Aug 19 '14 at 07:13
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    to be honest, this question has shown little or no effort. – Scary Wombat Aug 19 '14 at 07:13
  • to be honest, wombat, I'm really lost. I've tried every possible way to do this. The problem is, IP addresses do not necessarily contain [ and ]. – technophyle Aug 19 '14 at 07:22
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    "_I've tried every possible way to do this_" If you have tried **every possible way** then what makes you think we can do something different. – takendarkk Aug 19 '14 at 07:24
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    Don't play on words takendarkk. I just wanted to say I did all I could do(at least I think so), that's it. Is stackoverflow some kind of place where newbies aren't welcome at all? – technophyle Aug 19 '14 at 07:29

4 Answers4

2

Try to use RegExp approach with Pattern class. 

String str = "from abc.xyz.com (abc.xyz.com. [112.35.4.152]) by xx.yy.com with ESMTPS id ...";
Pattern pattern = Pattern.compile("\\d{1,3}\\.\\d{1,3}\\.\\d{1,3}\\.\\d{1,3}");
Matcher matcher = pattern.matcher(str);

if (matcher.find()) {
    System.out.println(matcher.group(0));
} else {
    System.out.println("No match.");
}

Prints:

112.35.4.152

DmitryKanunnikoff
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Use String.indexOf("") to locate your [ and ] indexes, supply those index values to String.substring() method , and extract the substring.

String Str="from abc.xyz.com (abc.xyz.com. [112.35.4.152]) by xx.yy.com with ESMTPS id";
System.out.println(Str.substring((Str.indexOf("[")+1), Str.indexOf("]")) );

Edit:- Please be more specific in your questions, I read in your comments that the ip address being enclosed in [] is not guranteed , in that case you need to use regex.

the pattern for IPV4 ip addresses is 

\\b(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\\b

Reference (Also you can find regex for IPV6 here)

Here is a small snippet that should work for you:-

 String ip_pattern ="\\b(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\\b";
 String input="from abc.xyz.com [112.35.4.152](abc.xyz.com. by xx.yy.com with ESMTPS id ";
 Pattern pattern = Pattern.compile(ip_pattern);
 Matcher matcher = pattern.matcher(input);

 if (matcher.find()) {
     System.out.println(matcher.group());
  }
 else{
     System.out.println("No ip found in given input");
  }
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Mustafa sabir
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0
"from abc.xyz.com (abc.xyz.com. [112.35.4.152]) by xx.yy.com with ESMTPS id ...".

By looking at your String, you can just get a substring start after [ and end with ]

String s = "from abc.xyz.com (abc.xyz.com. [112.35.4.152]) by xx
                                                    .yy.com with ESMTPS id ...";
int startIndex=s.indexOf("[");
int endIndex=s.indexOf("]");
System.out.println(s.substring(startIndex+1,endIndex));

Output:

112.35.4.152

Read about String#substring()

Ruchira Gayan Ranaweera
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-1

Try this

String IPADDRESS_PATTERN = 
    "(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)";

Pattern pattern = Pattern.compile(IPADDRESS_PATTERN);
Matcher matcher = pattern.matcher(ipString);
    if (matcher.find()) {
        return matcher.group();
    }
    else{
        return "0.0.0.0";
    }

Credits: Extract ip address from string in java

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Viswanath Lekshmanan
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