使用IndexOf（）查找字符串中字符的第n次出现

Question

I have a question regarding indexOf() . I am trying to program an EmailExtractor (Yes, this is a homework but I am not looking for code) which extracts the entire email address from a sentence that is input by a user.

For example - User Input: Mail us at abc@def.ghi.jk with your queries. The program will then display abc@def.ghi.jk from the above String . I understand indexOf() and substring() are required.

The idea I have now is to use indexOf() to locate the '@', and then search for the empty space just before the email address input by the user (nth).

My code is as follows:

System.out.println("This is an Email Address Extractor.\n");
System.out.print("Enter a line of text with email address: ");
String emailInput = scn.nextLine();
int spaceAt = emailInput.indexOf(" ");
for (int i = 1; i <= emailInput.indexOf("@"); i++){
    if (spaceAt < emailInput.indexOf("@")) {
        spaceAt = emailInput.indexOf(" ", spaceAt + 1);
    }
}

I understand and am aware of the problem in my code.

1) "Mail us at abc@def.ghi.jk with your queries".indexOf(" ") is 4, I am trying to get 10. However, the IF Condition I have input will cause it to skip to the next instance of indexOf() which is 25. (Because 10 < 14).

How do I go about avoiding this from happening?

Once again, I am not looking for purely the answer rather, I am trying to work around a solution. Thanks in advance!

Answer 1

How about finding the space before and after the @

int at = emailInput.indexOf('@');
int start = emailInput.lastIndexOf(' ', at) + 1;
int end = emailInput.indexOf(' ', at);
if (end == -1) end = emailInput.length();
String email = emailInput.substring(start, end);

Answer 2

you can use regular expressions instead of using indexOf() and substring()

^[_A-Za-z0-9-\\+]+(\\.[_A-Za-z0-9-]+)*@"+"[A-Za-z0-9-]+(\\.[A-Za-z0-9]+)*(\\.[A-Za-z]{2,})$"

use the Pattern and Matcher to validate the email id in your string. This may give you a clear view on it

Answer 3

One thing that you could do is call indexOf and test the result in the same iteration of the loop, instead of saving the check for the next iteration. Something like this in mostly pseudocode:

spaceAt := 0
indexOfAtSign := emailInput.indexOf('@');
while (spaceAt < indexOfAtSign) { // you never really used i anyway
    temp := index of next space
    if (temp > indexOfAtSign) break;
    else spaceAt = temp;
}

(There'd need to be a small bit of corner-case testing, but only for if there are no spaces either before the @ sign or after the @ sign)