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# Converting string to int

Greenhorn
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Hi

Further to my previous problem, I'm trying a solution that requires converting a string value to an int using the ff method;

int val = Integer.parseInt(s);

This however, is throwing a NumberFormatException exception.

My s is a string like "Dominique Ramoney"

Any ideas?

Ranch Hand
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My s is a string like "Dominique Ramoney"1

Your string above is not a vaild number. So the parseInt method fails. The String must be a valid number.

For e.g.

int val =Integer.parseInt("100");

Ranch Hand
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Originally posted by Dominique Ramoney:
Hi

Further to my previous problem, I'm trying a solution that requires converting a string value to an int using the ff method;

int val = Integer.parseInt(s);

This however, is throwing a NumberFormatException exception.

My s is a string like "Dominique Ramoney"

Any ideas?

I think , you should be use ::

Solution 1. Use String.trim() method for delete blank

Solution 2. Check input before parse (Number only)

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Or, you can catch the NumberFormatException, so that you know it is not a valid number.

Nick

Dominique Ramoney
Greenhorn
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Thanx guys

So it appears that since the trim suggestion didn't work, there is no way to represent a string such as "firstname lastname" as an integer?

Oh well, back to the drawing board.

Ranch Hand
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Dominique,

Why do you want to convert your name to an integer?

Jules

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Originally posted by Dominique Ramoney:
Thanx guys

So it appears that since the trim suggestion didn't work, there is no way to represent a string such as "firstname lastname" as an integer?

Do u mean to convert them into ASCII number? If so, we should approach another way, rather the technique given above...

Nicholas Cheung
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Why do you want to convert your name to an integer?

I guess he may pass all params into the method for coversion, and only those who can convert to numbers are those necessary data, as we dont have any ideas of whether a series of params are strings or numbers, unless we check it digit by digit before passing to the method.

Nick

Ko Ko Naing
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Originally posted by Nicholas Cheung:

I guess he may pass all params into the method for coversion, and only those who can convert to numbers are those necessary data, as we dont have any ideas of whether a series of params are strings or numbers, unless we check it digit by digit before passing to the method.

Nick

Nick, don't you think that it's ASCII numbers that the original poster wants... Then surely we need another approach to fulfil the requirement in this case...

somkiat puisungnoen
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we check it digit by digit before passing to the method.

You can check digit in yor data follow this ::

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there is no way to represent a string such as "firstname lastname" as an integer?

You could try getting ASCII value for each character. There is no ASCII value for a String.

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Dominique Ramoney
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Thanks for the advice everyone, I was really just trying to find a kludge but I've got my program working correctly now.

UR all a great help.

Ko Ko Naing
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Originally posted by Dominique Ramoney:
Thanks for the advice everyone, I was really just trying to find a kludge but I've got my program working correctly now.

UR all a great help.

It's good to know that ur program is working properly now... But you didn't clear the thing that we all doubt... Is it like trying to get the ASCII value of each character(as Pradeep mentioned) or finding a valid number in the string(as Nick mentioned)?

Anyway, we are glad to hear that you got your program working properly...

Ranch Hand
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Originally posted by Dominique Ramoney:
Thanx guys

So it appears that since the trim suggestion didn't work, there is no way to represent a string such as "firstname lastname" as an integer?

Oh well, back to the drawing board.

I'm not sure if this is the type of solution you're looking for... however, there's a method to generate an unique integer from a String and it's called "hashcode()".

Example:

String s = "abcde";
int i = s.hashcode();