Converting leading 0 String into Integer in java

Question

I have following string in my java class

String str="0000000000008";

Now I want to increment that so that the next value should be 0000000000009

For that purpose, I tried to cast this String str into Integer

Integer i=Integer.parseFloat(str)+1;

and when I print the value of i it prints only 17(as it removes the leading 0's from string at the time of cast).

How can I increment the String value, so that the leading 0's will remain, and the series will continue?

Andreas Dolk · Accepted Answer · 2012-03-02 11:44:40Z

7

Practical solution - use String.format:

str = String.format("%013d", Long.parseLong(str)+1);

answered Mar 2, 2012 at 11:44

Andreas Dolk

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2 Comments

Andreas Dolk Over a year ago

(didn't count the digits ;) it's only 13, not 17)

jarnbjo Over a year ago

You could also do str = String.format("%0" + str.length() + "d", Long.parseLong(str)+1); to keep the number of digits in the original string.

Tomasz Nurkiewicz · Accepted Answer · 2012-03-02 11:43:33Z

5

You are on the correct path. First parse to Long:

long cur = Long.parseLong("0000000000008");

increment and format back to String with leading 0s:

new java.text.DecimalFormat("0000000000000").format(cur + 1);

or alternatively:

String.format("%013d", Long.valueOf(cur));

answered Mar 2, 2012 at 11:43

Tomasz Nurkiewicz

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3 Comments

Arun Kumar Over a year ago

..Thanks for the answer. But the String str is not a constant. It may be changed during run-time. So, is there any other solution for this?

Tomasz Nurkiewicz Over a year ago

@ArunKumar: so, what is wrong with: Long.parseLong(knownAtRuntime) where knownAtRuntime is a String variable?

Arun Kumar Over a year ago

@ Tomasz Nurkiewicz -- You are right. Your answer will work for a constant. But my concern is that this will not work for a String variable whose value may be changed during execution of program.

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