Java how to see if String has an integer in it

Question

I'm trying to determine if my ticket number has a "9" then change the price to 5. This is the code I have. The code itself compiles, but doesn't work as intended. I imagine it's because I'm using a wrong operator. If anyone could let me know how I could change it for it to work that'd be greatly appreciated.

for(int x=0; x<Integer.toString(e1.getNumber()).length(); x++)
{
    if(Integer.toString(e1.getNumber()).charAt(x) == 9)
     {
            System.out.println("The price is $5");
      }
  }

Thank you! I just had to change 9 to '9'! Its working now!!!

The char for 9 is '9', now you are checking for some character that has the ASCII value 9 — Joakim Danielson, Oct 14 '19 at 17:29
Hi Milan. Rather than putting the solution as an edit in your original question, can you please add it as an anwser and select it as the correct answer? This will make it easier for others to know the solution if they need. — TheNeil, Oct 14 '19 at 18:28

score 0 · Answer 1 · answered Oct 14 '19 at 17:39

it depends on specific requirements. what if 9 is more then one times in string? what you do in such case? if you are sure that 9 appears once in your ticket price. just you next code.

if (yourString.contains("9")){
     yourString = yourString.replace("9", "5");
}

score 0 · Accepted Answer · answered Oct 15 '19 at 19:03

0

Just had to put '9' so it looks for a char!

answered Oct 15 '19 at 19:03

Milan Toth

29
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Java how to see if String has an integer in it

2 Answers2