error while parsing an integer

Question

This is my code :

  TextView headerVal = (TextView) nextChild.findViewById(R.id.textView3); 
  String tt = ((String)headerVal.getText()).replaceAll(",",""); //it was 1,000 and it became 1000           
    int longueur = tt.length();
    String aux2 = (String) tt.substring(0,longueur - 2);
    int contenu = (int) Integer.parseInt(aux2);// here is the error "unable to  parse 1000 as an integer".

PS: it works on the emulator but i got this error on my phone

Logcat:

04-04 13:04:15.210: E/AndroidRuntime(31718): FATAL EXCEPTION: main

04-04 13:04:15.210: E/AndroidRuntime(31718): java.lang.NumberFormatException: unable to parse '1 000' as integer

04-04 13:04:15.210: E/AndroidRuntime(31718):at java.lang.Integer.parse(Integer.java:433)

04-04 13:04:15.210: E/AndroidRuntime(31718):at java.lang.Integer.parseInt(Integer.java:422)

04-04 13:04:15.210: E/AndroidRuntime(31718):at java.lang.Integer.parseInt(Integer.java:382)
04-04 13:04:15.210:E/AndroidRuntime(31718):at com.example.appui.BarGraphActivity.setColumnParts(BarGraphActivity.java:408)
04-04 13:04:15.210: E/AndroidRuntime(31718):at com.example.appui.BarGraphActivity.access$2(BarGraphActivity.java:282)
04-04 13:04:15.210: E/AndroidRuntime(31718):at com.example.appui.BarGraphActivity$Desired_pension.onStopTrackingTouch(BarGraphActivity.java:162)

Probably a problem of [trimming](http://docs.oracle.com/javase/6/docs/api/java/lang/String.html#trim%28%29) — PCoder, Apr 04 '13 at 13:11
why you are typecasting 'int' in last statement after parsing value in Integer ? — Manish Lomte, Apr 04 '13 at 13:13
what is this statement returning exactly "String aux2 = (String) tt.substring(0,longueur - 2);" — Subburaj, Apr 04 '13 at 13:17
@ManishL it's the same with or without Subburaj because in the text view i had 1,000 & 2 letters so i used substring to use the number only — user2137817, Apr 04 '13 at 13:23

hd1 · Answer 1 · 2013-04-04T13:46:12.640

0

Use DecimalFormat as below:

 DecimalFormat format =(DecimalFormat) DecimalFormat.getInstance();
 format.applyPattern("#,###");
 try {
     integer = (Integer)format.parse(tt);
 } catch (ParseException e) {
     System.err.println(new Date()+": "+e.getMessage());
 }

edited Apr 04 '13 at 13:46

answered Apr 04 '13 at 13:10

hd1

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Post your entire code, it works for me, must be something silly – hd1 Apr 04 '13 at 13:23
can't post the whole code because it's too long but regarding the parse,it's the full code after that i use value i parsed for some operations – user2137817 Apr 04 '13 at 13:34

score 0 · Answer 2 · answered Apr 04 '13 at 13:15

Try this

public static void main(String[] args) {

        String test = ",100,0,0,";
        int newTest;
        test = test.replaceAll(",","");

        int longueur = test.length();
              // If the longueur length is greater than or equal to test.length() then it will return error, so put a check to see the longueur is not exceeding the test length.
        String aux2 = (String) test.substring(0,longueur - 2);

        newTest = Integer.parseInt(aux2);

        System.out.println(newTest);

    }

Result 100, if longueur - 1 Returns 1000

score 0 · Answer 3 · answered Apr 04 '13 at 13:18

0

You just use trim function after replace :-

String str2 = str.trim();

Above code remove all space into your string then convert string into integer:-

TextView headerVal = (TextView) nextChild.findViewById(R.id.textView3); 
String tt = ((String)headerVal.getText()).replaceAll(",",""); //it was 1,000 and it became 1000           

 String str2 = tt .trim();
int longueur = str2 .length();

Then do ur task what you want....

answered Apr 04 '13 at 13:18

duggu

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trim remove the spaces at the beginning or the end http://stackoverflow.com/questions/6932163/removing-spaces-from-string – user2137817 Apr 04 '13 at 13:33
@user2137817 then u have to check this link also – duggu Apr 05 '13 at 04:24

error while parsing an integer

3 Answers3