java.lang.NumberFormatException: For input string: "48?page=GapRules" > at > java.lang.NumberFormatException.forInputString(NumberFormatException.java:48) > >

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You try to parse a String that contains non-numeric characters to an int. The string must contain decimal characters only, optionally beginning with a + or - sign.

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Can occur when you try to convert a String to a numeric value but the String is not well formatted for the conversion.

Solutions on the web

via Oracle Community by 392 Guest, 1 year ago
For input string: "48?page=GapRules" > at > java.lang.NumberFormatException.forInputString(NumberFormatException.java:48) > >
via Google Groups by wshehab, 5 months ago
For input string: "00=B" > at > java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
via Google Groups by Alejandro Marcos, 1 year ago
For input > string: "8888 --address=0.0.0.0" > > at > java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) >
via Oracle Community by 3004, 1 year ago
error=2 at java.lang.Integer.parseInt(Integer.java:409)
via Oracle Community by 3004, 1 year ago
error=3 at java.lang.Integer.parseInt(Integer.java, Compiled Code)
via Google Groups by Phillip Read, 2 years ago
error=2 at java.lang.Integer.parseInt(Integer.java, Compiled Code)
java.lang.NumberFormatException: For input string: "48?page=GapRules" > at > java.lang.NumberFormatException.forInputString(NumberFormatException.java:48) > >
at java.lang.Integer.parseInt(Integer.java:456)

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