java.lang.NumberFormatException: For input > string: "8888 --address=0.0.0.0" > > at > java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) >

Searched on Google with the first line of a JAVA stack trace?

We can recommend more relevant solutions and speed up debugging when you paste your entire stack trace with the exception message. Try a sample exception.

Recommended solutions based on your search

Samebug tips

,

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.

Solutions on the web

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 Google Groups by wshehab, 3 months ago
For input string: "00=B" > at > java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
via Oracle Community by 392 Guest, 1 year ago
For input string: "48?page=GapRules" > at > java.lang.NumberFormatException.forInputString(NumberFormatException.java:48) > >
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 john, 2 years ago
error=0 at java.lang.Integer.parseInt(Integer.java:409)
java.lang.NumberFormatException: For input > string: "8888 --address=0.0.0.0" > > at > java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) >
at java.lang.Integer.parseInt(Integer.java:492)

Users with the same issue

Samebug visitor profile picture
Unknown user
Once, 1 year ago
2 times, 1 year ago
Samebug visitor profile picture
Unknown user
Once, 2 years ago
Samebug visitor profile picture
Unknown user
Once, 1 year ago
Once, 1 month ago
181 more bugmates

Know the solutions? Share your knowledge to help other developers to debug faster.