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Getting NumberFormat Exception

sandy mahajan
Greenhorn

Joined: Mar 27, 2008
Posts: 6
Hi,

String str1=request.getParameter("str1");
if(str1==null)
str1="";
using this code iam getting string(No Error).
But i want to convert that string into int & i code
int intstr1= Integer.parseInt(str1);
& i got following exception

java.lang.NumberFormatException: For input string: ""
java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
java.lang.Integer.parseInt(Integer.java:489)
java.lang.Integer.parseInt(Integer.java:518)
org.apache.jsp.jsp.Play_jsp._jspService(Play_jsp.java:109)
org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:94)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:324)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:292)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:236)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)
org.apache.jsp.jsp.DummyController_jsp._jspService(DummyController_jsp.java:84)
org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:94)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:324)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:292)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:236)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)

Please show me the right way.....Thanks.
K.Suresh Kumar
Ranch Hand

Joined: Nov 21, 2007
Posts: 41
Hi ,

you try convert null value to integer value.
So that you are getting NumberFormatException.
Your code :


yon can't convert null value to int value.

Regards,
Suresh Kumar.K
Muhammad Saifuddin
Ranch Hand

Joined: Dec 06, 2005
Posts: 1321



doesn't make sense, when you initialized st1 to "" after getting the value through request object.


Saifuddin..
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sandy mahajan
Greenhorn

Joined: Mar 27, 2008
Posts: 6
Hi,
I generated Random no. & retrives througth this code

String str1=request.getParameter("str1"); // getting String ex."34"
if(str1==null)
str1="";
but i want to convert to int for comparision & code

int intstr1= Integer.parseInt(str1);
but got NumberFormat exception

As per reply to me & i code

String str1=request.getParameter("str1"); // getting String ex."34"
if(str1==null)
str1="0";
int intstr1= Integer.parseInt(str1);
but same error occurs with str1="0" append with String

java.lang.NumberFormatException: For input string: "071 "
java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
java.lang.Integer.parseInt(Integer.java:477)
java.lang.Integer.parseInt(Integer.java:518)
org.apache.jsp.jsp.Play_jsp._jspService(Play_jsp.java:109)
org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:94)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:324)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:292)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:236)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)
org.apache.jsp.jsp.DummyController_jsp._jspService(DummyController_jsp.java:84)
org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:94)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:324)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:292)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:236)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)

I stuck at this point.
Please show me the way.........
Thanks
Herman Schelti
Ranch Hand

Joined: Jul 17, 2006
Posts: 387


There's a space behind the 071

Herman
Anubhav Anand
Ranch Hand

Joined: May 18, 2007
Posts: 341

Quick Tip: Please use code tags while posting code it makes a post much more readable and saves time.

You may also like to see how to ask questions on JavaRanch.
Jhakda Velu
Ranch Hand

Joined: Feb 26, 2008
Posts: 166
Hi
You can use trim() method to avoid extra spaces. But ensure that you have done the null check else a null pointer ex will be thrown.

Jhakda


If I become filthy rich, I'll sponsor research for painless dental treatment at Harvard Medical School. Thats why,I'm learning Java.I have 32 teeth, 22 are man made.
danjonila Jkatz
Greenhorn

Joined: Aug 12, 2013
Posts: 1
I have another question regarding my code using Java 1.4.2.

Here is my source code:





And here is the error I get:


Exception in thread "main" java.lang.NumberFormatException: For input string: ""
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
at java.lang.Integer.parseInt(Integer.java:470)
at java.lang.Integer.valueOf(Integer.java:554)
at WriteKong.sort(WriteKong.java:58)
at WriteKong.main(WriteKong.java:25)
Macintosh-d49a20e0a718:WriteKong arnykatz$ javac WriteKong.java
Macintosh-d49a20e0a718:WriteKong arnykatz$ java WriteKong
StudentID = 7493272
Course = English
MarkS =
Exception in thread "main" java.lang.NumberFormatException: For input string: ""
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
at java.lang.Integer.parseInt(Integer.java:470)
at java.lang.Integer.valueOf(Integer.java:554)
at WriteKong.sort(WriteKong.java:58)
at WriteKong.main(WriteKong.java:25)


Why do I get this error?
K. Tsang
Bartender

Joined: Sep 13, 2007
Posts: 2672
    
    9

@danjonila Your error is the same as the OP: passing in a "" (blank) to the Integer.valueOf() on line 58. Just like what the OP did, you need to check for null.


K. Tsang JavaRanch SCJP5 SCJD OCPJP7 OCPWCD5 OCPBCD5 OCPWSD5
Jesper de Jong
Java Cowboy
Saloon Keeper

Joined: Aug 16, 2005
Posts: 14483
    
  23

Welcome to JavaRanch, Danjonila.

Is there a special reason why you are using Java 1.4.2? That is a very old and outdated version of Java. I advise you to use Java 7, the current version. You can download the JDK for Java 7 on Oracle's website.


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