Exception : java.lang.NumberFormatException When convert String to long

Question

I am converting the string which have '1520056800` to long to have date. But I am getting NumberFormatException to convert this

Please help me.

long expiryDateMS = Long.parseLong(responseArray[0].replaceAll(" ", ""));
SimpleDateFormat simpleDateFormat = new SimpleDateFormat("yyyy-mm-dd");
Date date = new Date(expiryDateMS);

Stack trace

09-02 00:52:28.984: E/AndroidRuntime(12025): Caused by: java.lang.NumberFormatException: 1520056800
09-02 00:52:28.984: E/AndroidRuntime(12025):    at java.lang.Long.parse(Long.java:353)
09-02 00:52:28.984: E/AndroidRuntime(12025):    at java.lang.Long.parseLong(Long.java:344)
09-02 00:52:28.984: E/AndroidRuntime(12025):    at java.lang.Long.parseLong(Long.java:311)
09-02 00:52:28.984: E/AndroidRuntime(12025):    at com.example.astrill_openvpn.MainOnOffActivity.onCreate(MainOnOffActivity.java:99)
09-02 00:52:28.984: E/AndroidRuntime(12025):    at android.app.Instrumentation.callActivityOnCreate(Instrumentation.java:1047)
09-02 00:52:28.984: E/AndroidRuntime(12025):    at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:1615)

I'm 100% sure you have some invalid non-printable (so you don't see it) character in your string, if i use long expiryDateMS = Long.parseLong("1520 056800 ".replaceAll(" ", "")); it works, so please check responseArray[0] contents — BackSlash
– BackSlash, Commented Sep 1, 2013 at 19:31
Maybe you should break your chained call into two statements so you can see what you're trying to parse. You don't save anything by combining them, you know. But you greatly increase debugging difficulty. — Hot Licks
– Hot Licks, Commented Sep 1, 2013 at 19:33
@Android re-check my comment, i said invalid non-printable (so you don't see it) Loop through your string and print every character. Or maybe just print responseArray[0].replaceAll(" ", "").length(), if it will output something greater than 10 you have invalid characters in it. — BackSlash
– BackSlash, Commented Sep 1, 2013 at 19:34

Community · Accepted Answer · 2017-05-23 12:29:23Z

3

You probably have invisible or invalid characters in the string. Try this post for more details.

edited May 23, 2017 at 12:29

CommunityBot

11 silver badge

answered Sep 1, 2013 at 19:33

Jesko R.

82710 silver badges21 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Martijn Courteaux · Accepted Answer · 2013-09-01 19:32:56Z

0

This code should do the job. So I'm guessing you have got some encoding/special character problem. Try this to verify you have a real ASCII encoded number:

String str = responseArray[0].replaceAll(" ", "");
for (int i = 0; i < str.length(); ++i)
{
    char a = str.charAt(i);
    if (!('0' <= a && a <= '9')) System.out.println(a + " is not a valid digit!");
}

answered Sep 1, 2013 at 19:32

Martijn Courteaux

69.1k48 gold badges202 silver badges297 bronze badges

3 Comments

N Sharma Over a year ago

I checked and str is 1520056800. I think every digit in this is valid digit

N Sharma Over a year ago

I am using this SimpleDateFormat simpleDateFormat = new SimpleDateFormat("yyyy-mm-dd"); then it diplaying the date 1970-44-18..Its wrong. Please help to fix it

Martijn Courteaux Over a year ago

Use yyyy-MM-dd instead. mm is minute. MM is month.

Collectives™ on Stack Overflow

Exception : java.lang.NumberFormatException When convert String to long

2 Answers 2

Comments

3 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related