java.net.UnknownHostException: Invalid hostname for server: local

Question

What are the steps I should take to solve the error:

java.net.UnknownHostException: Invalid hostname for server: local

I added the new virtual host name at Android emulator but the result returns to

 java.net.UnknownHostException virtualhostname at 
  java.net.InetAddress.lookUpHostByName(InetAddress.java:506)

When I type my virtualhost URL on my PC, it works on display. Then again, when I ran on Emulator and check on Logcat, I couldn't be able to read or check the HTTP status if 200, 202, or an error code number. It simply returned to UnknownHostException

a bit more information would help us to solve the problem easier — clamp, Jun 26 '11 at 13:39
are you behind a Proxy server and configured the proxy settings? If so, you might need to exclude localhost and other hosts which don't need the Proxy server's help to resolve to a valid IP Address. — asgs, Jun 04 '13 at 07:52
I added the new virtual host name at Android emulator but the result returns to "java.net.UnknownHostException virtualhostname at java.net.InetAddress.lookUpHostByName(InetAddress.java:506)". When I type my virtualhost URL on my PC, it works on display. Then again, when I ran on Emulator and check on Logcat, I couldn't be able to read or check the HTTP status if 200, 202, or an error code number. It simply returned to "UnknownHostException" — David Dimalanta, Sep 20 '13 at 04:35

score 83 · Answer 1 · edited Feb 07 '22 at 21:38

83

I was having the same issue on my mac. I found the issue when I pinged my $HOSTNAME from terminal and it returned ping: cannot resolve myHostName: Unknown host.

To resolve:

Do echo $HOSTNAME on your terminal.
Whatever hostname it shows (lets say myHostName), try to ping it : ping myHostName. If it returns ping: cannot resolve myHostName: Unknown host then add an entry into your /etc/hosts file.
For that edit /etc/hosts file and add following:

127.0.0.1 myHostName

edited Feb 07 '22 at 21:38

Chaminda Bandara

2,067
2
28
31

answered Nov 01 '14 at 04:42

Shobhit Puri

25,769
11
95
124

1

Wonder how it works on some machines out of the box and on some we have to add the hosts entry – Skynet Oct 23 '17 at 11:35
@Skynet depends on the linux distribution I guess – Alex Dec 29 '21 at 11:50

score 72 · Accepted Answer · edited Dec 18 '17 at 23:53

72

What the exception is really saying is that there is no known server with the name "local". My guess is that you're trying to connect to your local computer. Try with the hostname "localhost" instead, or perhaps 127.0.0.1 or ::1 (the last one is IPv6).

From the javadocs:

Thrown to indicate that the IP address of a host could not be determined.

127.0.0.1or ::1 or "localhost" should always be the loopback interface, so if that doesn't work I'd be really surprised.

If there really is a server called "local" on your network - examine your DNS settings or add it to your hosts file.

edited Dec 18 '17 at 23:53

rogerdpack

62,887
36
269
388

answered Jun 26 '11 at 13:37

André Laszlo

15,169
3
63
81

2

On some implementations, a restart of the JVM might be necessary for gateway/dns changes to take effect, apparently. – André Laszlo May 03 '13 at 09:40

Ash · Answer 3 · 2014-07-04T15:32:58.333

16

java.net.UnknownHostException: Host is unresolved:

Thrown to indicate that the IP address of a host could not be determined.

This exception is also raised when you are connected to a valid wifi but router does not receive the internet. Its very easy to reproduce this:

Connect to a valid wifi
Now remove the cable from the router while router is pluged-in

You will observe this error!!

You can't really solve this, You can only notify the user gracefully. (something like - "Unable to make a connection")

edited Jul 04 '14 at 15:32

answered Sep 19 '12 at 21:32

Ash

1,391
15
20

1

It would be more general to say that the exception is raised when the host isn't found in DNS or when the DNS itself can't be found, e.g. for *any* network connectivity reason, or a misconfiguration. – user207421 Oct 18 '12 at 23:12
So how do you fix this problem? – cokedude Nov 28 '13 at 23:48
3

You can't really solve this, You can only notify the user gracefully. (something like - "Unable to make a connection") – Ash Jul 04 '14 at 15:33
What does it mean to have this problem in a school network but not my personal connection at home? – sodiumnitrate May 08 '15 at 16:30
@sodiumnitrate May be your school network requires some id/pwd authentication first.. Try opening any web url in browser and see if that is properly displayed – Ash May 09 '15 at 15:11

score 10 · Answer 4 · answered Nov 29 '15 at 07:51

This is not specific to the question, but this question showed up when I was Googling for the mentioned UnknownHostException, and the fix is not found anywhere else so I thought I'd add an answer here.

The exception that was continuously received was:

java.net.UnknownHostException:  google.com
    at java.net.AbstractPlainSocketImpl.connect(AbstractPlainSocketImpl.java:184)
    at java.net.SocksSocketImpl.connect(SocksSocketImpl.java:392)
    at java.net.Socket.connect(Socket.java:589)
    at java.net.Socket.connect(Socket.java:538)
    at java.net.Socket.<init>(Socket.java:434)
    at java.net.Socket.<init>(Socket.java:211)
    ...

No matter how I tried to connect to any valid host, printing it in the terminal would not help either. Everything was right.

The Solution

Not calling trim() for the host string which contained whitespace. In writing a proxy server the host was obtained from HTTP headers with the use of split(":") by semicolons for the HOST header. This left whitespace, and causes the UnknownHostException as a host with whitespace is not a valid host. Doing a host = host.trim() on the String host solved the ambiguous issue.

Thank you - this saved me. In my case the String host being passed to the function had a \n suffix, so running `.trim()` on that fixed it for me! — rayzinnz, Mar 17 '18 at 18:34
Thanks, I spent too much trying to figure out what the problem was until I realized I typed a blank space between two different IP addresses. — Sandoval0992, May 16 '18 at 21:00

score 7 · Answer 5 · answered Dec 04 '15 at 18:11

Your hostname is missing. JBoss uses this environment variable ($HOSTNAME) when it connects to the server.

[root@xyz ~]# echo $HOSTNAME
xyz

[root@xyz ~]# ping $HOSTNAME
ping: unknown host xyz

[root@xyz ~]# hostname -f
hostname: Unknown host

There are dozens of things that can cause this. Please comment if you discover a new reason.

For a hack until you can permanently resolve this issue on your server, you can add a line to the end of your /etc/hosts file:

127.0.0.1 xyz.xxx.xxx.edu xyz

Preventative maintenance: http://unix.stackexchange.com/questions/248616/what-is-the-best-way-to-keep-a-linux-servers-hostname-consistent — JustBeingHelpful, Dec 18 '15 at 06:28

Sidharth N. Kashyap · Answer 6 · 2014-06-20T11:07:17.570

5

This might happen due to various reasons

1) Check if you have VPN connected, you might get this error sometimes if yes

"Your hostname, localhost resolves to a loopback address: 127.0.0.1; using 10.xxx.1.193 instead (on interface cscotun0)"

2) Check your $HOSTNAME

3) try to ping $HOSTNAME on commandline and if it doesnt work, tweak the system settings to make your local host respond to pings

edited Jun 20 '14 at 11:07

answered May 28 '14 at 14:45

Sidharth N. Kashyap

299
2
6

score 4 · Answer 7 · answered Feb 07 '18 at 10:36

4

Try the following :

String url = "http://www.google.com/search?q=java";
URL urlObj = (URL)new URL(url.trim());
HttpURLConnection httpConn = 
(HttpURLConnection)urlObj.openConnection();
httpConn.setRequestMethod("GET");
Integer rescode = httpConn.getResponseCode();
System.out.println(rescode);

Trim() the URL

answered Feb 07 '18 at 10:36

rex roy

1,019
12
5

10

Trimming a hard-coded string with no spaces makes no sense. – Blake McBride Jun 08 '20 at 11:12

score 1 · Answer 8 · edited May 22 '13 at 07:12

1

Trying to connect to your local computer.try with the hostname "localhost" instead or perhaps ::/ - the last one is ipv6

edited May 22 '13 at 07:12

Volker E.

5,911
11
47
64

answered May 22 '13 at 06:47

joshwayne diestro

11
1

score 1 · Answer 9 · answered Sep 21 '16 at 05:06

Please try to set SPARK_LOCAL_IP environment variable to the ip address(can be localhost i.e. your own ip address) you want to connect. E.g.

$ export SPARK_LOCAL_IP=182.95.208.152

This way you will not be required to alter existing linux settings. Worked for me, hope helps you too.

Suraj Vaishnav · Answer 10 · 2018-08-20T06:53:33.670

0

Connect your mobile with different wifi connection with different service provider. I don't know the exact issue but i could not connect to server with a specific service provider but it work when i connected to other service provider. So try it!

edited Aug 20 '18 at 06:53

answered Aug 20 '18 at 06:44

Suraj Vaishnav

7,777
4
43
46

score 0 · Answer 11 · answered Sep 15 '19 at 19:37

0

I had this issue in my android app when grabbing an xml file the format of my link was not valid, I reformatted with the full url and it worked.

answered Sep 15 '19 at 19:37

vasmos

2,472
1
10
21

score 0 · Answer 12 · answered Nov 03 '19 at 17:42

0

If you are here because your emulator gives you that Exception, Go to Tools > AVD Manager in your android emulator and Cold boot your Emulator.

answered Nov 03 '19 at 17:42

Favour Felix Chinemerem

1,328
13
20

score 0 · Answer 13 · answered Jun 20 '22 at 09:09

0

I had the same issue.

Restart docker was the fix for me. For some reason it needed a restart, I don´t know why, but it worked.

answered Jun 20 '22 at 09:09

David Marciel

865
1
12
29

score -1 · Answer 14 · edited Jul 18 '17 at 21:17

-1

If your /etc/localhosts file has entry as below: Add hostname entry as below:

127.0.0.1 local host (add your hostname here)
::1 (add hostname here) (the last one is IPv6).

This should resolve the issue.

edited Jul 18 '17 at 21:17

Floern

33,559
24
104
119

answered Jul 18 '17 at 17:20

user8326951

1
1

java.net.UnknownHostException: Invalid hostname for server: local

14 Answers14

The exception that was continuously received was:

The Solution

Linked

Related