Android java.io.IOException: java.net.URISyntaxException:

Question

I'm getting an exception saying Java URI Syntax Exception "java.io.IOException: java.net.URISyntaxException: Invalid % sequence: %wl in query at index 88:" when i try to connect from my android application.

It seems to be throwing the exception where in the URL it says "%wl" and following is the URL. is there a work around for this.

http://192.168.111.111:9000/RB/db.svc/upd?LinkId=184617ED1F21&IPs=fe80::1a46:17ff:feed:1f21%wlan0,192.168.1.127,&MNo=0771111111&sPin=000&Status=0

URLEncoder.encode to encode url queryString

ρяσѕρєя K
– ρяσѕρєя K

2013-01-08 11:54:24 +00:00
Commented Jan 8, 2013 at 11:54 — ρяσѕρєя K
– ρяσѕρєя K, Commented Jan 8, 2013 at 11:54

Abubakkar · Accepted Answer · 2013-02-12 05:54:52Z

4

If you want to use % in your URL the first you need to do is to encode it.

So first you need to replace that % with %25 in your string ....1f21%wlan0... with .....1f21%25wlan0.... before connecting.

You can use the following code for encoding the URL in Java

String encodedUrl = java.net.URLEncoder.encode(<your_url>,"UTF-8");

Have a look at the below links for more information.

1.How to encode url in java

2.URL encoding character reference

UPDATE :

If you don't want to use URL encoder then you can try this out :

yourURL.replaceAll("%", "%25");

It is fine here to replace a single special character, but it would be a tedious task to do like this if you have many special characters that require proper URL encoding.

edited Feb 12, 2013 at 5:54

answered Jan 8, 2013 at 11:58

Abubakkar

15.7k8 gold badges58 silver badges84 bronze badges

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

5 Comments

Mr.Noob Over a year ago

so you mean first to replace and then to encode the url? and when I do it it gives me a malformed url exception

Abubakkar Over a year ago

if your are encoding it with URLEncoder then no need to replace, it will be automatically replaced.

Abubakkar Over a year ago

then there must be some problem with your actual url

Mr.Noob Over a year ago

Managed to work it out. I just replaced the "%" with "%25" manually and it worked. did not need to use the URLEncoder at all.

Nandagopal T Over a year ago

I replaced "%" with "%25", it worked fine. This is what i used , [myURL.replaceAll("%", "%25")]. Hope, it will be useful for someone. Thanks

Collectives™ on Stack Overflow

Android java.io.IOException: java.net.URISyntaxException:

1 Answer 1

5 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

5 Comments

Your Answer

Sign up or log in

Post as a guest

Related