How to resolve java.net.MalformedURLException?

Question

I am getting this error:

java.net.MalformedURLException: Protocol not found[java.lang.StringBuilder]

When the following line is getting executed:

url = new URL(urlString.toString());

urlString stores the following value:

http://maps.google.com/maps?f=d&hl=en&saddr=25.04202,121.534761&daddr=25.05202,121.554761&ie=UTF8&0&om=0&output=kml

What causes this Exception?

Try logging the urlString.toString() value as the parser complains that it starts with the value "java.lang.StringBuilder" — MahdeTo, Feb 23 '12 at 07:26

Snicolas · Accepted Answer · 2012-02-23T12:50:39.810

6

Chances are that you didn't clean after changing from

url = new URL( urlString );

to

url = new URL(urlString.toString());

You should log the value of the parameter passed to the constructor of URL. It's not what you think it should be.

urlString would print a value in the form of java.lang.StringBuilder@ thus throwing the exception if you try to build a url out of that.

But using to String will print the value of the content string built by the stringbuilder.

edited Feb 23 '12 at 12:50

answered Feb 23 '12 at 07:23

Snicolas

37,840
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yes when I print it shows url=[Ljava.lang.StringBuilder;@412b0fb0, but when I check its value while debugging it shows the proper URL. What do u mean by logging the value of the param to the constructor of URL? What else should I do? – ManJan Feb 23 '12 at 08:04
I meant log the value of the parameter *passed* to the constructor of URL. Did you clean the project. For sure, the problems comes from here. – Snicolas Feb 23 '12 at 11:20
Thank you so much Snicolas.. I figured out the problem. I was using AsyncTask and was not passing params correctly. Its now resolved :) – ManJan Feb 24 '12 at 01:40

score 3 · Answer 2 · answered Feb 16 '15 at 15:30

Here is an example with exception handling ...

String urlString = "https://www.example.com/";
URL url = null;

// handle Exception
try {
    url = new URL(urlString);
} catch (MalformedURLException e) {
    System.out.println("The URL is not valid.");
    System.out.println(e.getMessage());
}

// print
if (url != null) {
    System.out.println(url.toString());
}

score 0 · Answer 3 · answered Aug 31 '12 at 10:45

0

You can also get this exception if your network firewall blocking that URL.

answered Aug 31 '12 at 10:45

W00di

954
12
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score 0 · Answer 4 · answered Sep 27 '13 at 09:36

0

replace url with php file address placed over server instedof html

answered Sep 27 '13 at 09:36

raj

2,088
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How to resolve java.net.MalformedURLException?

4 Answers4