[SOLVED]problem with creating a URL

androidonkey · by **androidonkey** » Mon Feb 18, 2008 10:35 am

Hi All,

I'm having a problem creating a URL using a string variable. For example this approach returns a "Cannot be null" error due to the line URL theURL = new URL(OutboundURL):

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try{
 
    String OutboundURL;
 
    String Username = "someusername";
 
    OutboundURL = "http://www.somewebsite.com/someservice.php?un=" + Username;
 
    URL theURL = new URL(OutboundURL);
 
    InputStream in = theURL.openStream();
 
    int b = in.read();
 
    while(b != -1){                                                                             
 
            b = in.read();
 
     }
 
}
 
catch(MalformedURLException e){
 
}
 
catch(IOException e){
 
}
Parsed in 0.011 seconds,  using GeSHi 1.0.8.4

However the approach above works perfectly if I don't use a string variable:

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URL theURL = new URL("http://www.somewebsite.com/someservice.php?un=someusername");
Parsed in 0.011 seconds,  using GeSHi 1.0.8.4

My question is what's the difference between these two:
"http://www.somewebsite.com/someservice.php?un=someusername"
versus
OutboundURL = "http://www.somewebsite.com/someservice.php?un=" + Username

Thanks in advance.

plusminus · by **plusminus** » Mon Feb 18, 2008 12:25 pm

There should be no difference, especially because the code probably gets optimized to the exactly the same ...

Regards,
plusminus

androidonkey · by **androidonkey** » Mon Feb 18, 2008 6:39 pm

Plusminus,

I agree there shouldn't be but it seems to matter when I run it. But I found a way to make it work

Thanks.

plusminus · by **plusminus** » Mon Feb 18, 2008 7:41 pm

Hello androidonkey,

let the community know how you worked it out

Best Regards,
plusminus

androidonkey · by **androidonkey** » Tue Feb 19, 2008 12:32 am

Oh sorry about that. So here goes:

If you're creating a URL using more than one string variable you have to use String.valueOf() for example:

SomeURLString = "http://www.somewebsite//someservice?username=" + usernamestringvariable;

URL theURL = new URL(String.valueOf(SomeURLString));

I hope this helps - it took me a couple of hours to find out that I couldn't just plug a string variable in there like so:

URL theURL = new URL(SomeURLString);

Happy coding!

plusminus · by **plusminus** » Tue Feb 19, 2008 4:40 pm

Hello androidonkey,

fine to see that solution, but... does it make any sense, that it doesn't work the straight way

Regards,
plusminus

androidonkey · by **androidonkey** » Tue Feb 19, 2008 9:33 pm

Hi plusminus,

I'm not sure why it doesn't work the straightforward way but my theory is this:

If a string variable StringA is created from other string variables:
StringA = StringB + StringC + StringD;

I think StringA doesn't actually contain the values of StringB, StringC, and StringD. What it contains are the memory addresses which point to where the values of StringB, StringC, and StringD reside. So, a URL object probably requires a string literal and not pointers to memory addresses. Using String.valueof() will move over the actual values of StringB, C, D into StringA which then turns StringA into a literal.

Let me know what you think.

Thanks,
androidonkey

plusminus · by **plusminus** » Tue Feb 19, 2008 9:45 pm

Hello androidonkey,

the strings get concated. Each "+" creates a new object of the concated strings.

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StringA = StringB + StringC + StringD;
Parsed in 0.011 seconds,  using GeSHi 1.0.8.4

Perhaps the compiler does something wrong and when we use that function he cannot optimize

Regards,
plusminus

androidonkey · by **androidonkey** » Tue Feb 19, 2008 9:55 pm

Hey plusminus,

You're right - I agree that it's probably the compiler, which is why it took me so long to figure out the cause of this problem. I simply wasn't doing anything wrong but i'm glad i found this alternative. Should I report this as a bug to google?

Thanks,
androidonkey

[SOLVED]problem with creating a URL

[SOLVED]problem with creating a URL

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