Android - java.lang.ArrayIndexOutOfBoundsException

Question

I'm new to Java, but I have done some PSP coding in C before. This seems to be quite different.

Below is the stack trace for what I'm seeing:

java.lang.RuntimeException: An error occured while executing doInBackground()
enter code hereenter code here`at android.os.AsyncTask$3.done(AsyncTask.java:299)
enter code here`at java.util.concurrent.FutureTask$Sync.innerSetException(FutureTask.java:273)
at java.util.concurrent.FutureTask.setException(FutureTask.java:124)
at java.util.concurrent.FutureTask$Sync.innerRun(FutureTask.java:307)
at java.util.concurrent.FutureTask.run(FutureTask.java:137)
at android.os.AsyncTask$SerialExecutor$1.run(AsyncTask.java:230)
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1076)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:569)
at java.lang.Thread.run(Thread.java:856)
Caused by: java.lang.ArrayIndexOutOfBoundsException: length=1; index=1
at com.adev.abmp3.ViewSongDialog$FindLyrics.doInBackground(ViewSongDialog.java:173)
at com.adev.abmp3.ViewSongDialog$FindLyrics.doInBackground(ViewSongDialog.java:1)
at android.os.AsyncTask$2.call(AsyncTask.java:287)
at java.util.concurrent.FutureTask$Sync.innerRun(FutureTask.java:305)

And here is the patch of code:

@Override
protected Void doInBackground(Void... arg0) {
    String searchQuery = title.replaceAll("[^ \\w]", "");
    searchQuery = searchQuery.replace(" ", "+");
    String searchResult = httpRun("http://search.azlyrics.com/search.php?q="+searchQuery).toString();
    System.out.println("http://search.azlyrics.com/search.php?q="+searchQuery);
    if(!searchResult.contains("Sorry, no results") && !searchResult.equals("")) {
        String link = searchResult.split("<table width=100%>")[1];
        link = link.substring(link.indexOf("<a href=\""), link.indexOf("\" rel"));
        link = link.replace("<a href=\"", "");

        String lR = httpRun(link).toString();
        if(lR != "" && lR.contains("<!-- start of lyrics -->")) {
            lR = lR.substring(lR.indexOf("<!-- start of lyrics -->"), lR.indexOf("<!-- end of lyrics -->"));
            lR = lR.replace("<!-- start of lyrics -->", "");
            lyrics = lR;
        } else {
            fail = 1;
            lyrics = "Couldn't find lyrics";
        }
    } else {
        fail = 1;
        lyrics = "Couldn't find lyrics";
    }
    return null;
}

I've done some research, and have read that you can't print to the screen within doInBackground - but I'm not sure where to go from here. Any help would be appreciated.

EDIT: java.lang.StringIndexOutOfBoundsException in java.lang.String.startEndAndLength

Heres the new stack:

java.lang.RuntimeException: An error occured while executing doInBackground()
at android.os.AsyncTask$3.done(AsyncTask.java:299)
at java.util.concurrent.FutureTask$Sync.innerSetException(FutureTask.java:273)
at java.util.concurrent.FutureTask.setException(FutureTask.java:124)
at java.util.concurrent.FutureTask$Sync.innerRun(FutureTask.java:307)
at java.util.concurrent.FutureTask.run(FutureTask.java:137)
at android.os.AsyncTask$SerialExecutor$1.run(AsyncTask.java:230)
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1076)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:569)
at java.lang.Thread.run(Thread.java:856)
Caused by: java.lang.StringIndexOutOfBoundsException: length=7338; regionStart=537; regionLength=-538
at java.lang.String.startEndAndLength(String.java:593)
at java.lang.String.substring(String.java:1474)
at com.adev.abmp3.ViewSongDialog$FindLyrics.doInBackground(ViewSongDialog.java:174)
at com.adev.abmp3.ViewSongDialog$FindLyrics.doInBackground(ViewSongDialog.java:1)
at android.os.AsyncTask$2.call(AsyncTask.java:287)
at java.util.concurrent.FutureTask$Sync.innerRun(FutureTask.java:305)

omgz! check out this great library for parsing html called jsoup . will save you tons of time! — petey
– petey, Commented Feb 26, 2013 at 20:57
change lR != "" to !lR.equal("") and u should really check out jsoup.org — Shoshi
– Shoshi, Commented Feb 26, 2013 at 21:16

Gabe Sechan · Accepted Answer · 2013-02-26 20:56:40Z

3

Your split call is returning an array of size 1, but you're trying to get the 2nd item in it (with the [1]). You have to account for the possibility that your split string isn't in the input.

answered Feb 26, 2013 at 20:56

Gabe Sechan

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2 Comments

AmbitiousButRubbish Over a year ago

Ive never had to work with split arrays before. So by changing [1] to 2 it should fix it?

Gabe Sechan Over a year ago

No. The only valid index coming out of that call is 0- the split didn't actually split anything because the delimiter wasn't found. I think you're in for a lot of pain manipulating this one.

Brent Hronik · Accepted Answer · 2013-02-26 21:12:43Z

0

String link = searchResult.split("<table width=100%>")[1];

The error is occuring there. Indices start at 0, and your split is returning an array of size 1, so element 1 is out of bounds, and hence the exception is thrown. If you change it to String link = searchResult.split("<table width=100%>")[0]; the exception will no longer be thrown.

answered Feb 26, 2013 at 21:12

Brent Hronik

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3 Comments

AmbitiousButRubbish Over a year ago

Its still throwing out an error, let me try and get the stack

Brent Hronik Over a year ago

Are you sure? If you changed it to: String link = searchResult.split("<table width=100%>")[0]; the following error Caused by: java.lang.ArrayIndexOutOfBoundsException: length=1; index=1 wouldn't be the same, at the bare minimum it would change the index= value.

AmbitiousButRubbish Over a year ago

Sorry! I copied the wrong apk over. See the new stack, new line of code aswell...

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