Android Java, very simple code. but where goes wrong?

Question

I using a web services to get the value, make the long story short. I can get the value from webservices, for this case, the value for strVersion is 1(the for loop only return 1 record since there is only 1 record).

Now here is the problem, the result always return "Wrong: 1" instead of "OK: 1", come someone please tell me why? the datatype in MYSQL for this field is Varchar.

    AndroidHttpTransport aht = new AndroidHttpTransport(URL);
    aht.call(SOAP_ACTION, soapEnvelope);
    SoapObject resultString = (SoapObject) soapEnvelope.getResponse();

    final String[] strVersion = new String[resultString.getPropertyCount()];

    for(int i =0; i<resultString.getPropertyCount(); i++)
    {
        SoapObject array = (SoapObject) resultString .getProperty(i);
        strVersion[i] = array.getProperty(0).toString();                    
        if(strVersion[i] == "1")
        {
        Toast.makeText(getApplicationContext(), "OK: " + strVersion[i], Toast.LENGTH_SHORT).show();
        }
        else
        {
             Toast.makeText(getApplicationContext(), "Wrong: " + strVersion[i], Toast.LENGTH_SHORT).show();
        }
    }

score 4 · Accepted Answer · edited May 23 '17 at 12:04

4

Do String comparison using equals().

Change this:

if(strVersion[i] == "1")

to

if(strVersion[i].equals("1"))

Read this for more information.

== compares references,not the values. In your case, you want to check for the value equality, not the reference equality.

edited May 23 '17 at 12:04

Community

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answered Jun 20 '12 at 03:06

Kazekage Gaara

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been doing c# for a very long time. kinda confuse sometime. Thanks dude! :) – melvintcs Jun 20 '12 at 03:11

Android Java, very simple code. but where goes wrong?

1 Answers1