Android dynamic String Resources

Question

in my app I do a web request that returns some result code, e.g. 105. I have string resources that look like that

<string name="r105">O.K.</string>
<string name="r106">Something went wrong.</string>
<string name="r333">Fatal error.</string>

Now I want to do something like that

Toast.makeText(parent.getApplicationContext(),
        parent.getString(R.string.r+resultCode), Toast.LENGTH_LONG).show();

while r+resultCode is the resource identifier.

This does not work. Any idea how to do that?

In above code you used resource as r105 but, while accessing you written R.string.r Is it correct ?? — Yugandhar Babu
– Yugandhar Babu, Commented Dec 29, 2011 at 13:06

Lalit Poptani · Accepted Answer · 2011-12-29 13:32:35Z

16

Try this getResources().getIdentifier(name, defType, defPackage) in a simple way.

Toast.makeText(this, getResources().getIdentifier("r"+resultcode, "string", 
getPackageName()), Toast.LENGTH_LONG).show();

answered Dec 29, 2011 at 13:32

Lalit Poptani

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aspartame · Accepted Answer · 2011-12-29 13:38:52Z

4

You can do it using getResources().getIdentifier(name, defType, defPackage). Something like this:

// Assuming resultCode is an int, use %s for String
int id = getResources().getIdentifier(String.format("r%d", resultCode), 
                                      "string", getPackageName());
String result = getString(id);

edited Dec 29, 2011 at 13:38

answered Dec 29, 2011 at 13:18

aspartame

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1 Comment

aspartame Over a year ago

Ah, I thought resultCode was an int. Try %s for string or just do "r"+resultCode.

Yugandhar Babu · Accepted Answer · 2011-12-29 13:02:14Z

-1

Try as below, its working for me. Use parent.getApplicationContext() for you.

String str = getString(R.string.r)+resultCode;

        Toast.makeText(getApplicationContext(),
                str, Toast.LENGTH_LONG).show();

answered Dec 29, 2011 at 13:02

Yugandhar Babu

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3 Comments

DarkLeafyGreen Over a year ago

I think getString(R.string.r)+resultCode wont work because there is no resource called "r"

Yugandhar Babu Over a year ago

I given part of my code only, not entire thing. I used it in my project in xml file.

DarkLeafyGreen Over a year ago

I have updated my question, no it is not working because R.string.r fails, there is no resource r

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