Android Java Strings

Question

For Strings..

How can i Detect if the string "name" has the text "Hello" a toast will come up saying yes but if it doesn't a toast will come up saying no? my code:

        String ft;
    if(ft.contains("Hello")){

        Toast.makeText(main.this ,"Yes", Toast.LENGTH_SHORT).show();
        }
        else{
        Toast.makeText(main.this ,"No", Toast.LENGTH_SHORT).show();
        }

I get a error at ft on if(ft.contains("Hello")) {

"The local variable ft may not have been initialized"

Thanks Kartik it worked, question will this save when the user closes the application and when they reopen it will be changed to whatever i put? — Hailei Edwards
– Hailei Edwards, Commented Oct 5, 2011 at 6:37
Let me resay that how can i change the string to be false after they click on my button? — Hailei Edwards
– Hailei Edwards, Commented Oct 5, 2011 at 6:40
question will this save when the user closes the application and when they reopen it will be changed to whatever i put?--- No it will be destroyed when you exit from application. Either you use Static String or SharedPreferences(if they are too important for you) — Kartik Domadiya
– Kartik Domadiya, Commented Oct 5, 2011 at 6:41

Peter Knego · Accepted Answer · 2011-10-05 05:53:27Z

2

String text;
if(name.equals("Hello")){
    text = "Yes";
} else {
    text = "No";
}

Toast toast = Toast.makeText(context, text, duration);
toast.show();

You can also use shorthand expression instead of if:

String text = name.equals("Hello") ? "Yes" : "No";

answered Oct 5, 2011 at 5:53

Peter Knego

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6 Comments

shanet Over a year ago

Why not all in one line? Toast.makeText(context, (name.equals("Hello") ? "Yes" : "No"), duration).show();

Hailei Edwards Over a year ago

String ft; if(ft.equals("Hello")){ ft = "Yes"; } else { ft = "No"; } i get a error under the ft at ft.equals "The local variable ft may not have been initialized"

ngesh Over a year ago

This is gonna throw a compile time error... it is't right.. "text" should have atleast blank string value initially...

Peter Knego Over a year ago

@Hailei: should be name.equals("Hello") not ft

Hailei Edwards Over a year ago

My string name is FT not name

|

Kartik Domadiya · Accepted Answer · 2011-10-05 06:14:38Z

1

You declared String, but you didn't initialize. Initializing it is setting them equal to a value:

String ft;        // This is a declaration

String ft="";    // This is an initialization

Perhaps String ft = ""; is "declaration and initialization.

You get the error because you haven't initialized the variable, but you use them (e.g., ft.compare()) in if condition.

So try replacing String ft; with String ft="";

answered Oct 5, 2011 at 6:14

Kartik Domadiya

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Comments

ngesh · Accepted Answer · 2011-10-05 06:06:52Z

1

if(name.contains("Hello")){

Toast.makeText(activity.this ,"Yes", Toast.LENGTH_SHORT).show();
}
else{
Toast.makeText(activity.this ,"No", Toast.LENGTH_SHORT).show();
}

Updated:

String ft = "";

    String ft= name.equals("Hello") ? "Yes" : "No";
     Toast.makeText(activity.this ,ft, Toast.LENGTH_SHORT).show();

edited Oct 5, 2011 at 6:06

answered Oct 5, 2011 at 5:56

ngesh

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2 Comments

Hailei Edwards Over a year ago

The local variable ft may not have been initialized under if(ft.contains

Harry Joy Over a year ago

@HaileiEdwards: change String ft; to String ft=""; or String ft=null; whatever suites best in your case.

anujprashar · Accepted Answer · 2011-10-05 05:55:57Z

0

if(name.equals("Hello")){
    Toast.makeText(this, "Yes", Toast.LENGTH_LONG).show();
}else{
     Toast.makeText(this, "no", Toast.LENGTH_LONG).show();
}

answered Oct 5, 2011 at 5:55

anujprashar

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