What's wrong with this IF...AND statement?

Question

I have the code below in an Android method. When I debug the code the value of callType is "upload_latest" and the value of uploaded_date (defined as long) is 1424286105554, but code execution bypasses the "return false" statement and goes straight to the string declaration below it. I know I've missed something obvious but can't work out what it is - can anyone help?

if (callType == "upload_latest" && uploaded_date > 0) {
    return false;
}
String extension = "";

if `callType` is a `String` try `callType.equals("upload_latest")` — gsb, Feb 18 '15 at 19:23

score 1 · Answer 1 · edited May 23 '17 at 12:23

1

If callType is of type String (or any other non-primitive type), you need to use the .equals() method rather than the == operator. See this question for more information.

edited May 23 '17 at 12:23

Community

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answered Feb 18 '15 at 19:24

emerssso

2,376
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score 0 · Answer 2 · answered Feb 18 '15 at 19:23

0

To compare String values use .equals, == tests for reference equality.

callType.equals("upload_latest")

answered Feb 18 '15 at 19:23

Manuel Ramírez

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score 0 · Answer 3 · answered Feb 18 '15 at 19:25

0

Try something like this:

if ((callType.equals("upload_latest"))&& (uploaded_date > 0))) {
    return false;
}
String extension = "";

answered Feb 18 '15 at 19:25

apmartin1991

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score 0 · Answer 4 · answered Feb 18 '15 at 19:25

0

use callType.equals("upload_latest") instead of callType == "upload_latest"

The == operator checks to see of the strings are the same instance where .equals() checks to see if they represent the same character sequence.

answered Feb 18 '15 at 19:25

nerdy900

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What's wrong with this IF...AND statement?

4 Answers4