call function using if statement

Question

I want to be able to call a function within an if statement.

For example:

var photo = "yes";

if (photo=="yes") {

    capturePhoto();

}

else {
  //do nothing
};

This does nothing though. The function is clearly defined above this if statement.

Edit: Wow, downboated to hell! capturePhoto(); was just an example function that didn't really need any more explanation in this scenario?

What do you mean by "does nothing"? Does the function not get called or does it not do what you're expecting it to? What is the function -supposed- to do (posting the code would be useful!) that it doesn't? — Anthony Grist, Dec 05 '11 at 15:54
And what does `capturePhoto()` do? Unless your JS/browser install is totally hosed, or either of those `yes` strings are ninjas pretending to be yesses, there's no way that this code could NOT call capturePhoto. — Marc B, Dec 05 '11 at 15:54
A side note: you might want to use `true`/`false` boolean constants rather than `"yes"`/`"no"` strings. You could then format your `if` as `if (photo) capturePhoto();`. — Xion, Dec 05 '11 at 15:55
Why do people ask for help with their code, but then don't provide the code? — RightSaidFred, Dec 05 '11 at 15:56
What *Xion* said. Using `true` and `false` it making the code more readable and the execution is faster. If you are retrieving in your real code `"yes"/"no"` from user then just convert it to boolean `var photo = input === "yes";`. — kubetz, Dec 05 '11 at 15:57
I can't update RightSaidFred's comment enough. *POST YOUR DAMN CODE PEOPLE!* — Polynomial, Dec 05 '11 at 15:58
@RightSaidFred - It's clearly defined. Don't worry about it. — Jeff Camera, Dec 05 '11 at 16:01
LordSnoutimus: You've provided nothing of use in your question. If you really don't know if the `if` test should work, then you need to read a very basic programming tutorial. If you're having an issue in your actual code, then **you need to provide the code that is giving you the issue**. If you genuinely believe that the problem is with the `if`, and not your other code, then you need to learn some basic troubleshooting skills because they would lead you to the conclusion that the `if` will work properly. This question deserves far more downvotes than it has received. — RightSaidFred, Dec 05 '11 at 16:31

ComFreek · Accepted Answer · 2017-07-30T08:32:49.003

22

That should work. Maybe capturePhoto() has a bug? Insert an alert() or console.log():

var photo = "yes";
if (photo == "yes") {
 alert("Thank you StackOverflow, you're a very big gift for all programmers!");
 capturePhoto();
} else {
  alert("StackOverflow.com must help me!");
}

edited Jul 30 '17 at 08:32

answered Dec 05 '11 at 15:54

ComFreek

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It shouldn't be written '`It works!`' it should be written - '`thank you StackOverflow!`' ! – Aleks Mar 26 '13 at 14:48
1

@Aleks Why should the code thank StackOverflow if we haven't done anything except saying that `capturePhoto()` could have a bug? *BTW this is more than a year old ;)* – ComFreek Mar 26 '13 at 14:52
1

Because javascript doesn't report errors like we would want :) And even it is old more then a year, mentioning a word - `bug` stopped me for a second, and I figured a solution for my similar javascript problem not running a function. Even it is a year old I had to say it helped me +1 ;) – Aleks Mar 26 '13 at 15:14
1

@Aleks I'm glad that I could point you in the right direction :) Thank you, thank StackOverflow (see edit)! ;) – ComFreek Mar 26 '13 at 15:24
1

hehe :D it is absolutely true!! ;) thanks, have a happy programming ;) – Aleks Mar 26 '13 at 15:34

score 2 · Answer 2 · edited Dec 10 '13 at 13:54

2

I'm not seeing any problems here. I used this code and the function call worked. I kept your code and just added a function called capturePhoto().

Are you sure that the code you're using to call the function is firing?

var photo = "yes"; 
if (photo=="yes") 
{ 
    capturePhoto(); 
} 
else 
{ 
    //do nothing 
};
function capturePhoto() 
{ 
    alert("Pop up Message"); 
}

edited Dec 10 '13 at 13:54

BenMorel

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answered Dec 05 '11 at 16:10

TimWagaman

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score 1 · Answer 3 · answered Mar 24 '19 at 20:00

You should also consider using true and false instead of strings that could be manipulated depending on input.

If I had to correct the following code, then I should've done it like this;

var photo = true; // Will capture picture.
if (photo) { // 'true' is a truthy value.
    capturePhoto();
} else {
  // Do nothing
}

score 1 · Answer 4 · answered Dec 05 '11 at 15:55

You probably missed something, a quotation, a semicolon or something like that. I would recommend you to use a debugger like Firebug or even Google Chrome's Web Developer Tool. You will know what's wrong with your code and where it is wrong.

You may take a look at this live code that your code above works: http://jsfiddle.net/ZHbqK/

score 1 · Answer 5 · edited May 23 '17 at 12:02

The code looks fine to me (except you don't need the ; at the end of the last line). Check your error log; perhaps the browser thinks capturePhoto is not defined for some reason. You can also add alert statements to make sure the code is actually running:

var photo = "yes";

alert('Entering if statement');

if (photo=="yes") {
    alert('then');
    capturePhoto();
} else {
    alert('else');
    //do nothing
}

When you encounter a situation where it seems like a fundamental language feature is not working, get some more information about what is going on. It is almost never the platform's fault. It is occasionally a misunderstanding of how the feature works (e.g. why does parseInt('031') == 25 ?). It is usually a violation of an assumption you're making about the code that isn't holding up because of a problem elsewhere.

score 0 · Answer 6 · answered Dec 05 '11 at 15:58

The code that you posted does work.

I copied it and tested it.

Demo: http://jsfiddle.net/Guffa/vraPQ/

The only thing wrong with it that I can see is a semicolon after the closing bracket, but that is only a style problem. It will form an extra empty statement, but that doesn't cause any problems.

call function using if statement

6 Answers6

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