Function in an if statement

Question

I am trying to do an if statement if a function returns true and the else if it doesn't

Basically i need to show a html class in the return 1. I have try that below but I got an error. syntax error, unexpected 'if' (T_IF) This is what I have tried

    <div class="inner">
        <div class="stat" 
        <?php $access->getAccess() 
        if (getAccess == 1) {
        class="unpublished"
        }
        else {

    }
    ?>
    >
</div>
</div>

this is what the function looks like

 public function getAccess()
    {
        return $this->access;
    }

How can I do this right?

There are several issues with your syntax. Please check your use of brackets, semi-colon and $. Also, your logic doesnt make much sense. You might want to rethink whyt you are trying to do here and rewrite from scratch. — ToBe, Jan 30 '14 at 14:59

score 3 · Answer 1 · answered Jan 30 '14 at 14:58

3

Try with:

<div class="stat <?php echo $idea->getAccess() ? 'unpublished' : ''; ?>"></div>

answered Jan 30 '14 at 14:58

hsz

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1

You forgot an `echo` ;) – gen_Eric Jan 30 '14 at 15:00
@SarahJames What reason ? – hsz Jan 30 '14 at 15:39
If you look in the correct answer you are missing `== 1` because that was what I was trying to do in my fail if statement in my question. the function return 1 if its true and 0 if its a false. – Sarah James Jan 30 '14 at 15:49

score 2 · Accepted Answer · answered Jan 30 '14 at 15:00

2

You PHP code isn't echoing anything, plus you are missing semicolons.

Try to simplify this a bit:

<div class="stat <?=$idea->getAccess() == 1 ? 'unpublished' : ''?>">

P.S. You need to put all your classes into the same attribute.

answered Jan 30 '14 at 15:00

gen_Eric

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Thanks Rocket this works great. alot of the other answers were missing `== 1` and this one got it just right – Sarah James Jan 30 '14 at 15:30
@SarahJames: Do you mean if you go to `test.com#3546`, that the page would scroll to this `
`?
– gen_Eric Feb 03 '14 at 14:46

score 1 · Answer 3 · answered Jan 30 '14 at 14:57

1

To call a function you need to use brackets:

$access = getAccess();//Not getAccess

I note that you call another getAccess function too so you may have meant to do:

$access = $idea->getAccess() ;
if ($access == 1) {

Edit: Others have pointed out you are also missing a ;.

answered Jan 30 '14 at 14:57

Jim

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Best attempt of finding all of his problems so far. Still loads of them in his code though. – ToBe Jan 30 '14 at 15:01

score 1 · Answer 4 · answered Jan 30 '14 at 15:02

1

You need to conceptually separate PHP from your HTML. What is happening is that your PHP is being interpreted by your web server resulting in a HTML document that will be delivered to the client. What you want is to add a class to the div if an $idea has access, something like this:

<div class="inner">
    <div class="stat <?php echo $idea->getAccess() ? "unpublished" : ""?>">
    </div>
</div>

This is equivalent to a much more complex (and in my opinion harder to read):

<div class="inner">
    <div class="stat <?php 
       if ($idea->getAccess()) {
         echo "unpublished";
       }
      ?>">
    </div>
</div>

answered Jan 30 '14 at 15:02

cernunnos

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What error are you getting? check if the object you are using (in my answer that would be $idea) is an instance of the class that has the method "getAccess()", you can see the class of an object with a simple "echo get_class($idea);" – cernunnos Jan 30 '14 at 15:51
remember `class="unpublished"` is an HTML not php. `unpublished` is a css. – Sarah James Jan 30 '14 at 15:53
1

you cant have two class attributes (you can have more classes, but only 1 attribute) on a single element, you already have one class written on the html, "unpublished" will simply be added to your class if the condition is true, assuming that is what you want. – cernunnos Jan 30 '14 at 15:54

score 0 · Answer 5 · answered Jan 30 '14 at 14:58

0

You forgot a semi-colon:

getAccess();

answered Jan 30 '14 at 14:58

Ana Claudia

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Function in an if statement

5 Answers5