在另一个函数中调用函数

Question

I have a function printContent() which prints the arguments and logged() which checks if the user is logged in.

My point is to do something like this:

logged("printContent('TITLE', 'CONTENT', 1)", 1);

It doesn't work. It should printContent() if user is not logged in, but nothing is happening. If I'll try changing return() into print() it prints the text "printContent(text...)".

Here are these two functions:

function logged($echo = 0, $logout = 0)
{    
        if($_SESSION['user']) 
        {
            if($echo)
            {
                if(!($logout)) return $echo;
                else return false;
            }
            else return true;
        }
        else
        {  
            if($logout == 1) return $echo;
            else return false;
        }
}
function printContent($title, $content, $type = 0){
    if($type == 1){
        echo '<div class="right-box">';
        if($title) echo '<h3>'.$title.'</h3>';
        echo $content.'</div>';            
    }
    else {
        echo '<div class="left-box">';
        if($title) echo '<h2>'.$title.'</h2>';
        echo '<div class="left-box-content">'.$content.'</div></div>';
    }
}

Answer 1

It should printContent() if user is not logged in

You can try

if(!logged())
{
printContent('TITLE', 'CONTENT', 1);
}

Answer 2

Your logged() function is too complicated; the $echo and $logout parameters make the logic extremely hard to follow. You should simplify it to just this, doing one thing very well:

function isLoggedIn()
{
    return !empty($_SESSION['user']);
}

Then, the logic becomes quite simple afterwards:

if (!isLoggedIn()) {
    printContent('title', 'content', 1);
}

Play time

Being fancy, you could do this since 5.3, though it's a very contrived example of what you could accomplish with anonymous functions:

function ifLoggedIn($loggedIn, $loggedOut)
{
    return empty($_SESSION['user']) ? $loggedIn() : $loggedOut();
}

The $loggedIn and $loggedOut parameters are the callback parameters and get executed from inside the function, based on whether the user is logged in or not. To use it:

ifLoggedIn(function() {
}, function() {
    printContent('TITLE', 'CONTENT', 1);
});

Answer 3

What you seem to be looking for is a callback.

If I understand you correctly, you wish printContent only to execute, if the user is logged in, correct?

You may want to check this question for further info: How do I implement a callback in PHP?

Answer 4

you are trying to execute a plain string. Use

function1( function2() );

that is the same as

$tmp = function2();
function1($tmp);