What is the PHP shorthand for: print var if var exist

Question 1

We've all encountered it before, needing to print a variable in an input field but not knowing for sure whether the var is set, like this. Basically this is to avoid an e_warning.

<input value='<?php if(isset($var)){print($var);}; ?>'>

How can I write this shorter? I'm okay introducing a new function like this:

<input value='<?php printvar('myvar'); ?>'>

But I don't succeed in writing the printvar() function.

Question 2

For PHP >= 5.x:

My recommendation would be to create a issetor function:

function issetor(&$var, $default = false) {
    return isset($var) ? $var : $default;
}

This takes a variable as argument and returns it, if it exists, or a default value, if it doesn't. Now you can do:

echo issetor($myVar);

But also use it in other cases:

$user = issetor($_GET['user'], 'guest');

For PHP >= 7.0:

As of PHP 7 you can use the null-coalesce operator:

$user = $_GET['user'] ?? 'guest';

Or in your usage:

<?= $myVar ?? '' ?>

Question 3

If the variable you are checking would be in the global scope you could do:

array_key_exists('v', $GLOBALS)

Question 4

Try this expression:

typeof(variable) != "undefined" && variable !== null

This will be true if the variable is defined and not null, which is the equivalent of how PHP's isset works.

You can use it like this:

if(typeof(variable) != "undefined" && variable !== null) {
    bla();
}

Question 5

It's to fix a bug in old versions of Visual C++ (v6.0 and earlier). In the past, Visual C++ had broken scope rules about variables declared inside for statements:

// This compiles in old versions of Visual C++, but it is in fact INVALID C++
for(int i = 0; ...)
{
    ...
}

for(i = 0; ...)
{

}

In other words, Visual C++ gives i a scope as if it were declared outside the loop, and it lets you continue using it after the loop is done. This lead to code such as the above snippet. In more standards-compliant compilers, i is no longer in scope at the definition of the second for loop, so the compiler issues an error about i being undefined.

To fix this, some people used this macro (or very similar, equivalent macros):

#define for if(0) {} else for

This changes the for loop into this:

if(0)
{
}
else
    for(int i = 0; ...)
    {
        ...
    }

This puts the for loop into an extra level of scope, so that any variables declared in the for loop will be out of scope afterwards, regardless of Visual C++'s bug. This makes sure that the same code compiles correctly consistently in both Visual C++ and standards-compliant compilers, and that incorrect code does not compile correctly consistently.

Also note that if the macro were instead defined as so:

// DO NOT USE
#define for if(1) for

Then although that would have the same effect for some simple code, it would suddenly cause the following code to be compiled incorrectly:

if(foo)
    for(...)
    {
        ...
    }
else
    doSomething();

Because if you expand the macro, you get this:

if(foo)
    if(1)
        for(...)
        {
            ...
        }
    else
        doSomething();

And the else now matches up with the wrong if! So, the clever use of using if(0) {} else instead of if(1) avoids this problem.

As a final note, #define for if(0) {} else for does not cause infinite recursion, because the preprocessor will not recursively replace the macro which you are currently defining. It will only do one replacement in this case.

Question 6

There is none. Alas, you have to type out the full type name.

Edit: 7 years after being posted, type inference for local variables (with var) was added in Java 10.

Edit: 6 years after being posted, to collect some of the comments from below:

The reason C# has the var keyword is because it's possible to have Types that have no name in .NET. Eg:
```
var myData = new { a = 1, b = "2" };
```
In this case, it would be impossible to give a proper type to myData. 6 years ago, this was impossible in Java (all Types had names, even if they were extremely verbose and unweildy). I do not know if this has changed in the mean time.
var is not the same as dynamic. variables are still 100% statically typed. This will not compile:
```
var myString = "foo";
myString = 3;
```
var is also useful when the type is obvious from context. For example:
```
var currentUser = User.GetCurrent();
```
I can say that in any code that I am responsible for, currentUser has a User or derived class in it. Obviously, if your implementation of User.GetCurrent return an int, then maybe this is a detriment to you.

This has nothing to do with var, but if you have weird inheritance hierarchies where you shadow methods with other methods (eg new public void DoAThing()), don't forget that non-virtual methods are affected by the Type they are cast as.

I can't imagine a real world scenario where this is indicative of good design, but this may not work as you expect:

class Foo {
    public void Non() {}
    public virtual void Virt() {}
}

class Bar : Foo {
    public new void Non() {}
    public override void Virt() {}
}

class Baz {
    public static Foo GetFoo() {
        return new Bar();
    }
}

var foo = Baz.GetFoo();
foo.Non();  // <- Foo.Non, not Bar.Non
foo.Virt(); // <- Bar.Virt

var bar = (Bar)foo;
bar.Non();  // <- Bar.Non, not Foo.Non
bar.Virt(); // <- Still Bar.Virt

As indicated, virtual methods are not affected by this.

No, there is no non-clumsy way to initialize a var without an actual variable.

var foo1 = "bar";        //good
var foo2;                //bad, what type?
var foo3 = null;         //bad, null doesn't have a type
var foo4 = default(var); //what?
var foo5 = (object)null; //legal, but go home, you're drunk

In this case, just do it the old fashioned way:

object foo6;

What is the PHP shorthand for: print var if var exist

Answers

For PHP >= 5.x:

For PHP >= 7.0:

php,var,isset