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

Question

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.

Vestment

echo (isset($var) ? $var : '') is about as short as you can get and be syntactically correct. – Soapbark

without making a function for it I think @Marc B has provided the shortest way to do it... in a comment no less! – Radiolocation

Answer 1

For PHP >= 7.0:

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

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

Or in your usage:

<?= $myVar ?? '' ?>

For PHP >= 5.x:

My recommendation would be to create a issetor function:

function issetor(&$var, $default = null) {
    return isset($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');

Henpeck

Why are you doing a call by reference? Isn't value enough? Also, default is a keyword in PHP, it might be wise to not use that as a function name. It had weird results on my server. – Radiolocation

@afuzzyllama: Because I need to pass a potentially undefined variable. If I left out the reference it would give you a Undefined Variable Notice ;) – Henpeck

@afuzzyllama: Thanks for pointing out, renamed it to issetor (I think that's how they named it, when they wanted to implement it in PHP 6, but I'm not quite sure.) – Henpeck

One thing about this method - calling issetor as issetor($var['key']) will inject key into $var. In some cases, you might end up with unexpected behavior, particularly if you're iterating over keys later. – Emptyhanded

Not a big fan of this. PHP should throw an undefined variable warning when you pass it by reference as well. – Sidestep

In a lot of cases, I'd swap that for an !empty() instead. It's a very clever little line of code though, especially when combined with a sprintf() on the output. – Finger

the best definition for "shorhand" is in this answer with the ?? operator which I didn't knew until now :) – Trixy

Please can you tell me why you put & before $var like &$var? – Intersex

Answer 2

Another option:

<input value="<?php echo isset($var) ? $var : '' ?>">

Irritability

Yes, but it will also issue a E_NOTICE. Checking if it is set first fixes this. – Irritability

+1 and <input value="<?=isset($var) ? $var : '' ?>"> makes this even shorter. – Cayuse

Answer 3

The shortest answer I can come up with is <?php isset($var) AND print($var); ?>

Further details are here on php manual.

A simple alternative to an if statement, which is almost like a ternary operator, is the use of AND. Consider the following:

'; // This is an alternative isset( $value ) AND print( $value ); ?>

This does not work with echo() for some reason. I find this extremely useful!

Toaster

and is just an alias of the logical operator &&, but nobody uses it. Your code is the same as isset($var) && print($var); – Olnton

Answer 4

Better use a proper template engine - https://stackoverflow.com/q/3694801/298479 mentions two nice ones.

Here's your function anyway - it will only work if the var exists in the global scope:

function printvar($name) {
    if(isset($GLOBALS[$name])) echo $GLOBALS[$name];
}

Dawndawna

Sorry, looking for a solution for non-global vars. – Vestment

In this case.. get a template engine. – Dawndawna

Answer 5

<input value='<?php @print($var); ?>'>

Tapeworm

Error/warning/notice suppression is pretty dirty. – Dawndawna

It's dirty if it's suppressing an actual error. Is it really worth it to spend the time, or increase the complexity of the code when a simpler solution will suffice? As long as you are aware of what you are doing, and why the warning is being suppressed this is fine. – Tapeworm

There can be no errors in printing or echoing a variable. If the variable doesnt exist, it wont print. If there is something to print, php will take care for it. arychj is right. – Angio

Answer 6

You could do <?php echo @$var; ?>. Or <?= @$var ?>.

Pinsk

Error/warning/notice suppression is pretty dirty. – Dawndawna

@demian / thief: I'd agree if it was to supress a function's error, but frankly, supressing the "not defined" warning on a variable in an output situation is allowable – Soapbark

@Marc B: Except if you're going to do that, then why not just turn off warnings? – Marlborough

@mella: in some cases you don't have that level of control of the server, which is where the usual "ZOMG DON'T EVER USE SHORTTTAGS! THEY ARE TEH BAD!!!!!" breathelessly panicky warnings come from. – Soapbark

I agree 100%, but it's a legitimate answer to the question. – Pinsk

Using @ the code will still produce a warning but will just silence it, which tends to slow things down a bit. – Backhouse

@Marc B: Logically I agree, however, if someone more junior than you reads the code and says "ohh it's okay to use suppression", I guarantee you'll find it littered throughout the rest of the code that they write :) I think the overhead of writing in an isset or empty is much more preferable than bad habits :) – Lorelle

@demian: that's like saying we shouldn't have knives in the kitchen for dinner prep because they can be used to stab people in back alleys. Appropriate tools for appropriate jobs, and in the cases where you can't disable the warnings, suppression is the best bet, unless you want gunk up your code with repeated echo (isset($var) ? $var : '') constructs. – Soapbark

@Demian: Let me clear up that meme by citing from the default error handler in the PHP source code: case E_NOTICE: /* notices are no errors and are not treated as such like E_WARNINGS */ -- And the syntactic isset littering accomplishes little (micro optimization) over using the intended language construct for handling debug messages (=notices). – Summation

@mario: I'm not talking about optimization - I'm talking about best practices :) – Lorelle

@Mark B: If you're following current PHP OO practices, there's no reason that you can't have a static helper function with containing the otherwise redundant code. – Lorelle

Answer 7

There's currently nothing in PHP that can do this, and you can't really write a function in PHP to do it either. You could do the following to achieve your goal, but it also has the side effect of defining the variable if it doesn't exist:

function printvar(&$variable)
{
    if (isset($variable)) {
        echo $variable;
    }
}

This will allow you to do printvar($foo) or printvar($array['foo']['bar']). However, the best way to do it IMO is to use isset every time. I know it's annoying but there's not any good ways around it. Using @ isn't recommended.

For more information, see https://wiki.php.net/rfc/ifsetor.

Backhouse

Shouldn't that be if (isset($variable)) { – Kkt

Answer 8

This worked for me:

<?= isset($my_variable) ? $my_variable : $my_variable="Some Value"; ?>

Blanton

Answer 9

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

function printvar(&$val) {
    if (isset($val)) echo $val;
}

Marlborough

Won't work. Passing the var to the function will cause the notice. Not sure if it would work with a reference... might be worth a try. – Dawndawna

@ThiefMaster: Thought the same thing at the same time :-) – Marlborough

Answer 10

You could also use the following:

<input type="text" value="<?php echo @$var; ?>">//means if(isset($var)){ echo $var; }

Directorate

This answer has been given earlier, please abstain from posting double answers as it only adds distraction. – Goodish

Answer 11

$var;
(isset($var)) ? (print $var) : '';

$var = 'hello var';
(isset($var)) ? (print $var) : '';

Rau

Pretty complex... I mean a lot of opportunity to write syntax errors. – Vestment

I especially find the parentheses around the isset() and the print very adorable. – Horseman