You know how, in JavaScript, we can set a value to a variable if one doesn't, like this:
name = name || 'joe';This is quite common and very helpful. Another option is to do:
name || (name = 'joe');Well, in PHP, that doesn't work. What many do is:
if ( empty($name) ) $name = 'joe';Which works...but it's a bit verbose. My preference, at least for checking for empty strings, is:
$name = $name ?: 'joe';What's your preference for setting values if they don't already exist?
if(!isset($name)){ $name = 'joe'}
@zecho well, in case of default argument with = null, you don't need isset(...) check - you just need $name = $name ?: 'default'. I was talking bout that.
@everzet, you're not right again. When $var === null your statement generates notice 'Undefined variable' because doesn't make difference between null and undefined.
@zecho nope, here i'm absolutely right:
<?php
function printName($name = null) {
$name = $name ?: 'default';
echo $name."\n";
}
printName();
printName('test');will output:
default
test
$var === null means that variable is defined (its value is null).
try with:
error_reporting(E_ALL);
ini_set('display_errors', 1);also:
$var = null;
echo isset($var);
@zecho no, you try. No exceptions/notices/errors for me :)
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
function printName($name = null) {
$name = $name ?: 'default';
echo $name."\n";
}
printName();
printName('test');
do this outside a function/method. I won't explain you what's the difference in php between function parameter and a variable. Also put as first line in the function
echo (int)isset($name);to see the result when $name = null
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
$name = null;
var_dump(isset($name));
$name = $name ?: 'default';
echo $name."\n";
var_dump(isset($name));bool(false)
default
bool(true)
I don't get what you're trying to prove here, but please, explain me the difference in php between function parameter and variable.
@zecho also, i think you've started arguing about https://gist.github.com/3194444#gistcomment-381958 Please re-read it again, cuz it looks like we have misunderstanding here.
@everzet, sorry about misunderstanding, I didn't read properly this comment. The idea is that you can't use $name = $name ?: 'default' when variable is not initialized at all (even with $name = null ).
@zecho yep, and here is the source misunderstanding:
... even with
$name = null
$name = null initializes variable. With null value.
In
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
$name = null;
var_dump($name);var_dump($name); is a proper statement and will return:
NULL
Without any exception/notice. And
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
$name = null;
$name = $name ?: 'default';
echo $name;Will successfully output
default
Without any exception/notice. Because null is a proper value of any variable.
Just my 20 cents - null initialises the variable with nothing. It prepares the variable to be garbage collected in the next cycle.
It only does the following - puts it in the current scope so the interpreter is not considering it not existing. It's similar to what global $global_var; does.
var_dump( isset( $null_var ) ); will output bool(false) anyway.
Setting a variable to null is only useful for cases as with the one you circle around @everzet - having it as a function's default variable.
I think the best way is: $name = (isset($name)) ? $name : "joe"; very similar to name || (name = 'joe');
Jeffrey, guess what, you can do the same thing on PHP:
$myValue = $someOtherValue ?: true;
$item = isset($arr["super_long_key_name"]) ? $arr["super_long_key_name"] : "default";
$item =@ $arr["super_long_key_name"]) ?: "default";
php Echo number if the number is not zero.
$number ?: $number != '0'
isset($name) || $name = 'Joe';
or
!empty($name) || $name = 'Joe';
PHP7:
$name = $name ?? 'Joe';
How would you do it in an array?
$name = []
$new_array = [
'first_name' => $name['first'] ?? 'Joe'
]
I just used : if ( empty($name) ) $name = 'joe';
This is more a kind of "human readable", for me .
I used for setting the opacity of a php image processing app, so i can update the main app later, because i'm using a slider input on app test version but i still using the same URL for processing the images, with caused bug because of a missing variable (opacity) .
Thank you .
Here is the perfect solution I've found and been using for a while ..
$name = 'John';
echo $name ?: 'User has no name';
// result : John
This will print John if $name is not empty but if $name is empty, it will print out 'User has no name' instead, which acts like a default value.
$name = ''; // or $name = null;
echo $name ?: 'User has no name';
result : User has no name
Best way is always the ternary operators. Sorry to open this up again, but I spent a good 20 min testing out cases because this conversation did so many loops through multiple techniques but didn't appear to settle.
The form:
{var} = {case} ? {val} : {default};
and an example:
$values = [ 'test' => 'testing' ];
$target_val = isset($values['test']) ? $values['test'] : null;
I settled on this because it appeared to be the most versatile. You can set a default value and if your using a false evaluating value as the default its easy to see if the variable was set. a.k.a. in this example if I do:
if(!$target_val)
{
echo 'not set';
}
I know that the value is not set and if I want to check if it is set:
if(!!$target_val)
{
echo 'set';
}
And it may be a bit verbose but you can do it all in one swoop like this:
if(!($target_val = isset($values['test']) ? $values['test'] : null))
{
//Is not set case
}
//Continue normally because you know its set
And like has been said, you can shorthand the ternary and just use:
$target_val = $values['test'] ?: null;
But you will get a notice thrown on array value not set so I always just use isset and assign a value in the ternary to avoid silencing messages for no reason.
I know this is a very old thread but,
with php 7+ you can use something like the following:
$user = $user ?? "" ?: "User not set";$user assigned as |
$user will return |
|---|---|
undefined |
User not set |
"" |
User not set |
null |
User not set |
false |
User not set |
"John" |
John |
?? checks if the value is undefined or null
?: checks if the value is null or false or an empty string
RedSparr0w you can shorten this:
$user ?? $user = "User not set";
Finally, I prefer...
/**
* @param $value
* @param $default
* @return mixed
*/
function get_string($value, $default)
{
if (!empty($value)) {
return $value;
} else {
return $default;
}
}
Usages: get_string($name, "Unknown");
This all be over once PHP 7.4 arrives:
$name ??= 'joe';
Does it throw warnings?
This code:
<?php
error_reporting(E_ALL);
ini_set("display_errors",1);
echo("PHP Variable Init");
if ( empty($A) ) $A = 'joe';
var_dump($A);
$B = $B ?: 'joe';
var_dump($B);
if (!isset($C)) $C = 'joe';
var_dump($C);
isset($D) OR $D = 'joe';
var_dump($D);
$E = (isset($E))?$E:'joe';
var_dump($E);
!isset($F) && $F = 'joe';
var_dump($F);
$G || ($G = 'joe');
var_dump($G);
?>Will generate this:
string(3) "joe"
<br /><b>Notice</b>: Undefined variable: B in <b>/home/webben/domains/yocto.nu/private_html/projects/phpvarinit/index.php</b> on line <b>10</b><br />
string(3) "joe"
string(3) "joe"
string(3) "joe"
string(3) "joe"
string(3) "joe"
<br /><b>Notice</b>: Undefined variable: G in <b>/home/webben/domains/yocto.nu/private_html/projects/phpvarinit/index.php</b> on line <b>25</b><br />
string(3) "joe"
Since PHP 7.4 we can now easily just say:
$name ??= 'joe';
@JeffreyWay I like the last option as long as the var has a default set like in your example function to avoid the undefined variable message. Or shorten it to one line by doing it this way.
return $name ? $name : 'joe';
If the variable isn't set to '' by default it doesn't seem like there would be a way around using isset() to avoid the warning.