问题
I need to check if value is defined as anything, including null. isset
treats null values as undefined and returns false
. Take the following as an example:
$foo = null;
if(isset($foo)) // returns false
if(isset($bar)) // returns false
if(isset($foo) || is_null($foo)) // returns true
if(isset($bar) || is_null($bar)) // returns true, raises a notice
Note that $bar
is undefined.
I need to find a condition that satisfies the following:
if(something($bar)) // returns false;
if(something($foo)) // returns true;
Any ideas?
回答1:
IIRC, you can use get_defined_vars() for this:
$foo = NULL;
$vars = get_defined_vars();
if (array_key_exists('bar', $vars)) {}; // Should evaluate to FALSE
if (array_key_exists('foo', $vars)) {}; // Should evaluate to TRUE
回答2:
If you are dealing with object properties whcih might have a value of NULL you can use: property_exists() instead of isset()
<?php
class myClass {
public $mine;
private $xpto;
static protected $test;
function test() {
var_dump(property_exists($this, 'xpto')); //true
}
}
var_dump(property_exists('myClass', 'mine')); //true
var_dump(property_exists(new myClass, 'mine')); //true
var_dump(property_exists('myClass', 'xpto')); //true, as of PHP 5.3.0
var_dump(property_exists('myClass', 'bar')); //false
var_dump(property_exists('myClass', 'test')); //true, as of PHP 5.3.0
myClass::test();
?>
As opposed with isset(), property_exists() returns TRUE even if the property has the value NULL.
回答3:
See Best way to test for a variable's existence in PHP; isset() is clearly broken
if( array_key_exists('foo', $GLOBALS) && is_null($foo)) // true & true => true
if( array_key_exists('bar', $GLOBALS) && is_null($bar)) // false & => false
回答4:
I have found that compact
is a function that ignores unset variables but does act on ones set to null
, so when you have a large local symbol table I would imagine you can get a more efficient solution over checking array_key_exists('foo', get_defined_vars())
by using array_key_exists('foo', compact('foo'))
:
$foo = null;
echo isset($foo) ? 'true' : 'false'; // false
echo array_key_exists('foo', compact('foo')) ? 'true' : 'false'; // true
echo isset($bar) ? 'true' : 'false'; // false
echo array_key_exists('bar', compact('bar')) ? 'true' : 'false'; // false
Update
As of PHP 7.3 compact() will give a notice for unset values, so unfortunately this alternative is no longer valid.
compact() now issues an E_NOTICE level error if a given string refers to an unset variable. Formerly, such strings have been silently skipped.
回答5:
The following code written as PHP extension is equivalent to array_key_exists($name, get_defined_vars()) (thanks to Henrik and Hannes).
// get_defined_vars()
// https://github.com/php/php-src/blob/master/Zend/zend_builtin_functions.c#L1777
// array_key_exists
// https://github.com/php/php-src/blob/master/ext/standard/array.c#L4393
PHP_FUNCTION(is_defined_var)
{
char *name;
int name_len;
if (zend_parse_parameters(ZEND_NUM_ARGS() TSRMLS_CC, "s", &name, &name_len) == FAILURE) {
return;
}
if (!EG(active_symbol_table)) {
zend_rebuild_symbol_table(TSRMLS_C);
}
if (zend_symtable_exists(EG(active_symbol_table), name, name_len + 1)) {
RETURN_TRUE;
}
}
回答6:
You could use is_null and empty instead of isset(). Empty doesn't print an error message if the variable doesn't exist.
回答7:
Here some silly workaround using xdebug. ;-)
function is_declared($name) {
ob_start();
xdebug_debug_zval($name);
$content = ob_get_clean();
return !empty($content);
}
$foo = null;
var_dump(is_declared('foo')); // -> true
$bla = 'bla';
var_dump(is_declared('bla')); // -> true
var_dump(is_declared('bar')); // -> false
回答8:
is_null($bar)
returns true, since it has no values at all. Alternatively, you can use:
if(isset($bar) && is_null($bar)) // returns false
to check if $bar
is defined and will only return true if:
$bar = null;
if(isset($bar) && is_null($bar)) // returns true
来源:https://stackoverflow.com/questions/3803282/check-if-value-isset-and-null