为多个变量分配相同的值

use*_*248 103 php variables variable-assignment

如何在PHP中为多个变量分配相同的值?

我有类似的东西:

$var_a = 'A';
$var_b = 'A';
$same_var = 'A';
$var_d = 'A';
$some_var ='A';
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在我的情况下,我不能将所有变量重命名为具有相同的名称(这会使事情变得更容易),那么有没有办法以更紧凑的方式为所有变量分配相同的值?

Tim*_*per 224

$var_a = $var_b = $same_var = $var_d = $some_var = 'A';
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  • 这不能在Php类中使用,这是原始类型. (8认同)
  • @Eoin:在执行$ var_a = $ var_b = ... = new Class();时,所有变量将引用Class的相同实例。 (3认同)

Xeu*_*ron 6

添加到另一个答案。

$a = $b = $c = $d 实际上是指 $a = ( $b = ( $c = $d ) )

PHP默认int, string, etc.按值传递原始类型,按引用传递对象

这意味着

$c = 1234;
$a = $b = $c;
$c = 5678;
//$a and $b = 1234; $c = 5678;

$c = new Object();
$c->property = 1234;
$a = $b = $c;
$c->property = 5678;
// $a,b,c->property = 5678 because they are all referenced to same variable
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但是,您也可以使用关键字按值传递对象clone,但您必须使用括号。

$c = new Object();
$c->property = 1234;
$a = clone ($b = clone $c);
$c->property = 5678;
// $a,b->property = 1234; c->property = 5678 because they are cloned
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但是,您不能使用此方法通过关键字引用传递原始类型&

$c = 1234;

$a = $b = &$c; // no syntax error
// $a is passed by value. $b is passed by reference of $c

$a = &$b = &$c; // syntax error

$a = &($b = &$c); // $b = &$c is okay. 
// but $a = &(...) is error because you can not pass by reference on value (you need variable)

// You will have to do manually
$b = &$c;
$a = &$b;
etc.
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