一次为多个变量分配相同的值？

php variables variable-assignment

2022-08-30 06:58:08

如何一次为PHP中的多个变量分配相同的值？

我有这样的东西：

$var_a = 'A';
$var_b = 'A';
$same_var = 'A';
$var_d = 'A';
$some_var ='A';

在我的情况下，我无法将所有变量重命名为具有相同的名称（这将使事情变得更加容易），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？

答案 1

$var_a = $var_b = $same_var = $var_d = $some_var = 'A';

答案 2

添加到其他答案。

$a = $b = $c = $d实际上意味着$a = ( $b = ( $c = $d ) )

默认情况下，PHP 通过值传递基元类型，通过引用传递对象。int, string, etc.

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

但是，您也可以使用关键字按值传递对象，但您必须使用括号。clone

$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

但是，您不能使用此方法通过关键字通过引用传递基元类型&

$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.