如何一次为 PHP 中的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';
就我而言,我不能将所有变量重命名为相同的名称(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?
如何一次为 PHP 中的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';
就我而言,我不能将所有变量重命名为相同的名称(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?
$var_a = $var_b = $same_var = $var_d = $some_var = 'A';
添加到另一个答案。
$a = $b = $c = $d
实际上意味着$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
但是,您也可以通过值传递对象,使用关键字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
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.