你不能在一个PHP类中放入两个具有唯一参数签名的__construct函数。我想这样做:
class Student
{
protected $id;
protected $name;
// etc.
public function __construct($id){
$this->id = $id;
// other members are still uninitialized
}
public function __construct($row_from_database){
$this->id = $row_from_database->id;
$this->name = $row_from_database->name;
// etc.
}
}
PHP中最好的方法是什么?
我可能会这样做:
<?php
class Student
{
public function __construct() {
// allocate your stuff
}
public static function withID( $id ) {
$instance = new self();
$instance->loadByID( $id );
return $instance;
}
public static function withRow( array $row ) {
$instance = new self();
$instance->fill( $row );
return $instance;
}
protected function loadByID( $id ) {
// do query
$row = my_awesome_db_access_stuff( $id );
$this->fill( $row );
}
protected function fill( array $row ) {
// fill all properties from array
}
}
?>
然后,如果我想要一个学生,我知道ID:
$student = Student::withID( $id );
或者如果我有一个db行数组:
$student = Student::withRow( $row );
从技术上讲,您不需要构建多个构造函数,而只是静态帮助方法,但是这样可以避免构造函数中的大量意大利面条代码。
我知道我在这方面非常晚,但我提出了一个相当灵活的模式,应该允许一些真正有趣和通用的实现。
像往常一样设置类,使用您喜欢的任何变量。
class MyClass{
protected $myVar1;
protected $myVar2;
public function __construct($obj = null){
if($obj){
foreach (((object)$obj) as $key => $value) {
if(isset($value) && in_array($key, array_keys(get_object_vars($this)))){
$this->$key = $value;
}
}
}
}
}
When you make your object just pass an associative array with the keys of the array the same as the names of your vars, like so...
$sample_variable = new MyClass([
'myVar2'=>123,
'i_dont_want_this_one'=> 'This won\'t make it into the class'
]);
print_r($sample_variable);
The print_r($sample_variable); after this instantiation yields the following:
MyClass Object ( [myVar1:protected] => [myVar2:protected] => 123 )
Because we've initialize $group to null in our __construct(...), it is also valid to pass nothing whatsoever into the constructor as well, like so...
$sample_variable = new MyClass();
print_r($sample_variable);
Now the output is exactly as expected:
MyClass Object ( [myVar1:protected] => [myVar2:protected] => )
The reason I wrote this was so that I could directly pass the output of json_decode(...) to my constructor, and not worry about it too much.
This was executed in PHP 7.1. Enjoy!
在创建具有不同签名的多个构造函数时,我也面临着同样的问题,但不幸的是,PHP没有提供直接的方法来这样做。然而,我发现了一个技巧来克服它。希望也适用于你们所有人。
<?PHP
class Animal
{
public function __construct()
{
$arguments = func_get_args();
$numberOfArguments = func_num_args();
if (method_exists($this, $function = '__construct'.$numberOfArguments)) {
call_user_func_array(array($this, $function), $arguments);
}
}
public function __construct1($a1)
{
echo('__construct with 1 param called: '.$a1.PHP_EOL);
}
public function __construct2($a1, $a2)
{
echo('__construct with 2 params called: '.$a1.','.$a2.PHP_EOL);
}
public function __construct3($a1, $a2, $a3)
{
echo('__construct with 3 params called: '.$a1.','.$a2.','.$a3.PHP_EOL);
}
}
$o = new Animal('sheep');
$o = new Animal('sheep','cat');
$o = new Animal('sheep','cat','dog');
// __construct with 1 param called: sheep
// __construct with 2 params called: sheep,cat
// __construct with 3 params called: sheep,cat,dog
从PHP 8开始,我们可以使用命名参数:
class Student {
protected int $id;
protected string $name;
public function __construct(int $id = null, string $name = null, array $row_from_database = null) {
if ($id !== null && $name !== null && $row_from_database === null) {
$this->id = $id;
$this->name = $name;
} elseif ($id === null && $name === null
&& $row_from_database !== null
&& array_keys($row_from_database) === [ 'id', 'name' ]
&& is_int($row_from_database['id'])
&& is_string($row_from_database['name'])) {
$this->id = $row_from_database['id'];
$this->name = $row_from_database['name'];
} else {
throw new InvalidArgumentException('Invalid arguments');
}
}
}
$student1 = new Student(id: 3, name: 'abc');
$student2 = new Student(row_from_database: [ 'id' => 4, 'name' => 'def' ]);
通过适当的检查,可以排除无效的参数组合,这样创建的实例在构造函数的末尾是一个有效的实例(但错误只会在运行时检测到)。