php 解析错误：语法错误，意外的 '('，期待 ',' 或 ';' in
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Parse error: syntax error, unexpected '(', expecting ',' or ';' in
提问by crmepham
I am recieveing the following parse error:
我收到以下解析错误：
Parse error: syntax error, unexpected '(', expecting ',' or ';' in H:\Programs\USBWebserver v8.5\8.5\root\oopforum\func\register.class.php on line 7
解析错误：第 7 行的 H:\Programs\USBWebserver v8.5\8.5\root\oopforum\func\register.class.php 中的语法错误，意外的 '(', expecting ',' or ';'
which relates to the following line of code in my class:
这与我班级中的以下代码行有关：
private $random_name = rand(1000,9999).rand(1000,9999).rand(1000,9999).rand(1000,9999);
I can not see why this line of code would cause a parse error?
我看不出为什么这行代码会导致解析错误？
Here is some surrounding code:
这是一些周围的代码：
class register{
public $post_data = array();
private $dbh;
private $allowed_type = array('image/jpeg','image/png','image/gif');
private $random_name = rand(1000,9999).rand(1000,9999).rand(1000,9999).rand(1000,9999);
private $path = 'img/thumb_/'.$random_name. $_FILES['file']['name'];
private $max_width = 4040;
private $max_height = 4040;
private $max_size = 5242880;
private $temp_dir = $_FILES['file']['tmp_name'];
private $image_type = $_FILES['file']['type'];
private $image_size = $_FILES['file']['size'];
private $image_name = $_FILES['file']['name'];
private $image_dimensions = getimagesize($temp_dir);
private $image_width = $image_dimensions[0]; // Image width
private $image_height = $image_dimensions[1]; // Image height
private $error = array();
public function __construct($post_data, PDO $dbh){
$this->post_data = array_map('trim', $post_data);
$this->dbh = $dbh;
}
}
What is causing the parse error?
导致解析错误的原因是什么？
回答by nickb
You can't initialize member variables to anything that is not static, and you're trying to call a function.
您不能将成员变量初始化为任何非静态的东西，并且您正试图调用一个函数。
From the manual:
从手册：
This declaration may include an initialization, but this initialization must be a constant value--that is, it must be able to be evaluated at compile time and must not depend on run-time information in order to be evaluated.
这个声明可能包括一个初始化，但这个初始化必须是一个常量值——也就是说，它必须能够在编译时被评估，并且必须不依赖于运行时信息才能被评估。
The workaround is to set your variable in the constructor:
解决方法是在构造函数中设置变量：
private $random_name;
public function __construct() {
$this->random_name = rand(1000,9999).rand(1000,9999).rand(1000,9999).rand(1000,9999);
}
回答by Tyler Carter
You can't do this:
你不能这样做：
private $image_dimensions = getimagesize($temp_dir);
Functions, variables, and anything else not static can't be called to set class variables.
不能调用函数、变量和其他任何非静态的东西来设置类变量。
