本文介绍了语法错误,意外的 '.',期待 ')'的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
限时送ChatGPT账号..当我从另一个类调用静态变量时遇到问题.我得到这个漂亮的语法错误,其中 php 出乎意料的 '.'
I've got a problem when I'm calling a static var from another class. I get this pretty syntax error where php is unexpected the '.'
这里是我调用它的地方:
Here is where I'm calling it :
private $aLien = array(
"menu1" => array("Accueil","statique/".Variable_init::$langue."/accueil.html",0,0), //This line
"menu2" => array("Infos Pratiques","statique/".Variable_init::$langue."/info.html",0,0),
"menu3" => array("Faire une réservation","statique/".Variable_init::$langue."/reserver.html",0,0),
"menu4" => array("Pour Nous Joindre","statique/".Variable_init::$langue."/nousJoindre.html",0,0),
"menu5" => array("Plan du site","statique/".Variable_init::$langue."/plansite.html",0,0)
);
这是我来自另一个类的静态 var 声明:
And here is my static var declaration from another class:
class Variable_init implements iVariable_init{
public static $langue;
public static $id_choix;
public static $id_contenu;
推荐答案
http://docs.php/language.oop5.properties 说:
它们是通过使用关键字public、protected 或private 之一定义的,后跟一个普通的变量声明.这个声明可能包含一个初始化,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.
They are defined by using one of the keywords public, protected, or private, followed by a normal variable declaration. 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.
您的字符串连接不是恒定的.解析器不理解".operator 在初始化部分,因此打印 unexpected '.'
Your string concatenations are not constant. The parser doesn't "understand" the . operator in the initialization part and therefore prints unexpected '.'
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