解析错误:语法错误，意外的“."，预期为“，"或“;" [英] Parse error: syntax error, unexpected '.', expecting ',' or ';'

问题描述

这件事困扰着我很多.我收到解析错误:语法错误，意外的."，期望为，"或;"在这一行

This thing is bugging me a lot. I'm getting Parse error: syntax error, unexpected '.', expecting ',' or ';' at this line

public static $user_table = TABLE_PREFIX . 'users';

TABLE_PREFIX是由定义函数创建的常量

TABLE_PREFIX is a constant created by define function

推荐答案

静态类属性在编译时初始化.初始化静态类属性时，不能使用常量TABLE_PREFIX与字符串文字连接，因为在运行时才知道常量的值.而是在构造函数中对其进行初始化:

Static class properties are initialized at compile time. You cannot use a constant TABLE_PREFIX to concatenate with a string literal when initializing a static class property, since the constant's value is not known until runtime. Instead, initialize it in the constructor:

public static $user_table;

// Initialize it in the constructor 
public function __construct() {
  self::$user_table = TABLE_PREFIX . 'users';
}

// If you only plan to use it in static context rather than instance context 
// (won't call a constructor) initialize it in a static function instead 
public static function init() {
  self::$user_table = TABLE_PREFIX . 'users';
}

http://us2.php.net/manual/zh/language.oop5.static.php

与其他任何PHP静态变量一样，静态属性只能使用文字或常量进行初始化；不允许使用表达式.因此，尽管您可以将静态属性初始化为整数或数组(例如)，但不能将其初始化为另一个变量，函数返回值或对象.

Like any other PHP static variable, static properties may only be initialized using a literal or constant; expressions are not allowed. So while you may initialize a static property to an integer or array (for instance), you may not initialize it to another variable, to a function return value, or to an object.

针对PHP> = 5.6的更新

PHP 5.6带来了对表达式的有限支持:

Update for PHP >= 5.6

PHP 5.6 brought limited support for expressions:

在PHP 5.6和更高版本中，相同的规则与const表达式适用:只要可以在编译时对其进行求值，某些有限的表达式也是可能的.

In PHP 5.6 and later, the same rules apply as const expressions: some limited expressions are possible, provided they can be evaluated at compile time.

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