php - 在 php 中处理构造函数时出错

Question

运行以下 PHP 代码时出现以下错误

class tableData
{
    private $row1 = array("Kalle1", "address1", "postal code1", "1@email.se", "070111001", "08111001");
    private $row2 = array("kalle2", "address2", "postcode2", "2@email.se", "070111002", "08111002");
    private $row3 = array("kalle3", "address3", "postcode3", "3@email.se", "070111003", "08111003");
    private $row4 = array("kalle4", "address4", "postcode4", "4@email.se", "070111004", "08111004");
    private $rader = array(array($row1),array($row2),array($row3),array($row4));
}

错误对应变量$rader:

Parse error: syntax error, unexpected '$row1' (T_VARIABLE), expecting ')' in C:\xampp\htdocs\test_k.php on line 15

谁能帮我改正这个错误？

Answer 1

The PHP manual states :

This declaration [of class properties] 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.

private $rader = array(array($row1),array($row2),array($row3),array($row4));

所以你的 $row1 不是常量值，它需要运行时信息，即变量持有的值。

当 PHP 编译您的类时，在属性初始化时，它不知道您的 $row1 变量持有什么值。

您必须将该变量设置为空或 null(无论适合什么)，或者有一个 setter 方法来设置它并首先调用它，或者如果这始终是相同的数据，则使用构造函数。

基本上，您不能根据稍后在运行时来自对象的值来设置属性。