php - 解析错误 : Invalid numeric literal

Question

运行下面这段代码时出现以下错误:

代码:

<?php
    $a = array(00001, 00008, 00009, 00012);
    print_r($a);
?>

错误:

Parse error: Invalid numeric literal.

为什么会出现这个问题，我该如何解决？

Answer 1

这是因为在 PHP7 中处理整数(特别是八进制)的方式发生了变化(与 PHP5 不同)。

来自文档(来自 PHP7 迁移)

Invalid octal literals

Previously, octal literals that contained invalid numbers were silently truncated (0128 was taken as 012). Now, an invalid octal literal will cause a parse error.

来自整数的文档

Prior to PHP 7, if an invalid digit was given in an octal integer (i.e. 8 or 9), the rest of the number was ignored. Since PHP 7, a parse error is emitted.

将它们用作字符串或实际整数

$a = array(1, 8, 9, 12); // Integers
$a = array("00001", "00008", "00009", "00012"); // Strings

• PHP7 与 PHP5 的示例:https://3v4l.org/GRZF2
• http://php.net/manual/en/language.types.integer.php
• 手册:http://php.net/manual/en/migration70.incompatible.php