我该如何解决：使用未定义的常量段-假定为“段”

Question

我带了HTML网站，并将其变成了PHP网站。我目前处于测试阶段，但是在呈现的页面而不是内容上我一直收到“通知”。

我尝试用$将常量转换为变量，但是该通知仍然出现。但是，它从未定义的常量更改为未定义的变量。我将它们改回了常量。

我使用的是32位WinXP，因此，我将Xammp 1.8.2与PHP 5.4.31结合使用

有问题的数组：

    $navItems = array(
                      array(
                              slug => "index.php",
                              title => "Home"
                            ),
                      array(
                              slug => "about.php",
                              title => "About Me"
                            ),
                      array(
                              slug => "portfolio.php",
                              title => "Portfolio"
                            ),
                      array(
                              slug => "contact.php",
                              title => "Contact"
                            )
        );

我希望看到实际的页面内容。但是，我得到了：

注意：使用未定义的常量slug-在第5行的C：\ xampp \ htdocs \ zmglobal-it.com.php \ includes \ arrays.php中假定为“ slug”

注意：使用未定义的常量标题-在第6行的C：\ xampp \ htdocs \ zmglobal-it.com.php \ includes \ arrays.php中假定为“标题”

Answer 1

键值必须加引号，因此请使用“键名”或“键名”

''' { $navItems = array( array( "slug" => "index.php", "title" => "Home", ), array( "slug" => "about.php", "title" => "About Me", ), array( "slug" => "portfolio.php", "title" => "Portfolio", ), array( "slug" => "contact.php", "title" => "Contact", ) ); } '''

Answer 2

您的数组看起来不错，您可能只需要稍加修改，然后在所需的某些位置添加"或'，例如：

$navItems = array(
    array(
        "slug" => "index.php",
        "title" => "Home",
    ),
    array(
        "slug" => "about.php",
        "title" => "About Me",
    ),
    array(
        "slug" => "portfolio.php",
        "title" => "Portfolio",
    ),
    array(
        "slug" => "contact.php",
        "title" => "Contact",
    ),
);

或者也许：

$navItems = [
    [
        "slug" => "index.php",
        "title" => "Home",
    ],
    [
        "slug" => "about.php",
        "title" => "About Me",
    ],
    [
        "slug" => "portfolio.php",
        "title" => "Portfolio",
    ],
    [
        "slug" => "contact.php",
        "title" => "Contact",
    ],
];

var_dump($navItems);

输出

这是var_dump($navItems);的输出：

array(4) {
  [0]=>
  array(2) {
    ["slug"]=>
    string(9) "index.php"
    ["title"]=>
    string(4) "Home"
  }
  [1]=>
  array(2) {
    ["slug"]=>
    string(9) "about.php"
    ["title"]=>
    string(8) "About Me"
  }
  [2]=>
  array(2) {
    ["slug"]=>
    string(13) "portfolio.php"
    ["title"]=>
    string(9) "Portfolio"
  }
  [3]=>
  array(2) {
    ["slug"]=>
    string(11) "contact.php"
    ["title"]=>
    string(7) "Contact"
  }
}

Reference 1

What does the PHP error message “Notice: Use of undefined constant” mean?