为什么运行一下代码不报warning或者notice
<code>$data = null; var_dump($data['name']); </code>
结果：
<code>null </code>
为什么运行一下代码不报warning或者notice
<code>$data = null; var_dump($data['name']); </code>
结果：
<code>null </code>
这是一个很有意思的问题, 查了查资料, php.net上有人提出过
立即学习“PHP免费学习笔记（深入）”；
https://bugs.php.net/bug.php?id=68110
https://bugs.php.net/bug.php?id=54556
在后面的那页链接里看到了鸟哥@Laruence, 我想能最好深入解答这个问题, 就是他了.
<code>$arr = false; var_dump($arr['foo']['bar']['baz']); // NULL </code>
php手册上问题原文:
<code><?php $a['a'] = null; $a['b'] = array(); echo $a['a']['non-existent']; // DOES NOT throw an E_NOTICE error as expected. echo $a['b']['non-existent']; // throws an E_NOTICE as expected ?> </code>
I added this bug to bugs.php.net (https://bugs.php.net/bug.php?id=68110) however I made tests with php4, 5.4 and 5.5 versions and all behave the same way. This, in my point of view, should be cast to an array type and throw the same error. This is, according to the documentation on this page, wrong. From doc: "Note: Attempting to access an array key which has not been defined is the same as accessing any other undefined variable: an E_NOTICE-level error message will be issued, and the result will be NULL."
你检查下php.ini的配置呢 或者在输出前加error_reporting(-1);试下呢
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