$p = $p == '' ? "1" : $p;这样定义变量有什么问题？-PHP Tutorial-php.cn

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$p = $p == '' ? "1" : $p;这样定义变量有什么问题？

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Jun 20, 2016 pm 12:42 PM

为什么我在公司的运行就没问题，拷到自己电脑上，因为我的&p用作页数值，所以我每个功能都在$p第一个报错。

回复讨论(解决方案)

请问楼主 p = $p == '' ? "1" : $p; 你这是三元运算吗？

$p = $p == '' ? "1" : $p;
会有 Notice: Undefined variable: p 错误警告
原因是你没有屏蔽 E_NOTICE 级别错误，也就是说：你的程序是不健壮的

宽松的写法是
$p = @$p == '' ? "1" : $p;
严密的写法是
$p = ! isset($p) || $p == '' ? "1" : $p;

php7 可写作
$p = $p ?? "1";

谢谢2L。好不开心啊。$p问题解决了。
但是还有好多 Notice: Undefined 的问题。原来换个电脑差距这么大。

程序中 error_reporting(E_ALL ^ E_NOTICE);
php.ini 中 error_reporting = E_ALL ^ E_NOTICE
之后，Notice: Undefined 就没有了

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