Question

<?php echo isset($areas[ footer ]) ? $areas[ footer ] : null; ?>

如何改进?

Answer 1

请注意,你正在重复,而且以不实的条件,它将成为没有效果的“密码>。您可以这样说:<条码>空洞或<条码>;<条码>或<条码>未找到。其它选择是获得<代码> 发送<>>/代码>的回报价值:

$return = isset($areas[ footer ]) ? $areas[ footer ] : null;

if ($return)
{
  // $return contains footer info
}
else
{
  // footer was not set :(
}

Answer 2

视从区域获得多少美元,可能更清洁,将其分配给一个变量:

$footer = isset($areas[ footer ]) ? $areas[ footer ] : null;

然后,你可以使用脚步,不进行额外的盘问。

Answer 3

您也可以通过设定一个违约:

$footer = null;
if (isset($areas[ footer ])) {
  $footer = $areas[ footer ];
}

echo $footer;

Answer 4

无,这是处理这种产出的最简明方式。

Answer 5

“非常经常使用这种代码”

你们是否应该通过使用模版语言来彻底避免这一问题,或者将这种行为归入一种功能?

类似:

function get_area($area) {
    if... //your code
    return $area

Answer 6

一种较短的版本,即:

<?php !isset($areas[ footer ]) or print $areas[ footer ]; ?>

But i m not sure if it is faster or more elegant. What do you guys think?

Answer 7

echo $areas[ footer ];

简单,与原线完全相同。

Edit in reply to Felix This gives a notice, but unless you re supposed to turn this in as ultra-perfect homework, it doesn t really matter. You can either fill your code with isset calls or ignore small stuff like this. I d worry about it if I was working in say... Java, but pragmatically nobody is going to care if PHP code that works produces notices.

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