检查是否设置了变量，然后回显而不重复? [英] Check if variable is set and then echo it without repeating?

问题描述

是否有一种简洁的方法来检查是否设置了变量，然后在不重复相同变量名的情况下回显它?

Is there a concise way to check if a variable is set, and then echo it without repeating the same variable name?

代替此:

<?php
    if(!empty($this->variable)) {
        echo '<a href="', $this->variable, '">Link</a>';
    }
?>

我正在考虑使用这种C风格的伪代码:

I'm thinking about something in the lines of this C-style pseudocode:

<?php
    echo if(!empty($this->variable, '<a href="', %s, '">Link</a>'));
?>

PHP具有 sprintf ，但是它并没有完全满足我的期望.如果可以，我当然可以用它来做一个方法/函数，但是肯定有一种本机"做的方法吗?

PHP has sprintf, but it doesn't quite do what I was hoping for. If course I could make a method/function out of it, but surely there must be a way to do it "natively"?

更新: 如果我理解的话，三元运算还会重复$this->variable部分?

Update: Ternary operations would also repeat the $this->variable part, if I understood it?

echo (!empty($this->variable) ? '<a href="',$this->variable,'">Link</a> : "nothing");

推荐答案

您能找到的最接近的是使用简短形式的三元运算符(自PHP5.3起可用)

The closest you can get to what you are looking for is using short form of ternary operator (available since PHP5.3)

echo $a ?: "not set"; // will print $a if $a evaluates to `true` or "not set" if not

但这将触发未定义的变量"通知.您可以明显地用@

But this will trigger "Undefined variable" notice. Which you can obviously suppress with @

echo @$a ?: "not set";

还是，不是最优雅/最干净的解决方案.

Still, not the most elegant/clean solution.

因此，您可以期望的最干净的代码是

So, the cleanest code you can hope for is

echo isset($a) ? $a: '';

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