Question

我创建了一个函数如下：

//function for first name field
    function name ($fname) {

        //validate to see if field is empty
        if (empty($fname)) {
            return (false);
        } 

        $welcome_string = '<p> Welcome ' . $fname . '. We\'re glad you\'re here! Take a look around!</p>';
        return $welcome_string;

    }//end fname function

当函数通过检查时，我回显函数并获取welcome_string。但是，我需要在函数返回false时显示错误。我无法弄清楚如何做到这一点。下面是我调用函数的代码：

echo name($fname);

Answer 1

您要做的是添加一个检查：

$result = name($fname);

if ($result) {
    echo $result;
} else {
    echo "There was an error.";
}

使用三元组if，你可以用速记：

$result = name($fname);
echo $result ? $result : "There was an error.";

甚至更短而不引入新变量：

echo name($fname) ?: "There was an error.";

Answer 2

你可以这样做：

$result = name($fname); 
echo $result?$result:"error message";

Answer 3

<?php

function greeting($name)
{
    if (empty($name))
        return false;
    $welcome_string = 'Welcome ' . $name . ". We're glad you're here! Take a look around!\n";

    return $welcome_string;
}

foreach(['Rita', 'Sue', '', null, 0, '0'] as $name)
    echo ($message = greeting($name))
        ? $message
        : "Who are you?\n";

输出：

Welcome Rita. We're glad you're here! Take a look around!
Welcome Sue. We're glad you're here! Take a look around!
Who are you?
Who are you?
Who are you?
Who are you?

用于虚假php函数的echo语句

3 个答案: