Question

我正在使用Google Distance Matrix来计算出租车网站的两个位置之间的距离。我需要测试一个条件，以确定其中一个位置是否是伦敦机场，如果是，则返回一条消息，而不是计算行程的成本。

它在昨天和今天工作得很好......代码已经变得很苛刻了。如果接送和送货地点都是机场，它只返回信息，但如果只有一个，则计算全距离的费用，这违背了我的目的。下面是代码结构，我想知道当我启动strpos（）测试时，我是否在第39行使用了错误的嵌套条件语法....

if (isset($_POST['submitted'])):
        $origin = urlencode($_POST['origin']);
        $destination = urlencode($_POST['destination']);

        // Insert encoded url variables
        $url = "http://maps.googleapis.com/maps/api/distancematrix/json?origins=$origin&destinations=$destination&mode=driving&keyy={API KEY}";
        $json = file_get_contents($url); // get the data from Google Maps API

        $status = $result['rows'][0]['elements'][0]['status'];

    if (status is OK):
        // Calculate the distance
        $status = ...;
        $DistanceMetres = .....;
        $Distance = .....; // converted to yards

        // Calculate the Day Rate
        ....

        // Calculate the Night Rate
        ....

        // Calculate the Sunday Rate
        ....

        // Calculate the Christmas & NYE Rate
        ....

        // Set up variables for pick-up & drop-off locations
        $toairport = $result['destination_addresses'][0];
        $fromairport= $result['origin_addresses'][0];

        if (distance is the minimum distance) {
        echo the minimum cost of the trip

        } else { //ie if the distance is more than the minimum distance

            // Check to see if pick-up or drop-off destination is an airport
            if (strpos($toairport, 'Heathrow') || strpos($toairport, 'Gatwick') || strpos($toairport, 'London Luton') || strpos($toairport, 'London City Airport') || strpos($fromairport, 'Heathrow') || strpos($fromairport, 'Gatwick') || strpos($fromairport, 'London Luton') || strpos($fromairport, 'London City Airport') === false) {
                echo the cost of the trip

            // But if at least one location is an airport
            } else {
            echo a message saying a special flat rate is available for airport transfers
            }
        }

    else:
    echo that status is not okay
    endif;

else:

display input form

 endif;

Answer 1

我建议将检查作为一个函数来实现。例如

/**
 * Check if the location is an airport.
 *
 * @param string $location
 * @return bool
 */
function isAirport($location)
{
    $airportList = ['Heathrow', 'Gatwick', 'London Luton', 'London City Airport'];
    foreach ($airportList as $airport) {
        if (strpos($location, $airport) !== false) {
            return true;
        }
    }

    return false;
}

请注意，我们使用strpos运算符将false结果与!==进行比较，因为strpos可能会返回零==到false。< / p>

接下来，内部检查可能如下所示

if (isAirport($toairport) || isAirport($fromairport)) {
    echo 'a message saying a special flat rate is available for airport transfers'.PHP_EOL;
} else {
    echo 'the cost of the trip'.PHP_EOL;
}

请注意，我们检查$toairport是$ fromairport`变量是机场位置。我想这就是你希望脚本做的事情。

当我们使用这样的函数时，它非常容易阅读脚本并理解它在做什么。此外，它还可以通过添加新位置或改进逻辑来更轻松地修改脚本。例如，我们可能希望使检查用例不敏感。

嵌套的PHP条件语句结构问题

1 个答案: