Php 显示返回列表中的单个结果

我正在努力解决如何只显示第一个返回值（电话号码）。目前，该代码返回所有可用的数字，但我只想向客户显示一个

这是密码。我希望有人能帮忙。我是个新手，你看，我试过一些，但都没用

<?php
// Get the PHP helper library from twilio.com/docs/php/install
require_once('Services/Twilio.php'); // Loads the library

// Your Account Sid and Auth Token from twilio.com/user/account
$sid = "AC1f3ccf29618exxxxx"; 
$token = "{{Auth_Token}}"; 
$client = new Services_Twilio($sid, $token);

$numbers = $client->account->available_phone_numbers->getList('GB', 'Local', array(
        "Contains" => "44161"
    ));

foreach($numbers->available_phone_numbers as $number) { 
echo $number->phone_number;
}

?>

我只想显示第一个值，即

+441618503707

感谢您的帮助，我可能不会立即回复，因为这是目前正在进行的众多项目之一。尽管我的其他项目都在Infusionsoft上，但请放心，我不会太频繁地涉猎API/PHP

那么你可以使用这个

echo $numbers->available_phone_numbers[0]->phone_number;

它将包含数组中的第一个

更新

<?php
// Get the PHP helper library from twilio.com/docs/php/install
require_once('Services/Twilio.php'); // Loads the library

// Your Account Sid and Auth Token from twilio.com/user/account
$sid = "AC1f3ccf29618exxxxx"; 
$token = "{{Auth_Token}}"; 
$client = new Services_Twilio($sid, $token);

$numbers = $client->account->available_phone_numbers->getList('GB', 'Local', array(
        "Contains" => "44161"
    ));

echo $numbers->available_phone_numbers[0]->phone_number;

?>

Hi@showstepper，我试过了，它确实多次返回一个电话号码。有没有办法让它只显示一次？不要在foreach dude中使用它。在foreach外面使用它。嗨@showstepper，谢谢你在这方面的帮助，效果非常好！第二步，购买号码，帮助我花钱的穷人哈哈！

<?php
// Get the PHP helper library from twilio.com/docs/php/install
require_once('Services/Twilio.php'); // Loads the library

// Your Account Sid and Auth Token from twilio.com/user/account
$sid = "AC1f3ccf29618exxxxx"; 
$token = "{{Auth_Token}}"; 
$client = new Services_Twilio($sid, $token);

$numbers = $client->account->available_phone_numbers->getList('GB', 'Local', array(
        "Contains" => "44161"
    ));

echo $numbers->available_phone_numbers[0]->phone_number;

?>