PHP - 通过输入字段搜索 PHP 数组？答案

【问题标题】：PHP - Search through PHP array through input field?PHP - 通过输入字段搜索 PHP 数组？
【发布时间】：2018-03-24 08:46:24
【问题描述】：

<?php
$allBookings = [
    "IMD002" => [
        "client" => "James Holden",
        "paid"   => "yes",
        "email"  => "james.holden@gmail.com"
    ],
    "IMD003" => [
        "client" => "Harold P. Redman",
        "paid"   => "yes",
        "email"  => "HaroldPRedman@dayrep.com"
    ],
    "IMD004" => [
        "client" => "Marcus C. Nelson",
        "paid"   => "no",
        "email"  => "MarcusCNelson@armyspy.com"
    ]
];


// DONT EDIT ANYTHING BELOW HERE
if(!empty($_POST)) {
    $clientName = $booking['client'];
}
else {
    $clientName = null;
};
?>

<!-- DONT EDIT ANYTHING BELOW HERE -->
<div class="myApp">
    <form method="post">
        <input type="text" name="bookingReference" placeholder="Booking Reference">
        <input type="submit" value="Find Client">
    </form>
    
    <h2 id="result_name">Client name: <em><?php echo $clientName; ?></em></h2>
    
</div>

当我在输入字段（预订参考）中输入例如“IMD002”时，如何显示凭据？因为它应该显示在“IMD002”下的 php 中列出的数组。我假设它与 $_POST 声明有关？

非常感谢

【问题讨论】：

标签： php arrays post arraylist get

【解决方案1】：

$allBookings = [ ... ]之后

尝试添加类似的内容：

if(isset($_POST['bookingReference']) &&
   isset($allBookings[$_POST['bookingReference']])) {
    $booking = $allBookings[$_POST['bookingReference']];
}
else {
    $booking = ['client' => 'Invalid Ref/Not found'];
};

【讨论】：

【解决方案2】：

要搜索，请使用 $_POST['bookingReference'] 作为索引，$allBookings，您的数据数组；

试试：

$allBookings[$_POST['bookingReference']];

打印内容：

print_r( $allBookings[$_POST['bookingReference']] );

【讨论】：