如何按具有空值和多个条件的字段过滤 PHP 中的数组？

Question

我的 API 在 json_decode 之前调用 returns 以下 JSON 输出：

{
"projects": [
        {
            "project_id": 00000001,
            "name": "A title",
            "price": "0.99",
            "country": "US",
            "platform_types": [
                "android_phone",
                "ios_phone",
                "ios_tablet",
                "android_kindle",
                "android_tablet",
                "desktop"
            ],
            "comment": "A text of a comment"
        }
        {
            "project_id": 00000002,
            "name": "Another title",
            "price": "1.03",
            "country": "US",
            "platform_types": [
                "android_phone",
                "ios_phone",
                "ios_tablet",
                "android_kindle",
                "android_tablet",
                "desktop"
            ],
            "comment": "Another text of a comment"
        }
    ]
}

以下代码解析多级 json 并显示整个项目列表：

$json = file_get_contents($url, false, $context);
$result = json_decode($json, true);
foreach($result['projects'] as $project) {
    $project_id = $project['project_id'];
    $name = $project['name'];
    $price = $project['price'];
    $country = $project['country'];
    #no values for the fields, that's why commented
    #$android_phone = $project['platform_types']['android_phone'];
    #$ios_phone = $project['platform_types']['ios_phone'];
    #$ios_tablet = $project['platform_types']['ios_tablet'];
    #$android_kindle = $project['platform_types']['android_kindle'];
    #$android_tablet = $project['platform_types']['android_tablet'];
    #$desktop = $project['platform_types']['desktop'];
    $comment = $project['comment'];
    echo $project_id,'<br>',$name,'<br>',$price,'<br>',$country,'<br>',$comment,'<br>';
}

我得到以下输出：

00000001
A title
0.99
US
A text of a comment

00000002
Another title
1.03
US
Another text of a comment

问题是：

如何列出可用的设备类型（字段只有名称而没有值）？
如何根据特定条件过滤数组（价格必须等于或高于 1.00 美元）？
如何根据没有值的字段过滤元素（仅显示 Android 设备的项目）？

Answer 1

问题 1

要将每个项目的所有可用设备类型作为 json 列出，您应该访问它对应的树


foreach($result['projects'] as $project) {
    $project_device_types = $project['plateform_type'];
    echo '<pre>'.$project_device_types.'<pre>';
}

问题 2

过滤数组使用 [array_filter][1] in


$filtered = array_filter($result['projects'], function($project) {
    if($project['price'] >= 90) {
      return true
    }
});
// use it instead of the $result variable
foreach($filtered as $project) {
    $project_id = $project['project_id'];
    $name = $project['name'];
    $price = $project['price'];
}

问题 3

您可以按照第二个问题的确切模式进行操作，即使用 array_filter ，例如。假设您只需要 android 台设备

array_filter($result['projects'], function($project) {
    if(in_array($project['platform_types'], "Android")) { // check if the plateform type include android by using [in_array][1]
      return true
    }
});

NB: in_array is case sensitive so make sure that the capitalization is correct

如何按具有空值和多个条件的字段过滤 PHP 中的数组？

How to filter an array in PHP by fields with empty values and multiple criteria?

php

arrays

filtering

问题 1

问题 2

问题 3