Question

我是PHP的新手，我有一个非常基本的问题，但无法找到答案。这是一个JSONObject示例：

{
    "biome":"forest",
    "id":"51134535488",
    "animals":[
        {"species":"bear", "age":"8", "gender":"male", "family":"mamal"},
        {"species":"hawk", "age":"3", "gender":"female", "family":"bird"},
        {"species":"elk", "age":"5", "gender":"male", "family":"mamal"},
        {"species":"spider", "age":"0.3", "gender":"female", "family":"insect"}
    ]
}

其中有一个JSONArray，其中包含四个JSONObject。我将如何从JSONObject中仅获取JSONArray，然后进行foreach循环以获取所有内部行？我正在Lumen框架中工作，因此对Lumen的具体回答会非常感激（如果可以在Lumen中以其他方式完成）！

Answer 1

例如，如果您要显示四种：

$json = '{
    "biome":"forest",
    "id":"51134535488",
    "animals":[
        {"species":"bear", "age":"8", "gender":"male", "family":"mamal"},
        {"species":"hawk", "age":"3", "gender":"female", "family":"bird"},
        {"species":"elk", "age":"5", "gender":"male", "family":"mamal"},
        {"species":"spider", "age":"0.3", "gender":"female", "family":"insect"}
    ]
}';

foreach(json_decode($json)->animals as $animal) {
    echo $animal->species . "\n";
}

Answer 2

您可以在此处使用一些有用的行：

/* Set the json file directory */
$path = 'your path here';
/* here your json file name */
$jsonfile = 'youjsonfilename';
/* json decode */
$language = json_decode(file_get_contents($path . $jsonfile. '.json'));

然后，如果您的json文件如下所示：

{
    "test": {
        "test1"                 : "test2",
     }
}

您必须在php中编写此行以打印“ test2”，例如：

<?php echo $language->test->test1; ?>

Answer 3

如果您是从JSON字符串开始的，则意味着您的示例是PHP中的字符串变量，您可以执行以下操作：

$jsonString = '{"biome" : "forest", ...}';
$forest = json_decode($jsonString, true); // Passing true as a second argument converts the JSON string into an array, instead of a PHP object

$animals = $forest['animals']; // If you're sure animals is always an array, you can do the following for loop without any problems
foreach ($animals as $animal) {
    var_dump($animal);
}

如何从json对象检索json数组行？

3 个答案: