Question

我有以下代码，只想访问以下信息并打印Status

"Email": { "Status": "in-queue" }

我对PHP很陌生，所以我为任何错误道歉，为一堂课学习。

{
    "APIResponse": {
        "ResponseStatus": 1,
        "Email": {
            "EmailSid": "12893712893789",
            "SentEmails": "my@email.com",
            "Date": "2017-02-07 22:53:26",
            "Subject": "700message",
            "Status": "in-queue",
            "TotalEmailSent": 1,
            "TotalPrize": "0.0100",
            "ApiVersion": "2"
        }
    }
}

我知道如何在JavaScript中执行此操作，但对PHP来说似乎有点不同。

感谢所有人的帮助！：）

Answer 1

这是一个快速的PHP示例，介绍如何将JSON字符串转换为数组并显示值：

$json = '{ "APIResponse": { "ResponseStatus": 1, "Email": { "EmailSid": "12893712893789", "SentEmails": "my@email.com", "Date": "2017-02-07 22:53:26", "Subject": "700message", "Status": "in-queue", "TotalEmailSent": 1, "TotalPrize": "0.0100", "ApiVersion": "2" } } }'; $array = json_decode($json, true); echo $array['APIResponse']['Email']['Status'];

注意每个子节点如何成为关联数组中的子节点。您还可以执行print_r($array);以本机PHP关联数组格式查看整个结构。

Answer 2

<?php
 $json = '{
    "APIResponse": {
        "ResponseStatus": 1,
        "Email": {
            "EmailSid": "12893712893789",
            "SentEmails": "my@email.com",
            "Date": "2017-02-07 22:53:26",
            "Subject": "700message",
            "Status": "in-queue",
            "TotalEmailSent": 1,
            "TotalPrize": "0.0100",
            "ApiVersion": "2"
        }
    }
}';


// if this API response's JSON is being stored as a string type local variable decode it first.

$decoded_json = json_decode($json, true);

//Optional but helps when programing var_dump($decoded_json); Use this to help visualize the nested array variables you need to access.

// Use each array key to walk into and access the desired value

echo $decoded_json['APIResponse']['Email']['Status'];


 ?>

*根据评论中的问题编辑以解决错误：

虽然仍然确定如上所述首先解码Json，但下面应该允许您访问这些值。

echo $decoded_json['APIResponse']['Errors']['Error'][0]['Code'];

echo $decoded_json['APIResponse']['Errors']['Error'][0]['Message'];

最后，如果你遇到麻烦，还有更多建议。尝试从最外层开始一次访问一个阵列并继续工作，总是帮助我！

PHP访问JSON响应和打印

2 个答案: