将json解码为php变量

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我正在尝试将JSON结果的内容转换为php变量.我正在使用的代码是:

I am trying to get the contents of a JSON result into php variables. The code I am using is:

$request = json_decode(file_get_contents("api.sunrise-sunset/json?lat=51.507351&lng=-0.127758&date=today"), true);

JSON调用的输出如下所示:

The output of the JSON call looks like this:

array(2) { ["results"]=> array(10) { ["sunrise"]=> string(10) "4:21:35 AM" ["sunset"]=> string(10) "7:32:34 PM" ["solar_noon"]=> string(11) "11:57:04 AM" ["day_length"]=> string(8) "15:10:59" ["civil_twilight_begin"]=> string(10) "3:42:08 AM" ["civil_twilight_end"]=> string(10) "8:12:01 PM" ["nautical_twilight_begin"]=> string(10) "2:49:55 AM" ["nautical_twilight_end"]=> string(10) "9:04:14 PM" ["astronomical_twilight_begin"]=> string(10) "1:41:12 AM" ["astronomical_twilight_end"]=> string(11) "10:12:57 PM" } ["status"]=> string(2) "OK" }

如何将日出"时间和日落"时间转换为php变量$ SunRiseTime和$ SunSetTime；

How can I get the "sunrise" time and the "sunset" times into php variables $SunRiseTime and $SunSetTime;

非常感谢您抽出宝贵的时间.

Many thanks in advance for your time.

推荐答案

简单而直接.

<?php ini_set('display_errors', 1); $json = file_get_contents("api.sunrise-sunset/json?lat=51.507351&lng=-0.127758&date=today"); $array=json_decode($json,true); echo $SunRiseTime=$array["results"]["sunrise"]; echo $SunSetTime=$array["results"]["sunset"];

输出:4:21:35 AM 7:32:34 PM