将json解码为php变量 [英] decode json into php variables
本文介绍了将json解码为php变量的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
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:
$request = json_decode(file_get_contents("https://api.sunrise-sunset.org/json?lat=51.507351&lng=-0.127758&date=today"), true);
JSON调用的输出如下所示:
The output of the JSON call looks like this:
array(2) {
["results"]=> array(10) {
["sunrise"]=> string(10) "4:21:35 AM"
["sunset"]=> string(10) "7:32:34 PM"
["solar_noon"]=> string(11) "11:57:04 AM"
["day_length"]=> string(8) "15:10:59"
["civil_twilight_begin"]=> string(10) "3:42:08 AM"
["civil_twilight_end"]=> string(10) "8:12:01 PM"
["nautical_twilight_begin"]=> string(10) "2:49:55 AM"
["nautical_twilight_end"]=> string(10) "9:04:14 PM"
["astronomical_twilight_begin"]=> string(10) "1:41:12 AM"
["astronomical_twilight_end"]=> string(11) "10:12:57 PM"
}
["status"]=> string(2) "OK"
}
如何将日出"时间和日落"时间转换为php变量$ 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("https://api.sunrise-sunset.org/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
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