dsfdsfsdf45489
2016-10-05 08:40
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将JSON响应转换为PHP变量

I have a script (from GitHub) that uses the Google Calendar API to display upcoming events.

<script src="https://code.jquery.com/jquery-1.11.3.min.js"></script>
<script src="format-google-calendar.js"></script>
<script>
  formatGoogleCalendar.init({
    calendarUrl: 'https://www.googleapis.com/calendar/v3/calendars/7aqribjvrco83j69ufindoig3g@group.calendar.google.com/events?key=AIzaSyDls1AFdCYUTxYRWpYNUiLwOarKao3WV_c',
    past: true,
    upcoming: true,
    sameDayTimes: true,
    pastTopN: 20,
    upcomingTopN: 3,
    itemsTagName: 'li',
    upcomingSelector: '#events-upcoming',
    pastSelector: '#events-past',
    upcomingHeading: '<h2>Upcoming events</h2>',
    pastHeading: '<h2>Past events</h2>',
    format: ['*date*', ': ', '*summary*', ' &mdash; ', '*description*', ' in ', '*location*']
  });

What I want to do is store different parts of the data in a PHP variable per event, for instance:

<?php 
   $date= JSON Date
   <--! For next event !-->
   $date2= JSON Date2
   $date3= JSON Date3
   ?>

I only need 3 as it will only be displaying 3 upcoming events.

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1条回答 默认 最新

  • dongzhi2014 2016-10-05 08:44
    已采纳
        <?php
        $date= file_get_contents('url_for_date');
        $dateobj = json_decode($date);
        $date2= file_get_contents('url_for_date2');
        $date2obj = json_decode($date2);
        $date3= file_get_contents('url_for_date3');
        $date3obj = json_decode($date3);
        ?>
    
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