在 PHP 中使用 OpenWeatherMap 预报 API

Question

我正在尝试显示来自 openweathermap 的城市预报。但是我的 foreach 什么也没显示。怎么了？

<?php
  $url = "http://api.openweathermap.org/data/2.5/forecast?zip=85080,de&lang=de&APPID=MYKEY";

  $contents = file_get_contents($url);
  $clima = json_decode($contents, true);

  foreach($clima as $data) {
    echo $data->list->main->temp_min;
  }
?>

Answer 1

json_decode(string, true) 的结果是一个关联数组。

<?php

  $url = "http://api.openweathermap.org/data/2.5/forecast?zip=85080,de&lang=de&APPID=MYKEY";

  $contents = file_get_contents($url);
  $clima = json_decode($contents, true);

  foreach($clima['list'] as $data) {
    echo $data['main']['temp_min'];
  }

?>

如果要使用对象语法，请不要将 associative 设置为 true。

$clima = json_decode($contents);

foreach($clima->list as $data) {
  echo $data->main->temp_min;
}

Answer 2

您使用了 json_decode 参数 associative 为真。

所以 $data 更像是一个数组而不是对象。

基于 sample (similar to your url), you should access your values with square bracket syntax :

$data['main']['temp_min'];

Answer 3

让我们试试...

<?php
    $city    = 'Dhaka';
    $country = 'BD';
    $url     = 'http://api.openweathermap.org/data/2.5/forecast/daily?q=' . $city . ',' . $country . '&units=metric&cnt=7&lang=en&appid=c0c4a4b4047b97ebc5948ac9c48c0559';
    $json    = file_get_contents( $url );
    $data    = json_decode( $json, true );
    $data['city']['name'];
    // var_dump($data );
    
    foreach ( $data['list'] as $day => $value ) {
        echo 'Max temperature for day ' . $day
        . ' will be ' . $value['temp']['max'] . '<br />';
        echo '<img src="http://openweathermap.org/img/w/' . $value['weather'][0]['icon'] . '.png"
                    class="weather-icon" />';
    
    }

在 PHP 中使用 OpenWeatherMap 预报 API

Using OpenWeatherMap forecast API in PHP

php

api

json

openweathermap