Question

我试图建立一个小天气应用程序，我需要明天的主要天气状况。我已经搜索了StackOverflow几个小时，似乎无法让它工作。

    {"cod":"200","message":0.0026,"city":{"id":3413829,"name":"Reykjavik","coord":           
    {"lon":-21.895411,"lat":64.135483},"country":"IS","population":0,"sys":
    {"population":0}},"cnt":1,"list":[{"dt":1404349200,"temp":
    {"day":281.58,"min":281.18,"max":283.73,"night":283.73,"eve":281.38,"morn":282.15},"pressure":971.54,"humidity":100

但这里是我正在寻找的数组中的部分。

    ,"weather":[{"id":500,"main":"Rain","description":"light rain","icon":"10d"}],"speed":2.52,"deg":308,"clouds":92,"rain":2.5}]}

这里是php：

    $json_string = file_get_contents("http://api.openweathermap.org/data/2.5/forecast/daily?q=Reykjavik&cnt=1&mode=json");
    $jsonData = json_decode($json_string, true);
    $vedur = $jsonData["weather"]["main"];
    echo "Það er ".$vedur . "í Reykjavík";

Answer 1

可能是这样，你写了：

$jsonData["weather"]["main"][;

有一个额外的方括号应该是：

$jsonData["weather"]["main"];

Answer 2

你需要首先遍历列表数组以获取（我猜）所有日子，然后遍历那里的项目以显示每天的天气：

foreach ($jsonData['list'] AS $item) {
    foreach ($item['weather'] AS $weather) {
        var_dump($item['weather']); 
    }
}

或您的示例：

foreach ($jsonData['list'] AS $item) { 
    foreach ($item['weather'] AS $weather) {
        echo "Það er ".$weather['main'] . "í Reykjavík";
    }
}

Answer 3

 $json_string = file_get_contents("http://api.openweathermap.org/data/2.5/forecast/daily?q=Reykjavik&cnt=1&mode=json");
 $jsonData = json_decode($json_string, true);
 $vedur = $jsonData['list'][0]['weather'][0]['main'];
 echo "Það er ".$vedur . "í Reykjavík";

PHP从php中获取JSON数组中的值

3 个答案: