在PHP openweatherAPI中解析Json

在php中解析这个，我可以得到id名称、lon和lat，但我不能得到图标。。。 如何获得图标？ 我的代码是：

{
    "message": "accurate",
    "cod": "200",
    "count": 1,
    "list": [
        {
            "id": 2347078,
            "name": "Birim",
            "coord": {
                "lon": 9.997027,
                "lat": 10.062094
            },
            "main": {
                "temp": 307.488,
                "temp_min": 307.488,
                "temp_max": 307.488,
                "pressure": 968.5,
                "sea_level": 1022.03,
                "grnd_level": 968.5,
                "humidity": 61
            },
            "dt": 1402666090,
            "wind": {
                "speed": 1.41,
                "deg": 263.501
            },
            "sys": {
                "country": ""
            },
            "clouds": {
                "all": 0
            },
            "weather": [
                {
                    "id": 800,
                    "main": "Clear",
                    "description": "Sky is     Clear",
                    "icon": "01d"
                }
            ]
        }
    ]
}

---

$lf=$k['icon']$url=“”；好的，我明白了。。似乎我做对了，只是没有正确放置$url。。愚蠢的我：/n你的问题是

+$k['icon']

。

+

符号用于JavaScript连接，您应该使用PHP

而不是使用

进行连接。如果结构总是一样的话。您可以直接访问它

$t['list'][0]['weather'][0]['icon']
$lat = $_REQUEST['lat'];
$lng = $_REQUEST['lng'];
$response = file_get_contents("http://api.openweathermap.org/data/2.5/find?lat=$lat&lon=$lng&cnt=10");

$response = json_decode($response, true);
$t = $response['list'];
foreach ($t as $s) {
    $w = $s['weather']
    foreach ($w as $k) {
        $url="http://openweathermap.org/img/w/"+$k['icon'];
    }
    <img src="<?php echo $url?>" /></p>

  `$url="http://openweathermap.org/img/w/".$k['icon'];`