10

这是json文本:

{
"data": {
    "current_condition": [{
        "cloudcover": "75",
        "humidity": "63",
        "observation_time": "03:41 PM",
        "precipMM": "0.0",
        "pressure": "1020",
        "temp_C": "15",
        "temp_F": "59",
        "visibility": "16",
        "weatherCode": "116",
        "weatherDesc": [{
            "value": "Partly Cloudy"
        }],
        "weatherIconUrl": [{
            "value": "http:\/\/cdn.worldweatheronline.net\/images\/wsymbols01_png_64\/wsymbol_0002_sunny_intervals.png"
        }],
        "winddir16Point": "SSE",
        "winddirDegree": "160",
        "windspeedKmph": "7",
        "windspeedMiles": "4"
    }],
    "request": [{
        "query": "Northville, United States Of America",
        "type": "City"
    }],
    "weather": [{
        "date": "2013-09-24",
        "precipMM": "0.0",
        "tempMaxC": "20",
        "tempMaxF": "67",
        "tempMinC": "8",
        "tempMinF": "47",
        "weatherCode": "113",
        "weatherDesc": [{
            "value": "Sunny"
        }],
        "weatherIconUrl": [{
            "value": "http:\/\/cdn.worldweatheronline.net\/images\/wsymbols01_png_64\/wsymbol_0001_sunny.png"
        }],
        "winddir16Point": "ESE",
        "winddirDegree": "111",
        "winddirection": "ESE",
        "windspeedKmph": "10",
        "windspeedMiles": "6"
    }]
}

}

我正在尝试回显“temp_F”,但它不起作用。我无法弄清楚我做错了什么。我做到了这一点:

$url = file_get_contents("http://blahblahblahblah");
$arr = json_decode($url,true);

这就是一切都失败的地方。我已经完成了 var_dump,所以我知道数据在那里。但是我尝试过的每次“回声”尝试只会导致“阵列”显示在屏幕上。我尝试了以下许多变体:

echo $arr->{'data'}->{'current_condition[0]'}->{'temp_F'};

有人可以告诉我我做错了什么吗?谢谢!

4

4 回答 4

19

json_decode()withTRUE作为第二个参数给你一个关联数组。但是您目前正试图将其作为对象进行访问。

尝试以下操作:

echo $arr['data']['current_condition'][0]['temp_F'];
于 2013-09-24T16:04:52.240 回答
1

这不是您在 PHP 中访问数组的方式

$array['index']="value";

echo $array['index1']['index2']

对于您的示例:

echo $arr['data']['current_condition'][0]['temp_F']
于 2013-09-24T16:04:11.013 回答
0

您可以使用json将json结果检索到一个变量,然后使用变量信息在js中显示。

$.ajax({
    'type': 'GET',
    'url': 'abc.com,
    'dataType': 'json',
    success: function (data) {

    var response = data;
    // alert(response.data.current_condition)  //something like that
    // for (var i = 0; i < response.length; i++) { }

    }
于 2013-09-24T16:10:54.983 回答
0

您可以使用如下的 foreach 循环条件:

$url = file_get_contents("http://blahblahblahblah");
$arr = json_decode($url,true);
foreach($arr as $data)  {
   echo $data['current_condition'][0]['temp_F'];
   echo $data['request'][0]['query'];
   echo $data['weather'][0]['date'];
}
于 2021-12-23T12:55:37.547 回答