Question

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"
    }]
}

}

我正在尝试回应'temp_F'并且它无效。我无法弄清楚我做错了什么。我到目前为止：

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

这就是失败的地方。我已经完成了var_dump，所以我知道数据存在。但是我尝试过的每次“回声”尝试都会导致“数组”显示在屏幕上。我尝试过以下几种变体：

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

有人可以告诉我我做错了什么吗？谢谢！

Answer 1

以TRUE作为第二个参数的

json_decode()为您提供了一个关联数组。但是您目前正试图将其作为对象访问。

尝试以下方法：

echo $arr['data']['current_condition'][0]['temp_F'];

Answer 2

这不是你在PHP中访问数组的方式

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

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

对于你的例子：

echo $arr['data']['current_condition'][0]['temp_F']

Answer 3

您可以使用json将json结果检索到变量，然后使用变量信息以js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++) { }

    }

php和嵌套json：我如何访问这个元素？

3 个答案: