Question

我一直试图解析一些JSON来获取一段数据。我是JSON的新手，但我想我正在从字典中解析数据，这个数据本身是从字典中选择的。

我尝试使用以下PHP从下面的示例解析度量标准温度：

$temp_c = $parsed_json->{'hourly_forecast'}[0]->{'FCTTIME'}->{'temp'}->{'metric'};

我使用2作为定期API响应的第3个条目，但对于此摘录，它在技术上是0。

（长）样本是这样的：

{
  "response": {
  "version":"0.1",
  "termsofService":"http://www.wunderground.com/weather/api/d/terms.html",
  "features": {
  "hourly": 1
  }
    }
        ,
    "hourly_forecast": [
        {
        "FCTTIME": {
        "hour": "23","hour_padded": "23","min": "00","min_unpadded": "0","sec": "0","year": "2018","mon": "2","mon_padded": "02","mon_abbrev": "Feb","mday": "27","mday_padded": "27","yday": "57","isdst": "0","epoch": "1519772400","pretty": "11:00 PM GMT on February 27, 2018","civil": "11:00 PM","month_name": "February","month_name_abbrev": "Feb","weekday_name": "Tuesday","weekday_name_night": "Tuesday Night","weekday_name_abbrev": "Tue","weekday_name_unlang": "Tuesday","weekday_name_night_unlang": "Tuesday Night","ampm": "PM","tz": "","age": "","UTCDATE": ""
        },
        "temp": {"english": "34", "metric": "1"},
        "dewpoint": {"english": "21", "metric": "-6"},
        "condition": "Mostly Cloudy",
        "icon": "mostlycloudy",
        "icon_url":"http://icons.wxug.com/i/c/k/nt_mostlycloudy.gif",
        "fctcode": "3",
        "sky": "76",
        "wspd": {"english": "15", "metric": "24"},
        "wdir": {"dir": "ENE", "degrees": "75"},
        "wx": "Mostly Cloudy",
        "uvi": "0",
        "humidity": "60",
        "windchill": {"english": "24", "metric": "-4"},
        "heatindex": {"english": "-9999", "metric": "-9999"},
        "feelslike": {"english": "24", "metric": "-4"},
        "qpf": {"english": "0.0", "metric": "0"},
        "snow": {"english": "0.0", "metric": "0"},
        "pop": "21",
        "mslp": {"english": "30.21", "metric": "1023"}
        }
        ,
        {
        "FCTTIME": {

等

对于任何含糊不清或错误的道歉，我很乐意澄清任何事情。提前谢谢。

Answer 1

看起来FCTTIME部分是您的错误。 temp是FCTTIME的兄弟姐妹，而不是其子女。

此外，您不需要对这些键使用复杂（大括号）语法。对于那些无效的属性名称，例如以数字开头的字符串等，这才真正需要。

这应该有效：

$temp_c = $parsed_json->hourly_forecast[0]->temp->metric;

如何使用PHP从以下JSON解析JSON数据点

1 个答案: