从渥太华获取预测的查询不起作用:
SELECT * FROM weather.forecast WHERE woeid IN (SELECT woeid FROM geo.places WHERE text="Ottawa, Canada" limit 1)
结果:
"query": {
"count": 1,
"created": "2015-07-02T13:06:45Z",
"lang": "en-US",
"results": {
"channel": {
"title": "Yahoo! Weather - Error",
"description": "Yahoo! Weather Error",
"item": {
"title": "City not found",
"description": "\nInvalid Input /forecastrss?w=91982014\n"
}
}
}
}
自己执行,子查询得到正确的结果:
SELECT woeid FROM geo.places WHERE text="Ottawa, Canada" limit 1
结果:
"query": {
"count": 1,
"created": "2015-07-02T13:00:47Z",
"lang": "en-US",
"results": {
"place": {
"woeid": "91982014"
}
}
}
这适用于我尝试的所有其他城市,当我忽略“限制1”时,它也适用于渥太华,但返回的结果太多。我做错了还是API有问题?
答案 0 :(得分:0)
我认为问题在于,对于渥太华,有多个结果(多个WOEID),并且并非所有这些WOEID在weather.forecast表中都有相应的条目。我认为 是因为其中一些条目的级别不是我们预测的水平。
查询SELECT woeid FROM geo.places WHERE text="Ottawa, Canada"
会产生以下结果:
"results": {
"place": [
{
"woeid": "91982014"
},
{
"woeid": "29375164"
},
{
"woeid": "12483153"
}
]
}
如果您修改查询以包含地点类型SELECT woeid, placeTypeName.content FROM geo.places WHERE text="Ottawa, Canada"
,则可以看到结果包含城镇,县/区和岛屿,而城镇是结果中的第一项:
"results": {
"place": [
{
"woeid": "91982014",
"placeTypeName": "Town"
},
{
"woeid": "29375164",
"placeTypeName": "County/District"
},
{
"woeid": "12483153",
"placeTypeName": "Island"
}
]
}
在返回过多结果SELECT * FROM weather.forecast WHERE woeid IN (SELECT woeid FROM geo.places WHERE text="Ottawa, Canada")
的完整查询的结果中,您可以看到名为diagnostics
的部分,其中包含一个名为url
的数组,该数组提供的天气API网址因查询而调用:
http://weather.yahooapis.com/forecastrss?w=12483153
http://weather.yahooapis.com/forecastrss?w=29375164
http://weather.yahooapis.com/forecastrss?w=91982014
并且91982014的结果是错误:
您可以从结果SELECT * FROM weather.forecast WHERE woeid IN (SELECT woeid FROM geo.places WHERE text="Ottawa, Canada") AND title NOT LIKE "%Error%"
中删除错误案例,但仍会提供两个结果。
看起来你真正想要做的就是意识到加拿大有三个地方"渥太华"名称:安大略省的首府渥太华,安大略省的渥太华县/区,以及努勒维特境内的渥太华群岛。因此,您需要更具体地查询您的查询。
SELECT * FROM weather.forecast WHERE woeid IN (SELECT woeid FROM geo.places WHERE text="Ottawa, Ontario, Canada") AND title NOT LIKE "%Error%"
感觉很好,但它只能产生一个结果。