我在Google表格中使用JSON來查找Zip並找到該縣。我只想讓這個縣回來。我可以得到它返回每個值,所以ImportJSON函數正在工作。過濾同名的JSON數據
這是我的公式。我嘗試了所有參考的排列,但我不知道如何格式化它。
=ImportJSON(CONCATENATE("https://maps.googleapis.com/maps/api/geocode/json?address="92660), "results/address_components/long_name[3]", "noHeaders")
以下是來自Google地圖地理編碼API的JSON數據。我只想要縣名。在這個例子中,它是「橙縣」。
{
"results" : [
{
"address_components" : [
{
"long_name" : "92660",
"short_name" : "92660",
"types" : [ "postal_code" ]
},
{
"long_name" : "Newport Beach",
"short_name" : "Newport Beach",
"types" : [ "locality", "political" ]
},
{
"long_name" : "Orange County",
"short_name" : "Orange County",
"types" : [ "administrative_area_level_2", "political" ]
},
{
"long_name" : "California",
"short_name" : "CA",
"types" : [ "administrative_area_level_1", "political" ]
},
{
"long_name" : "United States",
"short_name" : "US",
"types" : [ "country", "political" ]
}
],
"formatted_address" : "Newport Beach, CA 92660, USA",
"geometry" : {
"bounds" : {
"northeast" : {
"lat" : 33.671823,
"lng" : -117.841337
},
"southwest" : {
"lat" : 33.6040739,
"lng" : -117.909447
}
},
"location" : {
"lat" : 33.6301328,
"lng" : -117.8721676
},
"location_type" : "APPROXIMATE",
"viewport" : {
"northeast" : {
"lat" : 33.671823,
"lng" : -117.841337
},
"southwest" : {
"lat" : 33.6040739,
"lng" : -117.909447
}
}
},
"place_id" : "ChIJRdSajSne3IAR8T4A2x-wgrE",
"types" : [ "postal_code" ]
}
],
"status" : "OK"
}
謝謝。這工作完美。事實證明,谷歌的地址並非全部相同,但這是另一個問題。 – Thomas