Google Maps API 准确访问未命名的键

Question

因此，我使用包含大约 5,000 个地址的电子表格并将其从 Google Maps Geolocation API 发送出去，以获取经纬度以及格式化的地址值，因为这里很多都是简写形式。

我遇到的问题是他们没有使用命名keys 供我使用，并且根据可用的位置移动位置。例如，这是一个完全常用的地址；

    {
   "results" : [
      {
         "address_components" : [
            {
               "long_name" : "27502",
               "short_name" : "27502",
               "types" : [ "street_number" ]
            },
            {
               "long_name" : "Antonio Parkway",
               "short_name" : "Antonio Pkwy",
               "types" : [ "route" ]
            },
            {
               "long_name" : "Ladera Ranch",
               "short_name" : "Ladera Ranch",
               "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" ]
            },
            {
               "long_name" : "92694",
               "short_name" : "92694",
               "types" : [ "postal_code" ]
            }
         ]

例如，这里是一个子前提内的位置：

{
   "results" : [
      {
         "address_components" : [
            {
               "long_name" : "117",
               "short_name" : "117",
               "types" : [ "subpremise" ]
            },
            {
               "long_name" : "3401",
               "short_name" : "3401",
               "types" : [ "street_number" ]
            },
            {
               "long_name" : "North Miami Avenue",
               "short_name" : "N Miami Ave",
               "types" : [ "route" ]
            },
            {
               "long_name" : "Wynwood",
               "short_name" : "Wynwood",
               "types" : [ "neighborhood", "political" ]
            },
            {
               "long_name" : "Miami",
               "short_name" : "Miami",
               "types" : [ "locality", "political" ]
            },
            {
               "long_name" : "Miami-Dade County",
               "short_name" : "Miami-Dade County",
               "types" : [ "administrative_area_level_2", "political" ]
            },
            {
               "long_name" : "Florida",
               "short_name" : "FL",
               "types" : [ "administrative_area_level_1", "political" ]
            },
            {
               "long_name" : "United States",
               "short_name" : "US",
               "types" : [ "country", "political" ]
            },
            {
               "long_name" : "33127",
               "short_name" : "33127",
               "types" : [ "postal_code" ]
            }
         ]

问题在于，在第一种情况下，address_components[0].short_name 将是我想要的 street_number，但在示例 2 中，address_components[0].short_name 实际上是批号。

以脚本方式，这导致无法准确构建我的 JSON，因为在某些区域我的 Zip Code 值将是街道名称，因为存在二级地址列表。

您会建议我如何应对这种情况，以确保我始终获得我所请求的数据，而不是当时恰好处于该位置的任何数据？

下面是一个例子，说明这会如何搞砸：

"731":  {
            "City":  "Noblesville",
            "Street":  "Norell Ln",
            "Type":  "dsg",
            "StreetNum":  "13157",
            "Zip":  "US",
            "State":  "Hamilton County",
            "Name":  "Noblesville ",
            "Region":  "Ohio Valley",
            "Long":  "-85.93149269999999",
            "StateAbr":  "Hamilton County",
            "Phone":  "3177761687",
            "Lat":  "39.9901088"
        }

情况并非总是如此，其中可能有 200-300 个是错误的，而其余的都是正确的，但必须猜测和希望是非常低效的。

为了包含我的一些脚本：

    $builtAddress = "$street $city $state $zip"

    Write-Verbose "Built Address: $builtAddress"

    $addressArray.Add($builtAddress) > $null

    $req = Invoke-WebRequest "https://maps.googleapis.com/maps/api/geocode/json?address=$builtAddress&key=AIzaSyDRo-UGY91_EiB2DeYzBU21-3FcaqIanPo" 
    $location = $req.Content | ConvertFrom-Json


    $lat = $location.results[0].geometry.location.lat
    $long = $location.results[0].geometry.location.lng
    $z = $location.results[0].address_components

    $name = $z.
    $phone = $phone
    $region = $region

    $objectProps = @{

    $store = @{
        Name = "$name"
        Lat = "$lat"
        Long = "$long"
        StreetNum = "$streetNum"
        Street = "$street"
        City = "$city"
        State = "$state"
        StateAbr = "$stateAb"
        Zip = "$zip"
        Phone = "$phone"
        Region = "$region"
        Type = "$type"

    }
}

Write-Verbose $store
$jsonObj = New-Object psobject -Property $objectProps | ConvertTo-Json -depth 100 | Out-File C:\Users\admin-dksc104694\Desktop\Map_Data\JSON\$outputFile -Append
Write-Output ',' | Out-File C:\Users\admin-dksc104694\Desktop\$outputFile -Append

    }

Answer 1

您可以根据类型进行过滤：

$StreetNumber = $z.Where({$_.types -contains 'street_number'})

然后在您的 $store 哈希表中，您可以使用点符号来提取值

$store = @{
    ...
    StreetNum = $StreetNumber.short_name
    ...
}