Google 映射 API 准确访问未命名键

Question

所以我正在制作一个包含大约 5,000 个地址的电子表格，并将其发送到 Google 地图地理定位 API 以获得 lat/long 以及许多格式化的地址值shorthand这里。

我遇到的问题是他们没有使用 named keys 供我使用，并且位置会根据可用的位置来回移动。例如，这是一个完全正常的地址；

    {
   "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" ]
            }
         ]

现在举例来说，这是一个位于次级房屋内的位置：

{
   "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" ]
            }
         ]

问题是在第一种情况下，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，这使得无法准确构建我的 JSON，因为在某些区域，我的 Zip Code 值将是街道名称，因为存在辅助地址列表。

你会建议我如何解决这个问题，以确保我始终准确地获得我请求的数据，而不是当时恰好在那个位置的任何数据？

下面是一个例子，说明这会把事情搞砸：

"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，可能其中 200-300 个是错误的，而其余的是正确的，但是必须猜测和希望是非常低效的。

为了包括我的一些脚本：

    $builtAddress = "$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筛选：

$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
    ...
}

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