Question

我正在尝试从另一个$ Table2更新$ Table1中的多个值。

说我有$ Table1（在这种情况下是导入的CSV文件）：

Model   ModelID   Blah
abc     0         Blah
ghi     0         Blah
mno     0         Blah

，我有$ Table2（在这种情况下，是从数据源获得的）：

name    id
abc     11
def     12
ghi     13
jkl     14
mno     15
pqr     16
etc.

我正在尝试从$ Table2。“ id”更新$ Table1。“ ModelID”中的值

WHERE $Table1."Model" = $Table2."name"

在SQL中，我会做类似的事情：

UPDATE $Table1
SET ModelID = $Table2."id"
WHERE $Table1."Model" = $Table2."name"

如何基于PowerShell中变量中各列的联接进行条件更新？

我在看：

-replace... (I can't seem to do conditional replaces based on joins)

Add-Member -MemberType NoteProperty "modelID" -Value ... (again, I can't seem to set the value based on joins)

foreach($item in $Table1)
{
    $Table1."ModelID" = $Table2."id" 
    where ?????
}.. (again, I can't seem to set the value based on joins)

我在这里超吃布丁吗？

Answer 1

这非常混乱，但似乎可以完成工作。

$Table2 = import-csv C:\temp\test.csv
$Table1 = import-csv C:\temp\Test55.csv

Foreach($item in $Table1){
    Foreach($tab2 in $Table2){
        If($tab2.name -match $item.model){
            $item.ModelID = $tab2.id
        }
    }
}

Answer 2

这是我认为可以提供帮助的代码。

foreach($i in $t1)
{
    foreach($j in $t2)
    {
        if($i.'model' -eq $j.'id')
        {
            $i.'modelid' = $j.'name'
            break
        }
    }
}

对于表1中的每个项目，在表2中查找模式，如果找到匹配项，则更改表1中的值。

Answer 3

使用哈希表从模型中查找ID的 dirty 变体。
使用此处的字符串作为源。

## Q:\Test\2018\11\23\SO_53440594.ps1
$table1 = @"
Model,ModelID,Blah
abc,0,Blah
ghi,0,Blah
mno,0,Blah
"@ | ConvertFrom-Csv

$Hashtable2 = @{}
@"
Name,Id
abc,11
def,12
ghi,13
jkl,14
mno,15
pqr,16
"@ | ConvertFrom-Csv | ForEach-Object {$Hashtable2.Add($_.Name,$_.Id)}

ForEach ($Row in $table1){
  if($Hashtable2.Containskey($Row.Model)){
    $Row.ModelID = $Hashtable2[$Row.Model]
  } else {
    "Model {0} not present in `$table2" -f $Row.Model
  }
}
$table1

示例输出：

Model ModelID Blah
----- ------- ----
abc   11      Blah
ghi   13      Blah
mno   15      Blah

Answer 4

这是另一种方法。它使用了“针对集合操作”技术[我认为是在v4中引入的]。当查找列表很大时，哈希表仍然是最快的方法。 [咧嘴]

$OneTable = @'
Model, ModelID, Blah
abc, 0, Blah
ghi, 0, Blah
mno, 0, Blah
'@ | ConvertFrom-Csv

# removed one line [ghi, 13] to allow for "no match" error test
$TwoTable = @'
Name, ID
abc, 11
def, 12
jkl, 14
mno, 15
pqr, 16
'@ | ConvertFrom-Csv

foreach ($OT_Item in $OneTable)
    {
    $Lookup = $TwoTable -match $OT_Item.Model
    if ($Lookup)
        {
        $OT_Item.ModelID = $Lookup.ID
        }
        else
        {
        Write-Warning ('No matching Model was found for [ {0} ].' -f $OT_Item.Model)
        }
    }

$OneTable

输出...

WARNING: No matching Model was found for [ ghi ].

Model ModelID Blah
----- ------- ----
abc   11      Blah
ghi   0       Blah
mno   15      Blah

Answer 5

这种方式...

$Data = @"
Model,ModelID,Blah
abc,0,Blah
ghi,0,Blah
mno,0,Blah
"@ | ConvertFrom-Csv

$Reference = @"
Name,Id
abc,11
def,12
ghi,13
jkl,14
mno,15
pqr,16
"@ | ConvertFrom-Csv

$Data | ForEach-Object {
    $Local:ThisModelKey = $_.Model
    if ($Reference.Name -contains $ThisModelKey) {
    $_.ModelID = (@($Reference | Where-Object { $_.Name -like $ThisModelKey } ))[0].ID
    }
}

结果是...

$Data

Model ModelID Blah
----- ------- ----
abc   11      Blah
ghi   13      Blah
mno   15      Blah

PowerShell-从另一个变量更新变量值

5 个答案: