Question

我有3个csv文件，它们只有1列长。我已经尝试了很多东西把它全部放在一个csv文件中，但我无法让它工作。当我输出它时，这一切最终都在一列中，这是我到目前为止所做的

#Putting Csv files to Array    
$CSV1 = @(gc $EmailPathCsv)
$CSV2 = @(gc $UserLPathCsv)
$CSV3 = @(gc $EmailPathCsv)

#
for ($i=0; $i -lt $CSV1.Count; $i++) 
    {
    $CSV4 += $CSV1[$i] + "," + $CSV2[$i] + "," + $CSV3[$i] + " "
    }

$csv4 | out-file -append $MergedCsvExport

Answer 1

你的循环正在将所有内容添加到$ CSV4中，每次循环$ CSV4越来越长。

然后打印一次。这就是为什么你得到一个很长的路线。每次循环尝试打印一次，每次都覆盖$ CSV4：

#Putting Csv files to Array    
$CSV1 = @(gc $EmailPathCsv)
$CSV2 = @(gc $UserLPathCsv)
$CSV3 = @(gc $EmailPathCsv)

#
for ($i=0; $i -lt $CSV1.Count; $i++) 
    {
    $CSV4 = $CSV1[$i] + "," + $CSV2[$i] + "," + $CSV3[$i] + " "
    out-file -InputObject $csv4 -Append -FilePath $MergedCsvExport
    }

Answer 2

使用格式字符串。

$CSV1 = @(gc $EmailPathCsv)
$CSV2 = @(gc $UserLPathCsv)
$CSV3 = @(gc $EmailPathCsv)

for ($i=0; $i -lt $CSV1.Count; $i++) 
 {
  '"{0}","{1}","{2}"' -f $CSV1[$i],$CSV2[$i],$CSV3[$i] |
   add-content $MergedCsvExport
 }

Answer 3

作为一个更有趣的答案：

$CSV1 = 1,2,3,4
$CSV2 = 10,11,12,13
$CSV3 = 'a','b','c','d'

$c = gv -Name CSV* | select -expand name | sort

(gv -Va $c[0])|%{$r=0}{@($c|%{(gv -Va $_)[$r]}) -join ',';$r++}

示例输出：

1, 10, a
2, 11, b
3, 12, c
4, 13, d

您可以将|Out-File "merged-data.csv"放在最后以保存到文件中。

它也适用于更多列，只需添加更多名为$CSV{something}的数组。

编辑：我想知道Get-Variable的输出是否是可预测的顺序，还是未指定？如果您不介意列顺序可能会更改，则会折叠为：

$CSV1 = 1,2,3,4
$CSV2 = 10,11,12,13
$CSV3 = 'a','b','c','d'

(gv CSV*)[0].Value|%{$r=0}{@((gv CSV*)|%{(gv -Va $_.Name)[$r]}) -join ',';$r++}

再次编辑：好吧，如果有人注意到并且好奇，并且有时间在他们手上，我已经将它全部扩展到解释它正在做什么：

# Search the local variable scope for variables named CSV*
# This will find $CSV1, $CSV2, etc.
# This means the number of columns
# isn't fixed, you can easily add more.
# Take their names, sort them. 
# Result: an array of strings "CSV1", "CSV2", ...
# for however many variables there are
$columnVariables = Get-Variable -Name "CSV*" | Select-Object -Expand Name | Sort-Object

# NB. if you remove $CSV3= from your code, it is
# still in memory from previous run. To remove
# it, use `Remove-Variable "CSV3"

# In pseudo-code, what the next part does is
# for row in range(data):
#    @(CSV1[row], CSV2[row], ... CSVn[row]) -join ','

# The outer loop over the number of columns
# is done by piping something of the right length
# into a foreach loop, but ignoring the content.
# So get the first column array content:
$firstColumn = (Get-Variable $columnVariables[0]).Value

# and pipe it into a loop.
# The loop uses ForEach {setup} {loop} pattern 
# to setup a row-counter before the loop body
$firstColumn | ForEach { $row=0 } {

    # Now we know the row we are on, we can get
    # $CSV1[row], $CSV2[row], ...
    # Take the column variable array "CSV1", "CSV2", ..
    # Loop over them
    $rowContent = $columnVariables | ForEach {

        # $_ a string of the name, e.g. "CSV1"
        # Use Get-Variable to convert it
        # into the variable $CSV1
        # with -ValueOnly to get the array itself
        # rather than details about the variable
        $columnVar = Get-Variable -ValueOnly $_

        # $columVar is now one of the $CSVn variables
        # so it contains, e.g. 1,2,3,4
        # Index into that for the current row
        # to get one item, e.g. 3
        # Output it into the pipeline
        ($columnVar)[$row]

        } # This is the end of the inner loop
          # The pipeline now contains the column
          # values/content making up a single row
          # 1
          # 10
          # 'a'

    # Back in the outer row loop, Take the row
    # content and make it a comma separated string
    # e.g. "1,10,a"
    # Output this into the pipeline
    @($rowContent) -join ','

    # Increment the row counter
    $row++
}

使用export-csv时遇到问题

3 个答案: