I have a csv file with a few columns. For example
col1;col2;col3;col4
text1;text2;text3;text4
text5;text6;text7;text8
text9;text10;text11;text12
I want to add text to column col3, like this:
col1;col2;col3;col4
text1;text2;append\text3;text4
text5;text6;append\text7;text8
text9;text10;append\text11;text12
So question is:
How to do this?
(I'm stuck whit that how I add data to each column in col3.)
Give this a whirl:
Import-Csv -Delim ';' cols.csv |
ForEach-Object {$_.col3 = "prepend\$($_.col3)";$_} |
Export-Csv cols2.csv -Delim ';' -NoTypeInformation
Use the -NoTypeInformation parameter to avoid this comment getting put at the top of your CSV:
#TYPE System.Management.Automation.PSCustomObject
However, if you don't mind the comment then you can leave off the -NoTypeInformation parameter.
You probably know how to read/write data. To change the data, send them to Foreach-Object, alter the data and pass the object further to Export-Csv.
Import-Csv d:\temp\so\csv1.txt -Delimiter ';' |
ForEach-Object { $_.col3 = 'append\' +$_.col3; $_ } |
Export-Csv d:\temp\so\csv2.txt -Delimiter ';'
Related
I have a csv file in path and i need to fetch the file and do Find & Replace $(dollar) value to ,(comma) using PowerShell script. please help me on the script .
CSV file
id|name|place
1|adam$|USA
2|john|USA
3|Jack$|England
Expected output : Need to do Find & Replace $(dollar) value to ,(comma) under name column
id|name|place
1|adam,|USA
2|john|USA
3|Jack,|England
Command Used
Import-csv D:\CSV\Customer.csv
ForEach-Object { $_.name -replace '$',',' }
i didn't got any error nor the expected changes. i need to update the values in existing CSV file
Below Answer by Zett42 works but
(Import-Csv D:\CSV\Customer.csv -Delimiter '|') |
Select-Object 'id',
#{ n = 'name'; e = { $_.name.Replace('$', ',') } },
'place' |
Export-Csv D:\CSV\Customer.csv -Delimiter '|'
but the tricky part is we are replacing the ,(Comma) value and rewriting the CSV files again, so CSV file creation part considers the value after ,(comma) as new column and move the value to column B instead of Column A.
Before :
After :
This should do the trick:
Import-Csv D:\CSV\Customer.csv -Delimiter '|' |
ForEach-Object {
$_.name = $_.name.Replace('$', ',')
$_ # Implicit output - gets passed to Export-Csv
} |
Export-Csv D:\CSV\Customer_new.csv -Delimiter '|' -NoTypeInformation
In your code, there is a pipe symbol | missing at the end of the Import-Csv line.
As the CSV uses non-standard delimiters, you have to specify them using -Delimiter.
Also, the -replace operator doesn't modify its arguments, so you are actually just outputting the replaced value. Also, '$' has special meaning in regular expression patterns, which -replace uses, so you have to backslash-escape it or use the String.Replace() method as I did in the code above. The String.Replace() method doesn't use RegEx.
In the ForEach-Object script block, the $_ at the end is necessary to output the current object again, so it gets send down the pipeline to Export-Csv. Otherwise Export-Csv wouldn't have anything to output.
Alternative code using Select-Object:
Import-Csv D:\CSV\Customer.csv -Delimiter '|' |
Select-Object 'id',
#{ n = 'name'; e = { $_.name.Replace('$', ',') } },
'place' |
Export-Csv D:\CSV\Customer_new.csv -Delimiter '|' -NoTypeInformation
To overwrite the existing file instead of creating a new file use the group operator () to collect all data in memory and close the old file before overwriting the file:
(Import-Csv D:\CSV\Customer.csv -Delimiter '|') |
Select-Object 'id',
#{ n = 'name'; e = { $_.name.Replace('$', ',') } },
'place' |
Export-Csv D:\CSV\Customer.csv -Delimiter '|' -NoTypeInformation
IMO this is somewhat cleaner as you don't need to modify the current pipeline object and don't need to explicitly forward it to the next command.
It uses a calculated property to generate the new value for the name column. In this case we don't need to assign the result back to $_.name because calculated properties automatically assign the output from the script block to the property.
I feel like I'm going about this the wrong way, but I've gotten myself half way there... I have a log file called DataSet.log that is formatted as:
First Dataset;24226382;2020-10-01 00:00;Second.Data.Set 1.0;0;Third.DataSet 1.0;2;Fourth.DataSet 1.0;0;Fifth.Dataset 1.0;0
First Dataset;24421469;2020-10-01 01:00;Second.Data.Set 1.0;0;Third.DataSet 1.0;4;Fourth.DataSet 1.0;0;Fifth.Dataset 1.0;0
First Dataset;24667838;2020-10-01 02:00;Second.Data.Set 1.0;0;Third.DataSet 1.0;6;Fourth.DataSet 1.0;0;Fifth.Dataset 1.0;0
First Dataset;24667839;2020-10-01 02:00;Second.Data.Set 1.0;0;Third.DataSet 1.0;1;Fourth.DataSet 1.0;0;Fifth.Dataset 1.0;0
I'm trying to convert this to a new CSV file and get it to display only the third and seventh columns. I've gotten it to display in PowerShell properly with:
Import-Csv .\DataSet.log -Header A,B,C,D,E,F,G,H,I,J,K -Delimiter ';' | Format-Table C,G
I've tried to export it as:
$testPush = Import-Csv .\DataSet.log -Header A,B,C,D,E,F,G,H,I,J,K -Delimiter ';' | Format-Table C,G
$testPush | Out-File .\test.csv
And it does create a new csv file, but it's only displaying the third column with a header of "C G." Also, that header is in A2, where A1 is blank, and A3 is populated with "- -" before the datetime from the third column populates the remaining rows... What the heck am I doing wrong?
Try this:
$testPush = Import-Csv .\DataSet.log -Header A,B,C,D,E,F,G,H,I,J,K -Delimiter ';' | Select-Object C,G
$testPush | Export-CSv -Path .\test.csv -NoTypeInformation
I have a csv file with two columns and multiple rows, which has the information of files with folder location and its corresponding size, like below
"Folder_Path","Size"
"C:\MSSQL\DATA\UsersData\FTP.txt","21345"
"C:\MSSQL\DATA\UsersData\Norman\abc.csv","78956"
"C:\MSSQL\DATA\UsersData\Market_Database\123.bak","1234456"
What i want do is remove the "C:\MSSQL\DATA\" part from every row in the csv and keep the rest of the folder path after starting from UsersData and all other data intact as this info is repetitive. So my csv should like this below.
"Folder_Path","Size"
"UsersData\FTP.txt","21345"
"UsersData\Norman\abc.csv","78956"
"UsersData\Market_Database\123.bak","1234456"
What i am running is as below
Import-Csv ".\abc.csv" |
Select-Object -Property #{n='Folder_Path';e={$_.'Folder_Path'.Split('C:\MSSQL\DATA\*')[0]}}, * |
Export-Csv '.\output.csv' -NTI
Any help is appreciated!
Seems like a job for a simple string replace:
Get-Content "abc.csv" | foreach { $_.replace("C:\MSSQL\DATA\", "") | Set-Content "output.csv"
or:
[System.IO.File]::WriteAllText("output.csv", [System.IO.File]::ReadAllText("abc.csv" ).Replace("C:\MSSQL\DATA\", ""))
This should work:
Import-Csv ".\abc.csv" |
Select-Object -Property #{n='Folder_Path';e={$_.'Folder_Path' -replace '^.*\\(.*\\.*)$', '$1'}}, Size |
Export-Csv '.\output.csv' -NoTypeInformation
I have a CSV file containing two columns:server name with domain and date
servername.domain.domain.com,10/15/2018 6:28
servername1.domain.domain.com,10/13/2018 7:28
I need to remove the fully qualified name so it only has the shortname and I need to keep the second column so it looks as is like below either by sending to a new CSV or somehow removing the domain inplace somehow. Basically I want the second column untouched but I need it to be included when creating a new CSV with the altered column 1.
servername,10/15/2018 6:28
servername1,10/13/2018 7:28
I have this:
Import-Csv "filename.csv" -Header b1,b2 |
% {$_.b1.Split('.')[0]} |
Set-Content "filename1.csv"
This works great, but the problem is the new CSV is missing the 2nd column. I need to send the second column to the new CSV file as well.
Use a calculated property to replace the property you want changed, but leave everything else untouched:
Import-Csv 'input.csv' -Header 'b1', 'b2' |
Select-Object -Property #{n='b1';e={$_.b1.Split('.')[0]}}, * -Exclude b1 |
Export-Csv 'output.csv' -NoType
Note that you only need to use the parameter -Header if your CSV data doesn't already have a header line. Otherwise you should remove the parameter.
If your input file doesn't have headers and you want to create the output file also without headers you can't use Export-Csv, though. Use ConvertTo-Csv to create the CSV text output, then skip over the first line (to remove the headers) and write the rest to the output file with Set-Content.
Import-Csv 'input.csv' -Header 'b1', 'b2' |
Select-Object -Property #{n='b1';e={$_.b1.Split('.')[0]}}, * -Exclude b1 |
ConvertTo-Csv -NoType |
Select-Object -Skip 1 |
Set-Content 'output.csv'
I need to convert a file with this format:
2015.03.27,09:00,1.08764,1.08827,1.08535,1.08747,8941
2015.03.27,10:00,1.08745,1.08893,1.08604,1.08762,7558
to this format
2015.03.27,1.08764,1.08827,1.08535,1.08747,1
2015.03.27,1.08745,1.08893,1.08604,1.08762,1
I started with this code but can't see how to achieve the full transformation:
Import-Csv in.csv -Header Date,Time,O,H,L,C,V | Select-Object Date,O,H,L,C,V | Export-Csv -path out.csv -NoTypeInformation
(Get-Content out.csv) | % {$_ -replace '"', ""} | out-file -FilePath out.csv -Force -Encoding ascii
which outputs
Date,O,H,L,C,V
2015.03.27,1.08745,1.08893,1.08604,1.08762,8941
2015.03.27,1.08763,1.08911,1.08542,1.08901,7558
After that I need to
remove the header (I tried -NoHeader which is not recognized)
replace last column with 1.
How to do that as simply as possible (if possible without looping through each row)
Update : finally I have simplified requirement. I just need to replace last column with constant.
Ok, this could be one massive one-liner... I'm going to do line breaks at the pipes for sanity reasons though.
Import-Csv in.csv -header Date,Time,O,H,L,C,V|select * -ExcludeProperty time|
%{$_.date = [datetime]::ParseExact($_.date,"yyyy.MM.dd",$null).tostring("yyMMdd");$_}|
ConvertTo-Csv -NoTypeInformation|
select -skip 1|
%{$_ -replace '"'}|
Set-Content out.csv -encoding ascii
Basically I import the CSV, exclude the time column, convert the date column to an actual [datetime] object and then convert it back in the desired format. Then I pass the modified object (with the newly formatted date) down the pipe to ConvertTo-CSV, and skip the first line (your headers that you don't want), and then remove the quotes from it, and lastly output to file with Set-Content (faster than Out-File)
Edit: Just saw your update... to do that we'll just change the last column to 1 at the same time we modify the date column by adding $_.v=1;...
%{$_.date = [datetime]::ParseExact($_.date,"yyyy.MM.dd",$null).tostring("yyMMdd");$_.v=1;$_}|
Whole script modified:
Import-Csv in.csv -header Date,Time,O,H,L,C,V|select * -ExcludeProperty time|
%{$_.date = [datetime]::ParseExact($_.date,"yyyy.MM.dd",$null).tostring("yyMMdd");$_.v=1;$_}|
ConvertTo-Csv -NoTypeInformation|
select -skip 1|
%{$_ -replace '"'}|
Set-Content out.csv -encoding ascii
Oh, and this has the added benefit of not having to read the file in, write the file to the drive, read that file in, and then write the file to the drive again.