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Choice options in a choice field column RRS feed

  • Question

  • Dear All,

    I have following script which inserts a new choice column in all the libraries of a site.

    Get-SPWeb -site  http://sp2010:8080/personal/rakhishma    -Limit All | foreach {

    $doclibtype = [Microsoft.SharePoint.SPBaseType]::DocumentLibrary

    $doclibs = $_.GetListsOfType($doclibtype)

    $doclibs.Count

    for ($i=0; $i -lt $doclibs.Count; $i++)
      
    {
             
     $doclib = $doclibs[$i]
              
    $doclib.Title
              
    $fieldtype = [Microsoft.SharePoint.SPFieldType]::Choice
    $doclib.Fields.Add(“Choice Column”,$fieldtype,$false)      
              
    $doclib.Update()
              
    $view = $doclib.Defaultview
              
    $view.viewFields.Add("Choice Column")
              
    $view.Update() 

    }  

    }

    Now, my question is how to insert choices like choice1, choice2 , choice3 into choice column using powershell...?

    Wednesday, July 11, 2012 8:57 AM

Answers

  • Sure Al.....This is the solution for the problem I mentioned :

    Get-SPWeb -site   http://eismosstest001:10101     -Limit All | foreach {

    $doclibtype = [Microsoft.SharePoint.SPBaseType]::DocumentLibrary

    $doclibs = $_.GetListsOfType($doclibtype)

    $doclibs.Count

    for ($i=0; $i -lt $doclibs.Count; $i++)

    {

    $doclib = $doclibs[$i]

    $doclib.Title

    $Choices = New-Object System.Collections.Specialized.StringCollection

    $Choices.Add("choice1")

    $Choices.Add("choice2")

    $Choices.Add("choice3")

    $Choices.Add("choice4")

    $Choices.Add("choice5")

    $Choices.Add("choice6")

    $fieldtype = [Microsoft.SharePoint.SPFieldType]::Choice

    $doclib.Fields.Add(“Choice column”,$fieldtype,$false,$false,$choices) 

    $doclib.Update()

    $view = $doclib.Defaultview

    $view.viewFields.Add("Choice column")

    $view.Update() 

    }  

    }

    • Marked as answer by Yan Li_ Tuesday, July 17, 2012 5:53 AM
    Thursday, July 12, 2012 5:27 AM

All replies

  • got the solution....please ignore this post
    Wednesday, July 11, 2012 2:50 PM
  • got the solution....please ignore this post

    Why don't you share your solution for others that might find it useful?


    Al Dunbar

    Wednesday, July 11, 2012 4:34 PM
  • Sure Al.....This is the solution for the problem I mentioned :

    Get-SPWeb -site   http://eismosstest001:10101     -Limit All | foreach {

    $doclibtype = [Microsoft.SharePoint.SPBaseType]::DocumentLibrary

    $doclibs = $_.GetListsOfType($doclibtype)

    $doclibs.Count

    for ($i=0; $i -lt $doclibs.Count; $i++)

    {

    $doclib = $doclibs[$i]

    $doclib.Title

    $Choices = New-Object System.Collections.Specialized.StringCollection

    $Choices.Add("choice1")

    $Choices.Add("choice2")

    $Choices.Add("choice3")

    $Choices.Add("choice4")

    $Choices.Add("choice5")

    $Choices.Add("choice6")

    $fieldtype = [Microsoft.SharePoint.SPFieldType]::Choice

    $doclib.Fields.Add(“Choice column”,$fieldtype,$false,$false,$choices) 

    $doclib.Update()

    $view = $doclib.Defaultview

    $view.viewFields.Add("Choice column")

    $view.Update() 

    }  

    }

    • Marked as answer by Yan Li_ Tuesday, July 17, 2012 5:53 AM
    Thursday, July 12, 2012 5:27 AM
  • Sure Al.....This is the solution for the problem I mentioned :

    Get-SPWeb -site   http://eismosstest001:10101     -Limit All | foreach {

    $doclibtype = [Microsoft.SharePoint.SPBaseType]::DocumentLibrary

    $doclibs = $_.GetListsOfType($doclibtype)

    $doclibs.Count

    for ($i=0; $i -lt $doclibs.Count; $i++)

    {

    $doclib = $doclibs[$i]

    $doclib.Title

    $Choices = New-Object System.Collections.Specialized.StringCollection

    $Choices.Add("choice1")

    $Choices.Add("choice2")

    $Choices.Add("choice3")

    $Choices.Add("choice4")

    $Choices.Add("choice5")

    $Choices.Add("choice6")

    $fieldtype = [Microsoft.SharePoint.SPFieldType]::Choice

    $doclib.Fields.Add(“Choice column”,$fieldtype,$false,$false,$choices) 

    $doclib.Update()

    $view = $doclib.Defaultview

    $view.viewFields.Add("Choice column")

    $view.Update() 

    }  

    }

    Thursday, July 12, 2012 5:27 AM