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The data to checked listbox is not correct

Posted on 2007-11-14
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Last Modified: 2010-04-23
       CheckedListBox1.DataSource = Me.BindingSource1

        CheckedListBox1.DisplayMember = Me.DataSet1.ITEM.ITM_TYPEColumn.ToString()


When starting the program all i get from the checked listbox is system.data.DataViewManagerListItemTypeDescriptor

Im connecting to a oracle database, and i dont understand why i dont get the Item types in the checked listbox when i start program :/

Please help
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Question by:leeds2000
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Expert Comment

by:Chumad
ID: 20282720
You need to also define the fields where the data comes from...

CheckedListBox1.DataTextField = "someField";
CheckedListBox1.DataValueField = "someotherField";
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Author Comment

by:leeds2000
ID: 20282738
uhm, im sorry, i dont quite understand that one, what should i use in the "someField" ?
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Author Comment

by:leeds2000
ID: 20282749
also, datatextfield is not a member of the checklistbox1 i get a error saying in visual studio, so i cant use the datatextfield or the datavaluefield
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Expert Comment

by:Chumad
ID: 20282765
Well, for DataTextField, that's what the user sees - the value in the list box. This needs to point to some field or property in DataSet1. Here, I'm going to assume you need to use "ITM_TYPEColumn"

CheckedListBox1.DataTextField = "ITM_TYPEColumn";

Likewise, DataValueField is going to point to the property/field in DataSet1 that contains the value - usually a primary key. You can probably use the same value here if you want "ITM_TYPEColumn"
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Expert Comment

by:Chumad
ID: 20282772
Try using DisplayMember instead:

CheckedListBox1.DisplayMember = "ITM_TYPEColumn";
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Author Comment

by:leeds2000
ID: 20282803
ok thanks, but i cant use checkedlistbox1.datatextfield, becouse of this error:

'datatextfield' is not a member of system.windows.forms.checkedlistbox

this error is the same for the checkedlistbox1.datavaluefield

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Expert Comment

by:Chumad
ID: 20282824
ok, I wasn't aware you were doing windows forms (vs web forms). Try this way:

CheckedListBox1.DisplayMember = "ITM_TYPEColumn"
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Author Comment

by:leeds2000
ID: 20282861
oki, well i tried it and i got the same result, is this code supposed to be instead of the .displaymember i already have in my code, or should i add it under my code?

I tried removing my version of the display member, and adding your version, but i got same result. i tried with just yours but i still got same, anything im doing wrong?
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Expert Comment

by:Chumad
ID: 20282891
Can you paste all your modified code in here (the version that has the suggested changes)?
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Author Comment

by:leeds2000
ID: 20283089
CheckedListBox1.DataSource = Me.BindingSource1

        CheckedListBox1.DisplayMember = Me.DataSet1.ITEM.ITM_TYPEColumn.ToString()

then i add your part of the code

CheckedListBox1.DisplayMember = "ITM_TYPEColumn"

ive also tried removing the CheckedListBox1.DisplayMember = Me.DataSet1.ITEM.ITM_TYPEColumn.ToString()

without any luck
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Accepted Solution

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leeds2000 earned 0 total points
ID: 20337175
The error was that i hadnt added a tableadapter to my form :/
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Author Comment

by:leeds2000
ID: 20365436
And my code worked as soon as i added this tableadapter, so always check if you have a tableadapter in visual studio 2005 if you cant get any data as soon as your form loads :)
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by:Vee_Mod
ID: 20391004
Closed, 500 points refunded.
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