How to bind TextBlock.Text in DataGridCell template using ItemsSource










3















I want to have DataGridCell with text and image.

Currently, my code looks like that



XAML:



<DataGrid Name="myDataGrid" CellStyle="StaticResource myCellStyle" />


Style:



<Style x:Key="myCellStyle" TargetType="DataGridCell">
<Setter Property="Template">
<Setter.Value>
<ControlTemplate>
<StackPanel Orientation="Horizontal">
<TextBlock Text="Binding"/>
<Image/>
</StackPanel>
</ControlTemplate>
</Setter.Value>
</Setter>
</Style>


C#:



myDataGrid.ItemsSource = myDataTable.DefaultView;


The question is:

How to bind text to a TextBlock using ItemsSource?










share|improve this question
























  • Who own's ItemsSource? Where does it reside?

    – ΩmegaMan
    Nov 13 '18 at 21:37











  • My mistake. I forgot about DataGrid name property which is set in my code. Please take a look on my example now.

    – Wojtman
    Nov 13 '18 at 21:43











  • Have you tried <TextBlock Text="Binding ColumnName"/> with ColumnName being the text column from your DataTable that you want to display?

    – Simon Evans
    Nov 13 '18 at 22:48












  • It doesn't work. I didn't mention about that, but I'm generating DataTable dynamically and my DataGrid almost always have different columns and rows. That's why i can't set style for specific column.

    – Wojtman
    Nov 13 '18 at 22:57















3















I want to have DataGridCell with text and image.

Currently, my code looks like that



XAML:



<DataGrid Name="myDataGrid" CellStyle="StaticResource myCellStyle" />


Style:



<Style x:Key="myCellStyle" TargetType="DataGridCell">
<Setter Property="Template">
<Setter.Value>
<ControlTemplate>
<StackPanel Orientation="Horizontal">
<TextBlock Text="Binding"/>
<Image/>
</StackPanel>
</ControlTemplate>
</Setter.Value>
</Setter>
</Style>


C#:



myDataGrid.ItemsSource = myDataTable.DefaultView;


The question is:

How to bind text to a TextBlock using ItemsSource?










share|improve this question
























  • Who own's ItemsSource? Where does it reside?

    – ΩmegaMan
    Nov 13 '18 at 21:37











  • My mistake. I forgot about DataGrid name property which is set in my code. Please take a look on my example now.

    – Wojtman
    Nov 13 '18 at 21:43











  • Have you tried <TextBlock Text="Binding ColumnName"/> with ColumnName being the text column from your DataTable that you want to display?

    – Simon Evans
    Nov 13 '18 at 22:48












  • It doesn't work. I didn't mention about that, but I'm generating DataTable dynamically and my DataGrid almost always have different columns and rows. That's why i can't set style for specific column.

    – Wojtman
    Nov 13 '18 at 22:57













3












3








3








I want to have DataGridCell with text and image.

Currently, my code looks like that



XAML:



<DataGrid Name="myDataGrid" CellStyle="StaticResource myCellStyle" />


Style:



<Style x:Key="myCellStyle" TargetType="DataGridCell">
<Setter Property="Template">
<Setter.Value>
<ControlTemplate>
<StackPanel Orientation="Horizontal">
<TextBlock Text="Binding"/>
<Image/>
</StackPanel>
</ControlTemplate>
</Setter.Value>
</Setter>
</Style>


C#:



myDataGrid.ItemsSource = myDataTable.DefaultView;


The question is:

How to bind text to a TextBlock using ItemsSource?










share|improve this question
















I want to have DataGridCell with text and image.

Currently, my code looks like that



XAML:



<DataGrid Name="myDataGrid" CellStyle="StaticResource myCellStyle" />


Style:



<Style x:Key="myCellStyle" TargetType="DataGridCell">
<Setter Property="Template">
<Setter.Value>
<ControlTemplate>
<StackPanel Orientation="Horizontal">
<TextBlock Text="Binding"/>
<Image/>
</StackPanel>
</ControlTemplate>
</Setter.Value>
</Setter>
</Style>


C#:



myDataGrid.ItemsSource = myDataTable.DefaultView;


The question is:

How to bind text to a TextBlock using ItemsSource?







c# wpf xaml datagrid






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 13 '18 at 21:43







Wojtman

















asked Nov 13 '18 at 21:21









WojtmanWojtman

768




768












  • Who own's ItemsSource? Where does it reside?

    – ΩmegaMan
    Nov 13 '18 at 21:37











  • My mistake. I forgot about DataGrid name property which is set in my code. Please take a look on my example now.

    – Wojtman
    Nov 13 '18 at 21:43











  • Have you tried <TextBlock Text="Binding ColumnName"/> with ColumnName being the text column from your DataTable that you want to display?

    – Simon Evans
    Nov 13 '18 at 22:48












  • It doesn't work. I didn't mention about that, but I'm generating DataTable dynamically and my DataGrid almost always have different columns and rows. That's why i can't set style for specific column.

    – Wojtman
    Nov 13 '18 at 22:57

















  • Who own's ItemsSource? Where does it reside?

    – ΩmegaMan
    Nov 13 '18 at 21:37











  • My mistake. I forgot about DataGrid name property which is set in my code. Please take a look on my example now.

    – Wojtman
    Nov 13 '18 at 21:43











  • Have you tried <TextBlock Text="Binding ColumnName"/> with ColumnName being the text column from your DataTable that you want to display?

    – Simon Evans
    Nov 13 '18 at 22:48












  • It doesn't work. I didn't mention about that, but I'm generating DataTable dynamically and my DataGrid almost always have different columns and rows. That's why i can't set style for specific column.

    – Wojtman
    Nov 13 '18 at 22:57
















Who own's ItemsSource? Where does it reside?

– ΩmegaMan
Nov 13 '18 at 21:37





Who own's ItemsSource? Where does it reside?

– ΩmegaMan
Nov 13 '18 at 21:37













My mistake. I forgot about DataGrid name property which is set in my code. Please take a look on my example now.

– Wojtman
Nov 13 '18 at 21:43





My mistake. I forgot about DataGrid name property which is set in my code. Please take a look on my example now.

– Wojtman
Nov 13 '18 at 21:43













Have you tried <TextBlock Text="Binding ColumnName"/> with ColumnName being the text column from your DataTable that you want to display?

– Simon Evans
Nov 13 '18 at 22:48






Have you tried <TextBlock Text="Binding ColumnName"/> with ColumnName being the text column from your DataTable that you want to display?

– Simon Evans
Nov 13 '18 at 22:48














It doesn't work. I didn't mention about that, but I'm generating DataTable dynamically and my DataGrid almost always have different columns and rows. That's why i can't set style for specific column.

– Wojtman
Nov 13 '18 at 22:57





It doesn't work. I didn't mention about that, but I'm generating DataTable dynamically and my DataGrid almost always have different columns and rows. That's why i can't set style for specific column.

– Wojtman
Nov 13 '18 at 22:57












2 Answers
2






active

oldest

votes


















2














You must do couple of things to fix it



First, set 'AutoGenerateColumns' to true



<DataGrid CellStyle="StaticResource myCellStyle" AutoGenerateColumns="True">


Next in your cell style



<Style x:Key="myCellStyle" TargetType="DataGridCell">
<Setter Property="Template">
<Setter.Value>
<ControlTemplate TargetType="x:Type DataGridCell">
<StackPanel Orientation="Horizontal">
<TextBlock Text="Binding RelativeSource=RelativeSource TemplatedParent,
Path=Content.Text"/>
<Image/>
</StackPanel>
</ControlTemplate>
</Setter.Value>
</Setter>
</Style>


Hope this helps.






share|improve this answer


















  • 1





    That is exactly what I needed. Thank you very much!

    – Wojtman
    Nov 14 '18 at 5:32


















0














This assumes that myDataTable.DefaultView is a list of some objects(class instances).



So when the grid gets its ItemsSource set, it will then will present each row with one of the items from the list. So the binding to be specified will be a property/properties on the class.



So taking from your example template, if it is bound to a hypothetical list of person classes with FirstName and LastName on the class instance, one could set the template to bind to use those properties for each row like this:



 <StackPanel Orientation="Horizontal">
<TextBlock Text="Binding FirstName"/>
<TextBlock Text="Binding LastName"/>
<Image/>
</StackPanel>





share|improve this answer























  • Unfortunatelly, myDataTable is made from strings (not custom objects with properties) and that's the problem. I have no idea how bind single string to my TextBlock.

    – Wojtman
    Nov 13 '18 at 23:02











  • Create a Data Transfer Object class (DTO) with the appropriate properties and take the datatable data and new up a list of the DTOs from it. Program the DTO class's constructor to get specific columns from the datatable and load target properties from that. Then bind as shown but have the grid use the new List<DTOClass>.

    – ΩmegaMan
    Nov 13 '18 at 23:03












  • I made as u said and now if style is implemented then grid do not show data but if style is not implemented then every cell have value like "myNamespace.myClass". So maybe there is something wrong with this property in my style <Setter Property="Template"> ?

    – Wojtman
    Nov 13 '18 at 23:33











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














You must do couple of things to fix it



First, set 'AutoGenerateColumns' to true



<DataGrid CellStyle="StaticResource myCellStyle" AutoGenerateColumns="True">


Next in your cell style



<Style x:Key="myCellStyle" TargetType="DataGridCell">
<Setter Property="Template">
<Setter.Value>
<ControlTemplate TargetType="x:Type DataGridCell">
<StackPanel Orientation="Horizontal">
<TextBlock Text="Binding RelativeSource=RelativeSource TemplatedParent,
Path=Content.Text"/>
<Image/>
</StackPanel>
</ControlTemplate>
</Setter.Value>
</Setter>
</Style>


Hope this helps.






share|improve this answer


















  • 1





    That is exactly what I needed. Thank you very much!

    – Wojtman
    Nov 14 '18 at 5:32















2














You must do couple of things to fix it



First, set 'AutoGenerateColumns' to true



<DataGrid CellStyle="StaticResource myCellStyle" AutoGenerateColumns="True">


Next in your cell style



<Style x:Key="myCellStyle" TargetType="DataGridCell">
<Setter Property="Template">
<Setter.Value>
<ControlTemplate TargetType="x:Type DataGridCell">
<StackPanel Orientation="Horizontal">
<TextBlock Text="Binding RelativeSource=RelativeSource TemplatedParent,
Path=Content.Text"/>
<Image/>
</StackPanel>
</ControlTemplate>
</Setter.Value>
</Setter>
</Style>


Hope this helps.






share|improve this answer


















  • 1





    That is exactly what I needed. Thank you very much!

    – Wojtman
    Nov 14 '18 at 5:32













2












2








2







You must do couple of things to fix it



First, set 'AutoGenerateColumns' to true



<DataGrid CellStyle="StaticResource myCellStyle" AutoGenerateColumns="True">


Next in your cell style



<Style x:Key="myCellStyle" TargetType="DataGridCell">
<Setter Property="Template">
<Setter.Value>
<ControlTemplate TargetType="x:Type DataGridCell">
<StackPanel Orientation="Horizontal">
<TextBlock Text="Binding RelativeSource=RelativeSource TemplatedParent,
Path=Content.Text"/>
<Image/>
</StackPanel>
</ControlTemplate>
</Setter.Value>
</Setter>
</Style>


Hope this helps.






share|improve this answer













You must do couple of things to fix it



First, set 'AutoGenerateColumns' to true



<DataGrid CellStyle="StaticResource myCellStyle" AutoGenerateColumns="True">


Next in your cell style



<Style x:Key="myCellStyle" TargetType="DataGridCell">
<Setter Property="Template">
<Setter.Value>
<ControlTemplate TargetType="x:Type DataGridCell">
<StackPanel Orientation="Horizontal">
<TextBlock Text="Binding RelativeSource=RelativeSource TemplatedParent,
Path=Content.Text"/>
<Image/>
</StackPanel>
</ControlTemplate>
</Setter.Value>
</Setter>
</Style>


Hope this helps.







share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 14 '18 at 0:43









RajNRajN

3,66521021




3,66521021







  • 1





    That is exactly what I needed. Thank you very much!

    – Wojtman
    Nov 14 '18 at 5:32












  • 1





    That is exactly what I needed. Thank you very much!

    – Wojtman
    Nov 14 '18 at 5:32







1




1





That is exactly what I needed. Thank you very much!

– Wojtman
Nov 14 '18 at 5:32





That is exactly what I needed. Thank you very much!

– Wojtman
Nov 14 '18 at 5:32













0














This assumes that myDataTable.DefaultView is a list of some objects(class instances).



So when the grid gets its ItemsSource set, it will then will present each row with one of the items from the list. So the binding to be specified will be a property/properties on the class.



So taking from your example template, if it is bound to a hypothetical list of person classes with FirstName and LastName on the class instance, one could set the template to bind to use those properties for each row like this:



 <StackPanel Orientation="Horizontal">
<TextBlock Text="Binding FirstName"/>
<TextBlock Text="Binding LastName"/>
<Image/>
</StackPanel>





share|improve this answer























  • Unfortunatelly, myDataTable is made from strings (not custom objects with properties) and that's the problem. I have no idea how bind single string to my TextBlock.

    – Wojtman
    Nov 13 '18 at 23:02











  • Create a Data Transfer Object class (DTO) with the appropriate properties and take the datatable data and new up a list of the DTOs from it. Program the DTO class's constructor to get specific columns from the datatable and load target properties from that. Then bind as shown but have the grid use the new List<DTOClass>.

    – ΩmegaMan
    Nov 13 '18 at 23:03












  • I made as u said and now if style is implemented then grid do not show data but if style is not implemented then every cell have value like "myNamespace.myClass". So maybe there is something wrong with this property in my style <Setter Property="Template"> ?

    – Wojtman
    Nov 13 '18 at 23:33
















0














This assumes that myDataTable.DefaultView is a list of some objects(class instances).



So when the grid gets its ItemsSource set, it will then will present each row with one of the items from the list. So the binding to be specified will be a property/properties on the class.



So taking from your example template, if it is bound to a hypothetical list of person classes with FirstName and LastName on the class instance, one could set the template to bind to use those properties for each row like this:



 <StackPanel Orientation="Horizontal">
<TextBlock Text="Binding FirstName"/>
<TextBlock Text="Binding LastName"/>
<Image/>
</StackPanel>





share|improve this answer























  • Unfortunatelly, myDataTable is made from strings (not custom objects with properties) and that's the problem. I have no idea how bind single string to my TextBlock.

    – Wojtman
    Nov 13 '18 at 23:02











  • Create a Data Transfer Object class (DTO) with the appropriate properties and take the datatable data and new up a list of the DTOs from it. Program the DTO class's constructor to get specific columns from the datatable and load target properties from that. Then bind as shown but have the grid use the new List<DTOClass>.

    – ΩmegaMan
    Nov 13 '18 at 23:03












  • I made as u said and now if style is implemented then grid do not show data but if style is not implemented then every cell have value like "myNamespace.myClass". So maybe there is something wrong with this property in my style <Setter Property="Template"> ?

    – Wojtman
    Nov 13 '18 at 23:33














0












0








0







This assumes that myDataTable.DefaultView is a list of some objects(class instances).



So when the grid gets its ItemsSource set, it will then will present each row with one of the items from the list. So the binding to be specified will be a property/properties on the class.



So taking from your example template, if it is bound to a hypothetical list of person classes with FirstName and LastName on the class instance, one could set the template to bind to use those properties for each row like this:



 <StackPanel Orientation="Horizontal">
<TextBlock Text="Binding FirstName"/>
<TextBlock Text="Binding LastName"/>
<Image/>
</StackPanel>





share|improve this answer













This assumes that myDataTable.DefaultView is a list of some objects(class instances).



So when the grid gets its ItemsSource set, it will then will present each row with one of the items from the list. So the binding to be specified will be a property/properties on the class.



So taking from your example template, if it is bound to a hypothetical list of person classes with FirstName and LastName on the class instance, one could set the template to bind to use those properties for each row like this:



 <StackPanel Orientation="Horizontal">
<TextBlock Text="Binding FirstName"/>
<TextBlock Text="Binding LastName"/>
<Image/>
</StackPanel>






share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 13 '18 at 22:58









ΩmegaManΩmegaMan

16.1k54260




16.1k54260












  • Unfortunatelly, myDataTable is made from strings (not custom objects with properties) and that's the problem. I have no idea how bind single string to my TextBlock.

    – Wojtman
    Nov 13 '18 at 23:02











  • Create a Data Transfer Object class (DTO) with the appropriate properties and take the datatable data and new up a list of the DTOs from it. Program the DTO class's constructor to get specific columns from the datatable and load target properties from that. Then bind as shown but have the grid use the new List<DTOClass>.

    – ΩmegaMan
    Nov 13 '18 at 23:03












  • I made as u said and now if style is implemented then grid do not show data but if style is not implemented then every cell have value like "myNamespace.myClass". So maybe there is something wrong with this property in my style <Setter Property="Template"> ?

    – Wojtman
    Nov 13 '18 at 23:33


















  • Unfortunatelly, myDataTable is made from strings (not custom objects with properties) and that's the problem. I have no idea how bind single string to my TextBlock.

    – Wojtman
    Nov 13 '18 at 23:02











  • Create a Data Transfer Object class (DTO) with the appropriate properties and take the datatable data and new up a list of the DTOs from it. Program the DTO class's constructor to get specific columns from the datatable and load target properties from that. Then bind as shown but have the grid use the new List<DTOClass>.

    – ΩmegaMan
    Nov 13 '18 at 23:03












  • I made as u said and now if style is implemented then grid do not show data but if style is not implemented then every cell have value like "myNamespace.myClass". So maybe there is something wrong with this property in my style <Setter Property="Template"> ?

    – Wojtman
    Nov 13 '18 at 23:33

















Unfortunatelly, myDataTable is made from strings (not custom objects with properties) and that's the problem. I have no idea how bind single string to my TextBlock.

– Wojtman
Nov 13 '18 at 23:02





Unfortunatelly, myDataTable is made from strings (not custom objects with properties) and that's the problem. I have no idea how bind single string to my TextBlock.

– Wojtman
Nov 13 '18 at 23:02













Create a Data Transfer Object class (DTO) with the appropriate properties and take the datatable data and new up a list of the DTOs from it. Program the DTO class's constructor to get specific columns from the datatable and load target properties from that. Then bind as shown but have the grid use the new List<DTOClass>.

– ΩmegaMan
Nov 13 '18 at 23:03






Create a Data Transfer Object class (DTO) with the appropriate properties and take the datatable data and new up a list of the DTOs from it. Program the DTO class's constructor to get specific columns from the datatable and load target properties from that. Then bind as shown but have the grid use the new List<DTOClass>.

– ΩmegaMan
Nov 13 '18 at 23:03














I made as u said and now if style is implemented then grid do not show data but if style is not implemented then every cell have value like "myNamespace.myClass". So maybe there is something wrong with this property in my style <Setter Property="Template"> ?

– Wojtman
Nov 13 '18 at 23:33






I made as u said and now if style is implemented then grid do not show data but if style is not implemented then every cell have value like "myNamespace.myClass". So maybe there is something wrong with this property in my style <Setter Property="Template"> ?

– Wojtman
Nov 13 '18 at 23:33


















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