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how to show the image within the grid.

edited September 19

I have figured out to show the json and the image using util.horizontalrun. I would prefer to show the image where it shows the url in the datagrid (screenshotfilename). how do i do it?

`void Main()
{
string screenshotobject = @{ 'eventName': 'EMA20Pullback', 'eventTime': '2020-11-16T02:29:34', 'ExecutionId': null, 'bartime': '0001-01-01T00:00:00', 'Instrument': 'EURUSD', 'priceAtEvent': 0.0, 'id': 0, 'screenshotFileName': 'https://i.imgur.com/4ZMy9o7.png'}";

ScreenshotObject picture = JsonConvert.DeserializeObject(screenshotobject);
Util.HorizontalRun(true, picture, Util.Image(picture.screenshotFileName)).Dump();
// picture.Dump();
// Util.Image ( picture.screenshotFileName).Dump("from URI");

}

// You can define other methods, fields, classes and namespaces here

public class ScreenshotObject
{
public string eventName;
public DateTime eventTime;
public string ExecutionId;

public DateTime bartime;
public string Instrument;
public double priceAtEvent;
public int id;
public string screenshotFileName;
public ScreenshotObject(DateTime dt, String eventname, string instrument)
{
    this.eventName = eventname;
    this.eventTime = dt;
    this.Instrument = instrument;

}

}

`

Comments

  • Add the following property to your ScreenshotObject class:

    public object Screenshot => Util.Image (screenshotFileName);
    

    If you update to the latest beta, you can also specify a size in which to display the image:

    public object Screenshot => Util.Image (screenshotFileName, Util.ScaleMode.ResizeTo (300));
    
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