About WPF brush problem

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How to draw a rectangular gradient fill,WPF always round a path gradient, how to become a rectangular path gradient, We can use GDI + to fill out PathGradientBrush brush effects, WPF how to do it?

Posted 2-Jun-13 20:25pm

Berton2012

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Sergey Alexandrovich Kryukov · Accepted Answer · 2013-06-02T20:29:00

Solution 1

No, you can't use GDI+. WPF has nothing do to with it, it is based on DirectX. And no, it does not "always round", this is your fantasy.
I have no idea what did you try (and it's not nice of you to not tell us about it), but you should have used this:
http://msdn.microsoft.com/en-us/library/system.windows.media.lineargradientbrush%28v=vs.110%29.aspx[^].

—SA

Posted 2-Jun-13 20:29pm

Sergey Alexandrovich Kryukov

Comments

Berton2012 3-Jun-13 3:02am

Thank you very much for your help, but I don't think you understand what I want, I want to effect the following links, hope you can see and can help me again, thank you!

http://bbs.csdn.net/topics/390475347

Sergey Alexandrovich Kryukov 3-Jun-13 3:14am

This picture shows exactly what I explained, linear gradient, isn't this obvious? Is this what you want to show, or something you don't?!
If this is what you want to show, please use linear gradient brush and accept my answer formally (green button). If not, explain properly.

If I didn't understand it, this is because you did not specify things accurately.

If you cannot see how to implement the picture you referenced with linear gradient, your problem is vision and logic. I hope this is not the case.

—SA

Berton2012 3-Jun-13 5:00am

Hello friends, can continue to help me? Thank you!

Berton2012 3-Jun-13 4:03am

Thank you very much for your reply, I wanted to explain it to you I want results, please check the link again:
http://bbs.csdn.net/topics/390475347?page=1#post-394678835

Those pictures are done with GDI+.

Sergey Alexandrovich Kryukov 3-Jun-13 9:32am

Yes, yes, yes, linear gradient. Can't you see it?
—SA

Berton2012 3-Jun-13 22:40pm

Thank you, I'll try, can you give me an example?

Sergey Alexandrovich Kryukov 3-Jun-13 22:54pm

Example of what? Gradient brush? Here:

http://msdn.microsoft.com/en-us/library/system.windows.media.lineargradientbrush.aspx

OK, your picture is explained simply.

First, center box: this is 4 triangles of painted with 4 identical gradients oriented in 4 different directions, bright point in the center, black on outside.

Now, two outsides rings, each is done out of 2 rings: 8 trapezoidal regions, each painted with the same gradient with 4 different orientation, adjacent fragments have opposite directions of gradient.

Each gradient may have 2-3 intermediate points, how to say exactly, you can experiment with picture just a bit. Also, I think there are 4 diagonal highlights and 2 shades on each of half-diagonals. They are thin, so from the first glance, it's hard to say how are they rendered, probably some thing areas with gradient in 4 diagonal directions.

All gradients are linear. The full code would be a good couple of pages, would not belong here.

Can you see it now?

—SA

Berton2012 4-Jun-13 1:37am

I think it should be RadialGradientBrush instead of LinearGradientBrush, LinearGradientBrush is not possible to do it. Thank you.

Sergey Alexandrovich Kryukov 4-Jun-13 2:35am

Look, this is not a place to convince you. I answered in detail, but you are not giving any arguments, just repeating same things. Is it a big deal to try it out? If you think you know how to do it, do it.
—SA

Berton2012 4-Jun-13 4:55am

OK!To look at me GDI+ do with the code, I hope you can understand my meaning.How to do with WPF?
Code:(The use of WinForm)

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;
using System.Drawing.Drawing2D;

namespace DDPractice
{
public partial class BrushFrm : Form
{
public BrushFrm()
{
InitializeComponent();
}

public Brush bs;

private void BrushFrm_Paint(object sender, PaintEventArgs e)
{
Graphics g = e.Graphics;

GraphicsPath path1 = new GraphicsPath();
GraphicsPath path2 = new GraphicsPath();
RectangleF rect1 = new RectangleF(50f, 50f, 120f, 120f);
RectangleF rect2 = new RectangleF(250f, 50f, 120f, 120f);

path1.AddRectangle(rect1); //Path using rectangle filling
path2.AddEllipse(rect2); //Path using Circular filling
PathGradientBrush brush1 = new PathGradientBrush(path1);
PathGradientBrush brush2 = new PathGradientBrush(path2);

Color[] colors = new Color[] { Color.Black, Color.White, Color.Black, Color.White, Color.Black, Color.White };
float[] pos = new float[] { 0.0f, 0.2f, 0.4f, 0.6f, 0.8f, 1.0f };

ColorBlend blend = new ColorBlend();
blend.Positions = pos;
blend.Colors = colors;
brush1.InterpolationColors = blend;
brush2.InterpolationColors = blend;

g.SmoothingMode = SmoothingMode.AntiAlias;
g.PixelOffsetMode = PixelOffsetMode.HighQuality;
g.FillRectangle(brush1, rect1);
g.FillPath(brush1, path1);

g.FillRectangle(brush2, rect2);
g.FillPath(brush2, path2);

brush1.Dispose();
brush2.Dispose();
g.Dispose();
}
}
}

Sergey Alexandrovich Kryukov 4-Jun-13 9:00am

I agree, so what? I'm not saying that my description is the only way to draw the similar picture.
You told me that you want it in WPF. But it does not have PathGradientBrush. But look at your own words. You were talking about RadialGradientBrush, which hardly can be used.

So, I described how to do it in linear gradient. It's really quite possible; the image is based on squares.

Look at these discussions:
http://social.msdn.microsoft.com/Forums/en-US/wpf/thread/209aa8c3-055a-4c05-923a-bab934535399
https://groups.google.com/forum/?fromgroups#!topic/wpf-disciples/E1i30Epc1A4

See also this CodeProject article:
http://www.codeproject.com/Articles/20018/Gradients-made-easy

Just some food for thought.

—SA

Berton2012 4-Jun-13 22:47pm

Thank you very much for your sincere help, it seems that this is a complex issue.

Blessing

Sergey Alexandrovich Kryukov 4-Jun-13 23:03pm

Well, not too complex; I guess not you can do it.
You are very welcome.
Good luck, call again.
—SA