Degrafa 换色器

发布于 2024-07-11 00:22:22 字数 685 浏览 12 评论 0原文

<degrafa:LinearGradientFill id="bluedream">
    <degrafa:GradientStop color="#6ab5d0"/>
    <degrafa:GradientStop color="#388aae"/>
</degrafa:LinearGradientFill>

<degrafa:GeometryComposition graphicsTarget="{[bgCanvas]}">
    <degrafa:RoundedRectangle id="color_preset" fill="{bluedream}"/>
</degrafa:GeometryComposition>

我对 degrafa 代码有问题，我为 color_preset.fill 定义了一组不同的渐变，当用户在组合框中选择不同的颜色时，它会动态变化。

我将 fill="{bluedream}" 替换为 fill="using_variable" 并导致编译错误消息：Initializer for 'fill': values of type com. degrafa.core.IGraphicsFill 无法用文本表示。

有没有办法使用此代码作为换色器？

原文

<degrafa:LinearGradientFill id="bluedream">
    <degrafa:GradientStop color="#6ab5d0"/>
    <degrafa:GradientStop color="#388aae"/>
</degrafa:LinearGradientFill>

<degrafa:GeometryComposition graphicsTarget="{[bgCanvas]}">
    <degrafa:RoundedRectangle id="color_preset" fill="{bluedream}"/>
</degrafa:GeometryComposition>

I have issue with degrafa code which I have defined a set of different gradients for color_preset.fill to be dynamic change when user select different color in the combobox.

I replaced fill="{bluedream}" with fill="using_variable" and lead to error compiled message: Initializer for 'fill': values of type com.degrafa.core.IGraphicsFill cannot be represented in text.

Is there a solution to use this code as a color changer?

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み零 2024-07-18 00:22:22

示例中的 fill 属性指向 ID 为“bluedream”的 LinearGradientFill。您可以用不同的渐变（或实心或其他填充）替换填充，也可以更改渐变填充本身的颜色：

<degrafa:LinearGradientFill id="bluedream">
    <degrafa:GradientStop color="{your_combobox.selectedItem}"/>
    <degrafa:GradientStop color="{your_other_combobox.selectedItem}"/>
</degrafa:LinearGradientFill>

<degrafa:GeometryComposition graphicsTarget="{[bgCanvas]}">
     <degrafa:RoundedRectangle id="color_preset" fill="{bluedream}"/>
</degrafa:GeometryComposition>

我还没有运行该代码，但它应该可以工作。这个想法是用下拉列表的 selectedItem 更改 GradientStop 的颜色（前提是它是一个字符串）。
此示例执行的操作非常相似，但使用颜色选择器而不是下拉菜单：
http://degrafa.org/source/CS4IconPreviewer/CS4IconPreviewer.html

The fill property in your example is pointing to the LinearGradientFill with id "bluedream". You can either replace the fill with a different gradient (or solid or other fill) or change the colors of the gradient fill itself:

<degrafa:LinearGradientFill id="bluedream">
    <degrafa:GradientStop color="{your_combobox.selectedItem}"/>
    <degrafa:GradientStop color="{your_other_combobox.selectedItem}"/>
</degrafa:LinearGradientFill>

<degrafa:GeometryComposition graphicsTarget="{[bgCanvas]}">
     <degrafa:RoundedRectangle id="color_preset" fill="{bluedream}"/>
</degrafa:GeometryComposition>

I haven't run that code, but it should work. The idea is to change the color of the GradientStop with your dropdown's selectedItem (provided that is a string).
This example does something very similar, but with a color picker instead of a dropdown:
http://degrafa.org/source/CS4IconPreviewer/CS4IconPreviewer.html

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