如何从线性渐变获取当前颜色？ [英] How to get current color from Linear Gradient?

问题描述

我有一个Seek Bar，值范围从1到10.ThUMB停在1,2,3,4,5 ... 10。

I have an Seek Bar whose values ranges from 1 to 10. The THUMB stops at 1,2,3,4,5 ... 10.

背景颜色如果SeekBar是线性渐变[颜色从红色开始，然后黄色，最后绿色]。如何获取缩略图所在位置的当前颜色？

The background color if SeekBar is Linear Gradient [Colors Start from RED, then YELLOW and lastly GREEN]. How to get the current color where thumb is positioned?

推荐答案

pskink的建议是对的。您可以使用 ArgbEvaluator 达到此目标。

pskink's suggestion is right. You can use an ArgbEvaluator to achieve this goal.

让我们假设这是您的 SeekBar ：

 C1                                  C2                                  C3
 +-------|-------|-------|-------|---+---|-------|-------|-------|-------+                          
 1       2       3       4       5       6       7       8       9       10

您有10个 Thumb 位置（来自 1 - 10的数字），3种颜色（ + 符号表示颜色的位置， C1 C2 和 C3 代表颜色的名称）。

You've got 10 Thumb positions (numbers from 1 - 10), 3 colors (+ sign indicates the position of the color, and C1, C2 and C3 represent the name of the color).

C1 和 C2 之间的距离（以及 C2 和 C3 ）可以分为9个部分。这9个可代表您的 Thumb 位置：

The distance between C1 and C2 (as well as between C2 and C3) can be devided into 9 pieces. Those 9 pieces can represent your Thumb positions:

 C1                                  C2                                  C3
 +-------|-------|-------|-------|---+---|-------|-------|-------|-------+ 
 |       |       |       |       |   |   |       |       |       |       |                    
0/9     2/9     4/9     6/9    8/9  9/9  |       |       |       |       |
                                     |   |       |       |       |       |
                                    0/9  1/9    3/9     5/9     7/9     9/9

因此， SeekBar 的值可以通过以下方式计算：

Therefore the values of your SeekBar can be calculated this way:

int c1 = 0xFFFF0000; // ARGB representation of RED
int c2 = 0xFFFFFF00; // ARGB representation of YELLOW
int c3 = 0xFF00FF00; // ARGB representation of GREEN
ArgbEvaluator evaluator = new ArgbEvaluator();

int thumb1 = (int) evaluator.evaluate(0f,      c1, c2); // 0f/9f = 0f
int thumb2 = (int) evaluator.evaluate(2f / 9f, c1, c2);
int thumb3 = (int) evaluator.evaluate(4f / 9f, c1, c2);
int thumb4 = (int) evaluator.evaluate(6f / 9f, c1, c2);
int thumb5 = (int) evaluator.evaluate(8f / 9f, c1, c2);
int thumb6 = (int) evaluator.evaluate(1f / 9f, c2, c3);
int thumb7 = (int) evaluator.evaluate(3f / 9f, c2, c3);
int thumb8 = (int) evaluator.evaluate(5f / 9f, c2, c3);
int thumb9 = (int) evaluator.evaluate(7f / 9f, c2, c3);
int thumb10 = (int) evaluator.evaluate(1f,     c2, c3); // 9f/9f = 1f

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