从R.color检索颜色编程 [英] Retrieve color programmatically from R.color

问题描述

我有一个包含许多TextViews，一个TextView的应包含根据所检索到的数据的不同的背景颜色一个ListView。

I have a ListView which contains many TextViews, and one TextView should contain a different background color depending on the data that is retrieved.

由于我不想硬$ C C I用于R.color设置我的色彩的颜色。这工作不错，但我必须手动检查每一种颜色，因为我注意到能够得到的颜色像一个HashMap中。所以，我的第一次尝试是这样的：

Because i dont want to hardcode the colors i used R.color to set my Colors. That works nice, but i have to check manually for every color, because i am note able to get the colors like an HashMap. So my first Try was this:

    switch(line) {
    case "1":
        lineColor = context.getResources().getColor(R.color.line1);
    case "2":
        lineColor = context.getResources().getColor(R.color.line2);
    ....
    ....
    }

这似乎远离清洁code，所以我尝试了不同的方法使用字符串数组：

This seems far away from clean code, so i tried a different approach by using String-Arrays:

<string-array name="line_color_names">
    <item>1</item>
    <item>2</item>
    ....
</string-array>

<string-array name="line_color_values">
    <item>#e00023</item>
    <item>#ef9ec1</item>
    ....
</string-array>

在我AdapterClass我刚刚创建了一个HashMap中，并把这个字符串数组在一起：

In my AdapterClass i just created a HashMap and put this String-Arrays together:

    String[] line_color_names = context.getResources().getStringArray(
            R.array.line_color_names);
    String[] line_color_values = context.getResources().getStringArray(
            R.array.line_color_values);

    lineColors = new HashMap<String, String>();
    for (int i = 0; i < line_color_names.length; i++) {
        lineColors.put(line_color_names[i], line_color_values[i]);
    }

所以我的问题是：这是实现这一目标的唯一途径，或有另一种，最好通过直接从R.color服用颜色

So my Question is: Is this the only way to achieve this or is there another, ideally by taking colors directly from R.color?

在此先感谢！

推荐答案

您可以通过使用资源名称取色ID（ R.color.foo ），并在解决问题运行时间：

You can get color ID by using resource name (R.color.foo) and resolve it at runtime:

public int getColorIdByResourceName(String name) {
  int color;
  try {
    Class res = R.color.class;
    Field field = res.getField( name );
    int color = field.getInt(null);
  } catch ( Exception e ) {
    e.printStackTrace();
  }
  return color;
}

然后

int colorId  = getColorIdByResourceName("foo21");
int colorVal = getResources().getColor(getColorIdByResourceName("foo21"));

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