首页 \ 问答 \ Android attr到xml drawables(Android attr to xml drawables)

Android attr到xml drawables(Android attr to xml drawables)

 TL; DR我正在寻找public static Drawable getDrawableFromAttribute(Context context, String attrName) 。  
 
 我正在寻找一种方法，加载动态drawable，这是我的风格定义自定义属性。 那是我的配置  
 attr.xml  
<resources>
   <attr name="custom_image" type="reference">
</resources>
 
 styles.xml  
<resources>
   <style name="demo">
      <item name="custom_image">@drawable/fancy_picture</item>
   </style>
</resources>
 
 fancy_picture名为/res/drawables/fancy_pictures.xml 。  
 现在，我希望有人输入字符串“custom”和“image”， ImageView应该在其中显示fancy_picture。  
 做这个的最好方式是什么？ 如果我使用XML-Layout文件，我可以写  
<ImageView
    ...
    android:src="?custom_image"
    ...
    />
 
 我没有在我的样式xml中使用declare-styleable ，如果可能的话我想完全忽略它们。 

TL;DR I looking for an implementation of public static Drawable getDrawableFromAttribute(Context context, String attrName). 
 
I'm searching for a way, to load dynamic drawables, which are defined in my style with custom attributes. That's my configuration 
attr.xml 
<resources>
   <attr name="custom_image" type="reference">
</resources>
 
styles.xml 
<resources>
   <style name="demo">
      <item name="custom_image">@drawable/fancy_picture</item>
   </style>
</resources>
 
The fancy_picture is a named /res/drawables/fancy_pictures.xml. 
Now, I want someone to enter the string "custom" and "image" and the ImageView should show the fancy_picture in it. 
What is the best way to do this? If I use a XML-Layout file, I could write 
<ImageView
    ...
    android:src="?custom_image"
    ...
    />
 
I didn't use declare-styleable in my style xml and I would like to ignore them completely, if possible. 

原文：https://stackoverflow.com/questions/29654275

更新时间：2023-06-29 18:06

最满意答案

 在你的情况下，变量$ fruit将具有值“pear”，而$ color =“green”因为你总是重新定义它  
 <?php
 $fruits = array();
 $fruits[] = ["name" => "Apple", "colour"=>"red"];
 $fruits[] = ["name" => "Banana", "colour"=>"yellow"];
 $fruits[] = ["name" => "Pear", "colour"=>"green"];
 ?>
 
 第一种方式：  
<?php
include 'fruits.php';
$num=count($fruits);
$i = 0;
while ($i < $num){
echo "<p>My ".$fruits[$i]["name"]." is ".$fruits[$i]["colour"].".</p>";
$i++;
}
?>
 
 第二种方式（ 更容易 ）：  
<?php 
include 'fruits.php';
foreach ($fruits as $fruit){
  echo "<p>My ".$fruit["name"]." is ".$fruit["colour"].".</p>";
}

?>
 
 其中$ fruit是数组$ fruits的元素 

in your case, variable $fruit will have a value "pear", and $colour = "green" Because you always redefine it 
 <?php
 $fruits = array();
 $fruits[] = ["name" => "Apple", "colour"=>"red"];
 $fruits[] = ["name" => "Banana", "colour"=>"yellow"];
 $fruits[] = ["name" => "Pear", "colour"=>"green"];
 ?>
 
1st way: 
<?php
include 'fruits.php';
$num=count($fruits);
$i = 0;
while ($i < $num){
echo "<p>My ".$fruits[$i]["name"]." is ".$fruits[$i]["colour"].".</p>";
$i++;
}
?>
 
2nd way (much easier): 
<?php 
include 'fruits.php';
foreach ($fruits as $fruit){
  echo "<p>My ".$fruit["name"]." is ".$fruit["colour"].".</p>";
}

?>
 
where $fruit is element of the array $fruits

Android attr到xml drawables(Android attr to xml drawables)

最满意答案

相关问答

daloRadius不工作，找不到DB.PHP文件(daloRadius Is not working, DB.PHP file not found)[2022-03-13]

php db访问所有文件而不包含(php db access in all files without include)[2023-07-18]

PHP文件为软DB？(PHP file as soft DB?)[2023-09-08]

Php - mysqli多次连接到db(Php - mysqli connect to db several times)[2022-12-27]

PHP - DB2许可证文件位置(PHP - DB2 license file location)[2022-12-17]

如何从另一个.php使用$ db到使用OOP的另一个.php的类？(How to use $db from another .php to another .php's class using OOP?)[2023-11-24]

如何使用PHP和Mysql DB下载文件(How do I download a file using PHP and Mysql DB)[2021-12-11]

PHP OO - 我应该如何处理多个类中的软错误？(PHP OO - How should I handle soft errors in multiple classes?)[2023-03-06]

使用php从mysql db创建一个xml文件(using php to create an xml file from a mysql db)[2022-05-04]

SQLite3与php 5.4没有创建db文件(SQLite3 with php 5.4 no db file created)[2022-05-19]

相关文章

最新问答