Android 在 XML 中连接多个字符串？
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Concatenate multiple strings in XML?
提问by Lukap
I go through this How to concatenate multiple strings in android XML?and in the end there are comments that
我经历了这个如何在 android XML 中连接多个字符串？最后有评论说
For clarity, Its works:
For clarity, Its works:
<string name="title">@string/app_name</string>.– Andrzej Du?
<string name="title">@string/app_name</string>.–安杰伊杜？
I made my own example but it doesn't works. So does Andrzejwrong or I am doing something wrong in my code.
我做了我自己的例子，但它不起作用。所以确实安杰错误或我在我的代码做错了什么。
R.strings.bbbshould contains "bbb aaa"but instead of "bbb aaa"it contains "bbb @strings/aaa"
R.strings.bbb应该包含"bbb aaa"但不是"bbb aaa"它包含"bbb @strings/aaa"
<string name="aaa">aaa</string>
<string name="bbb">bbb @strings/aaa</string>
Query:
询问：
Is it possible to do some concatenation only in xml, without source code changes?
是否可以只在 xml 中进行一些连接，而无需更改源代码？
Reason why I don't want to edit in code because I use this strings in xml/preferences.xml
我不想在代码中编辑的原因，因为我在 xml/preferences.xml
For Example:
例如：
<ListPreference android:key="key_bbb" android:title="@string/bbb"....
<ListPreference android:key="key_bbb" android:title="@string/bbb"....
If you know what I mean, here there is no possibility to use something like this
如果你知道我的意思，这里不可能使用这样的东西
String title = res.getString(R.string.title, appName);
采纳答案by Lalit Poptani
No I don't think you can concatenate.
不，我认为你不能连接。
<string name="aaa">aaa</string>
<string name="bbb">bbb @string/aaa</string>
Output - bbb @string/aaa
输出 - bbb @string/aaa
If you do,
如果你这样做，
<string name="aaa">aaa</string>
<string name="bbb">@string/aaa bbb</string> -> This won't work it
will give compilation error
Because here it will search for a String with reference @string/aaa bbbwhich does not exists.
因为在这里它将搜索一个@string/aaa bbb不存在的带有引用的字符串。
Problem in your case was, you where using @strings/aaawhich should be @string/aaa
在您的情况下，问题是，您@strings/aaa应该在哪里使用@string/aaa
回答by Savelii Zagurskii
No, you can't concatenate strings in XML but you can define XML resources.
不，您不能连接 XML 中的字符串，但您可以定义 XML 资源。
<?xml version="1.0" encoding="utf-8"?>
<!DOCTYPE resources [
<!ENTITY appname "MyAppName">
<!ENTITY author "MrGreen">
]>
<resources>
<string name="app_name">&appname;</string>
<string name="description">The &appname; app was created by &author;</string>
</resources>
The original answer was posted here.
原来的答案张贴在这里。
回答by Dharmendra
In XML only this is not possible but using the java code you can use the String.format()method.
仅在 XML 中这是不可能的，但使用 java 代码可以使用该String.format()方法。
<string name="aaa">aaa</string>
<string name="bbb">bbb %1$s</string>
In java code
在java代码中
String format = res.getString(R.string.bbb);
String title = String.format(format, res.getString(R.string.aaa));
So title will be a full string after concatenation of two strings.
所以标题将是两个字符串连接后的完整字符串。
回答by Boy
You can concatenate the resources in gradle.build:
您可以连接 gradle.build 中的资源：
resValue "string", "TWITTER_CALLBACK", "twitter_callback_" + applicationId
回答by batsheva
回答by César Mu?oz
Yes, you can do so without having to add any Java/Kotlin code, just XML, by using this library: https://github.com/LikeTheSalad/android-string-referencewhich does that at buildtime.
是的，您可以这样做而无需添加任何 Java/Kotlin 代码，只需使用此库：https: //github.com/LikeTheSalad/android-string-reference，它在构建时执行此操作。
For your case, you'd have to set up your strings like this:
对于您的情况，您必须像这样设置字符串：
<string name="aaa">Some text</string>
<string name="template_bbb">bbb ${aaa}</string>
And then the library will create the following at buildtime:
然后库将在构建时创建以下内容：
<string name="bbb">bbb Some text</string>
Disclaimer: I'm the author of this library.
免责声明：我是这个库的作者。
回答by netpork
Kotlin version:
科特林版本：
strings in the xml:
xml中的字符串：
<string name="school_girl">Gogo Yubari</string>
<string name="assistant">Sofie Fatale</string>
<string name="boss">O-Ren Ishii</string>
<string name="list">%s, %s, %s</string>
and then in the code:
然后在代码中：
val killBillV1 = getString(R.string.list).format(
Locale.US,
getString(R.string.school_girl),
getString(R.string.assistant),
getString(R.string.boss)
)
回答by Waqar Aslam
How about this technique
这个技术怎么样
Use an array
使用数组
<string name="app_link">https://myapp.com/</string>
<array name="update_dialog_msg">
<item>Update can be downloaded manually by visiting this link:\n</item>
<item>@string/app_link</item>
</array>
And then make a method in java class as below
然后在java类中创建一个方法如下
public String getStringArray(int resStringArray)
{
String str = "";
String[] strArray = mContext.getResources().getStringArray(resStringArray);
for (String s: strArray)
{
str += s;
}
return str;
}
Now call it from any java class as
现在从任何 java 类调用它作为
mDialog.setMessage(getStringArray(R.array.update_dialog_msg);
// OR
textView.setText(getStringArray(R.array.update_dialog_msg);
回答by Ejrr1085
No, but is posible in xslt file with concat function:
不，但可以在带有 concat 函数的 xslt 文件中使用：
<html>
<body>
<ul>
<xsl:for-each select="stock">
<xsl:if test="starts-with(@symbol, 'C')">
<li>
<xsl:value-of select="concat(@symbol,' - ', name)" />
</li>
</xsl:if>
</xsl:for-each>
</ul>
</body>
</html>
http://www.java2s.com/Tutorial/XML/0100__XSLT-stylesheet/concatfunction.htm
http://www.java2s.com/Tutorial/XML/0100__XSLT-stylesheet/concatfunction.htm
