blocks|key|297438|text|您可以使用Uri中的parse静态方法|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|297439|//...
import+android.net.Uri;
//...

Uri+myUri+=+Uri.parse("http://stackoverflow.com")|code-block|syntax|javascript|297440|entityMap^0|5|3|A|5|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

You can use the <code>parse</code> static method from <code>Uri</code>
<pre><code>//...
import android.net.Uri;
//...

Uri myUri = Uri.parse(&quot;http://stackoverflow.com&quot;)
</code></pre>

blocks|key|299790|text|您将如何处理URI？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|299791|例如，如果您打算将其与HttpGet一起使用，则可以在创建HttpGet实例时直接使用该字符串。|299792|HttpGet+get+=+new+HttpGet("http://stackoverflow.com");|code-block|syntax|javascript|299793|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

What are you going to do with the URI?

If you're just going to use it with an HttpGet for example, you can just use the string directly when creating the HttpGet instance.

<pre><code>HttpGet get = new HttpGet("http://stackoverflow.com");
</code></pre>

blocks|key|3027747|text|我只是在使用java.net+包。您可以在此处执行以下操作：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|BOLD|entityRanges|data|3027748|...
import+java.net.URI;
...

String+myUrl+=+"http://stackoverflow.com";
URI+myURI+=+new+URI(myUrl);|code-block|syntax|javascript|3027749|entityMap^0|6|8|F|1|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]|$9|Q|A|R|B|D]]|E|@]|F|$]]|$1|G|3|H|5|I|7|S|8|@]|E|@]|F|$J|K]]|$1|L|3|-4|5|6|7|T|8|@]|E|@]|F|$]]]|M|$]]

I am just using the <code>java.net</code> package.
Here you can do the following:
<pre><code>...
import java.net.URI;
...

String myUrl = &quot;http://stackoverflow.com&quot;;
URI myURI = new URI(myUrl);
</code></pre>

blocks|key|3027763|text|您可以使用将字符串解析为Uri，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3027764|Uri+myUri+=+Uri.parse("http://stackoverflow.com");|code-block|syntax|javascript|3027765|以下是如何在隐式意图中使用新创建的Uri的示例。以便在用户电话上的浏览器中查看。|3027766|//+Creates+a+new+Implicit+Intent,+passing+in+our+Uri+as+the+second+paramater.
Intent+webIntent+=+new+Intent(Intent.ACTION_VIEW,+myUri);

//+Checks+to+see+if+there+is+an+Activity+capable+of+handling+the+intent
if+(webIntent.resolveActivity(getPackageManager())+!=+null){
++++startActivity(webIntent);
}|3027767|NB:机器人和是有区别的。|offset|length|style|BOLD|3027768|entityMap^0|0|0|0|0|0|3|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

You can parse a String to a Uri by using <a href="https://developer.android.com/reference/android/net/Uri.html#parse(java.lang.String)" rel="noreferrer">Uri.parse()</a> as shown below:

<pre><code>Uri myUri = Uri.parse("http://stackoverflow.com");
</code></pre>

The following is an example of how you can use your newly created Uri in an implicit intent. To be viewed in a browser on the users phone.

<pre><code>// Creates a new Implicit Intent, passing in our Uri as the second paramater.
Intent webIntent = new Intent(Intent.ACTION_VIEW, myUri);

// Checks to see if there is an Activity capable of handling the intent
if (webIntent.resolveActivity(getPackageManager()) != null){
 startActivity(webIntent);
}
</code></pre>

NB: There is a difference between Androids <a href="https://developer.android.com/reference/java/net/URI.html" rel="noreferrer">URI</a> and <a href="https://developer.android.com/reference/android/net/Uri.html" rel="noreferrer">Uri</a>.

blocks|key|3027770|text|如果URI没有完全按照其标准编码，java.net.URI中的Java解析器将会失败。例如，尝试解析：http://www.google.com/search?q=cat%7Cdog。垂直条形图将抛出异常。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3027771|使用urllib可以轻松地将字符串转换为java.net.URI。它将对URL进行预处理和转义。|3027772|assertEquals("http://www.google.com/search?q=cat%257Cdog",
++++Urls.createURI("http://www.google.com/search?q=cat%7Cdog").toString());|code-block|syntax|javascript|3027773|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/EricEdens/urllib^0|H|C|1F|12|0|K|C|2|6|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@$9|12|A|13|1|14]]|E|$]]|$1|H|3|I|5|J|7|15|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

Java's parser in <code>java.net.URI</code> is going to fail if the URI isn't fully encoded to its standards. For example, try to parse: <code>http://www.google.com/search?q=cat|dog</code>. An exception will be thrown for the vertical bar.

<a href="https://github.com/EricEdens/urllib" rel="nofollow noreferrer">urllib</a> makes it easy to convert a string to a <code>java.net.URI</code>. It will pre-process and escape the URL. 

<pre><code>assertEquals("http://www.google.com/search?q=cat%7Cdog",
 Urls.createURI("http://www.google.com/search?q=cat|dog").toString());
</code></pre>

blocks|key|3242119|text|如果你正在使用+Kotlin+和Kotlin+android扩展，那么有一个很好的方法可以做到这一点。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|3242120|val+uri+=+myUriString.toUri()|code-block|syntax|javascript|3242121|要将Kotlin扩展(、KTX、)添加到项目中，请在应用程序模块的build.gradle中添加以下内容|3242122|++repositories+{
++++google()
}

dependencies+{
++++implementation+'androidx.core:core-ktx:1.0.0-rc01'
}|3242123|entityMap^0|7|8|0|0|B|5|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

If you are using Kotlin and Kotlin android extensions, then there is a beautiful way of doing this.

<pre><code>val uri = myUriString.toUri()
</code></pre>

To add Kotlin extensions (KTX) to your project add the following to your app module's build.gradle

<pre><code> repositories {
 google()
}

dependencies {
 implementation 'androidx.core:core-ktx:1.0.0-rc01'
}
</code></pre>

blocks|key|3027786|text|你也可以这样做|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3027787|对于http|3027788|var+response+=+await+http.get(Uri.http("192.168.100.91",+"/api/fetch.php"));|code-block|syntax|javascript|3027789|或|3027790|对于https|3027791|var+response+=+await+http.get(Uri.https("192.168.100.91",+"/api/fetch.php"));|3027792|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|V|8|@]|9|@]|A|$G|H]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

you can do this too
for http
<pre><code>var response = await http.get(Uri.http(&quot;192.168.100.91&quot;, &quot;/api/fetch.php&quot;));
</code></pre>
or
for https
<pre><code>var response = await http.get(Uri.https(&quot;192.168.100.91&quot;, &quot;/api/fetch.php&quot;));
</code></pre>

blocks|key|299827|text|import+java.net.URI;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|299828|下面也适用于我：|unstyled|299829|URI+uri+=+URI.create("http://stackoverflow.com");|299830|或|299831|URI+uri+=+new+URI("http://stackoverflow.com");|299832|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|P|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|M|3|-4|5|F|7|T|8|@]|9|@]|A|$]]]|N|$]]

<pre><code>import java.net.URI;
</code></pre>
Below also works for me :
<pre><code>URI uri = URI.create(&quot;http://stackoverflow.com&quot;);
</code></pre>
OR
<pre><code>URI uri = new URI(&quot;http://stackoverflow.com&quot;);
</code></pre>

How can I convert a String to a Uri in Java (Android)? i.e.:

<pre><code>String myUrl = "http://stackoverflow.com";
</code></pre>

myUri = ???;

Convert String to Uri

Android

Kotlin

Java

如何在Java (Android)中将字符串转换为Uri？即：String myUrl = "http://stackoverflow.com";myUri =？；

问将字符串转换为Uri
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