blocks|key|506415|text|使用URL.toURI()+(Android+doc)方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|506416|示例：|506417|URL+url+=+new+URL("http://www.google.com");+//Some+instantiated+URL+object
URI+uri+=+url.toURI();|code-block|syntax|javascript|506418|请确保处理相关异常，如URISyntaxException。|506419|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/6/docs/api/java/net/URL.html#toURI%2528%2529|1|http://developer.android.com/reference/java/net/URL.html#toURI%2528%2529|2|http://docs.oracle.com/javase/6/docs/api/java/net/URISyntaxException.html^0|2|B|2|B|0|F|B|1|0|0|0|B|I|B|I|2|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]|$9|16|A|17|1|18]]|E|$]]|$1|F|3|G|5|6|7|19|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|1A|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1B|8|@$9|1C|A|1D|B|C]]|D|@$9|1E|A|1F|1|1G]]|E|$]]|$1|O|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]|W|$5|R|S|T|E|$U|X]]|Y|$5|R|S|T|E|$U|Z]]]]

Use <a href="http://docs.oracle.com/javase/6/docs/api/java/net/URL.html#toURI%28%29" rel="noreferrer"><code>URL.toURI()</code></a> (<a href="http://developer.android.com/reference/java/net/URL.html#toURI%28%29" rel="noreferrer">Android doc</a>) method.

Example:

<pre><code>URL url = new URL("http://www.google.com"); //Some instantiated URL object
URI uri = url.toURI();
</code></pre>

Make sure to handle relevant exception, such as <a href="http://docs.oracle.com/javase/6/docs/api/java/net/URISyntaxException.html" rel="noreferrer"><code>URISyntaxException</code></a>.

blocks|key|506353|text|我想你的答案可以从这里找到..|type|unstyled|depth|inlineStyleRanges|entityRanges|data|506354|Uri.Builder.build()在普通URL上工作得很好，但在支持端口号时却失败了。|offset|length|style|CODE|506355|我发现让它支持端口号的最简单的方法是让它首先解析给定的URL，然后再使用它。|506356|Uri.Builder+b+=+Uri.parse("http://www.yoursite.com:12345").buildUpon();

b.path("/path/to/something/");
b.appendQueryParameter("arg1",+String.valueOf(42));

if+(username+!=+"")+{
++b.appendQueryParameter("username",+username);
}

String+url+=+b.build().toString();+|code-block|syntax|javascript|506357|来源：http://twigstechtips.blogspot.com/2011/01/android-create-url-using.html|506358|entityMap|0|LINK|mutability|MUTABLE|url|http://twigstechtips.blogspot.com/2011/01/android-create-url-using.html^0|0|0|J|0|0|0|3|1Z|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@$D|10|E|11|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|12|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|13|8|@]|9|@]|A|$M|N]]|$1|O|3|P|5|6|7|14|8|@]|9|@$D|15|E|16|1|17]]|A|$]]|$1|Q|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

I think your answer can be found from here..

<code>Uri.Builder.build()</code> works quite well with normal URLs, but it fails with port number support.

The easiest way that I discovered to make it support port numbers was to make it parse a given URL first then work with it.

<pre><code>Uri.Builder b = Uri.parse("http://www.yoursite.com:12345").buildUpon();

b.path("/path/to/something/");
b.appendQueryParameter("arg1", String.valueOf(42));

if (username != "") {
 b.appendQueryParameter("username", username);
}

String url = b.build().toString(); 
</code></pre>

Source : <a href="http://twigstechtips.blogspot.com/2011/01/android-create-url-using.html">http://twigstechtips.blogspot.com/2011/01/android-create-url-using.html</a>

blocks|key|506551|text|来自How+to+create+a+Uri+from+a+URL?|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|506552|Uri+uri+=++Uri.parse(+"http://www.facebook.com"+);|code-block|syntax|javascript|506553|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/8240139/how-to-create-a-uri-from-a-url^0|2|V|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

From <a href="https://stackoverflow.com/questions/8240139/how-to-create-a-uri-from-a-url">How to create a Uri from a URL?</a>

<pre><code>Uri uri = Uri.parse( "http://www.facebook.com" );
</code></pre>

blocks|key|506523|text|注意，在安卓中，Uri不同于Java+Uri。下面是如何避免使用硬编码的字符串，同时，只使用http+URL字符串的路径部分编码以符合RFC2396的Uri：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|506524|示例Url字符串：|506525|String+thisUrl+=+"http://lapi.transitchicago.com/api/1.0/ttarrivals.aspx?key=[redacted]&mapid=value"|code-block|syntax|javascript|506526|方法：|506527|private+Uri.Builder+builder;
public+Uri+getUriFromUrl(String+thisUrl)+{
++++URL+url+=+new+URL(thisUrl);
++++builder+=++new+Uri.Builder()
++++++++++++++++++++++++++++.scheme(url.getProtocol())
++++++++++++++++++++++++++++.authority(url.getAuthority())
++++++++++++++++++++++++++++.appendPath(url.getPath());
++++return+builder.build();
}|506528|要处理查询字符串，您需要按照描述的here解析url.getQuery()，然后将其提供给构建器。appendQueryParameter()。|offset|length|506529|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/13592567/923920^0|0|0|0|0|0|H|4|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|11|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|13|8|@]|9|@$O|14|P|15|1|16]]|A|$]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

Note that in Android, Uri's are different from Java URI's. Here's how to avoid using hardcoded strings, and at the same time, create a Uri with just the path portion of the http URL string encoded to conform to RFC2396:

Sample Url String: 

<pre><code>String thisUrl = "http://lapi.transitchicago.com/api/1.0/ttarrivals.aspx?key=[redacted]&amp;mapid=value"
</code></pre>

method:

<pre><code>private Uri.Builder builder;
public Uri getUriFromUrl(String thisUrl) {
 URL url = new URL(thisUrl);
 builder = new Uri.Builder()
 .scheme(url.getProtocol())
 .authority(url.getAuthority())
 .appendPath(url.getPath());
 return builder.build();
}
</code></pre>

To handle query strings you will need to parse the url.getQuery() as described <a href="https://stackoverflow.com/a/13592567/923920">here</a> and then feed that into builder. appendQueryParameter().

blocks|key|506431|text|try+{
++++uri+=+new+URI(url.toString());
}+catch+(URISyntaxException+e)+{
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|506432|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>try {
 uri = new URI(url.toString());
} catch (URISyntaxException e) {
}
</code></pre>

blocks|key|506439|text|URI+uri+=+null;
URL+url+=+null;

//+Create+a+URI
try+{
++++uri+=+new+URI("www.abc.com");
}+catch+(URISyntaxException+e)+{
}

//+Convert+an+absolute+URI+to+a+URL
try+{
++++url+=+uri.toURL();
}+catch+(IllegalArgumentException+e)+{
++++//+URI+was+not+absolute
}+catch+(MalformedURLException+e)+{
}

//+Convert+a+URL+to+a+URI
try+{
++++uri+=+new+URI(url.toString());
}+catch+(URISyntaxException+e)+{
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|506440|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>URI uri = null;
URL url = null;

// Create a URI
try {
 uri = new URI("www.abc.com");
} catch (URISyntaxException e) {
}

// Convert an absolute URI to a URL
try {
 url = uri.toURL();
} catch (IllegalArgumentException e) {
 // URI was not absolute
} catch (MalformedURLException e) {
}

// Convert a URL to a URI
try {
 uri = new URI(url.toString());
} catch (URISyntaxException e) {
}
</code></pre>

I would like to <code>use intent.setData(Uri uri)</code> to pass data obtained from a URL. In order to do this, I need to be able to create a Uri from a URL (or from a byte[] I read from the URL, or a <code>ByteArrayInputStream</code> I create from the <code>byte[]</code>, etc). However, I cannot figure out how this is supposed to be done. 

So, is there anyway to create a Uri from data obtained from a URL without first writing the data to a local file?

How to create a android.net.Uri from a java.net.URL?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我想通过use intent.setData(Uri uri)来传递从网址获得的数据。为此，我需要能够从URL (或从URL读取的byte[]，或从byte[]创建的ByteArrayInputStream等)创建Uri。然而，我不知道应该如何做到这一点。那么，有没有办法在不先将数据写入本地文件的情况下，根据从URL获得的数据创建Uri呢？

问如何从java.net.URL创建android.net.Uri？
EN

回答 6

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问如何从java.net.URL创建android.net.Uri？EN

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问如何从java.net.URL创建android.net.Uri？
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