blocks|key|4502255|text|您不能使用File.ReadAllText读取资源文件。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4502256|相反，您需要使用Assembly.GetManifestResourceStream打开资源流。|4502257|您也不把路径传递给它，而是将它传递给名称空间。该文件的命名空间将是Assemblies默认名称空间%2B文件所在项目中的文件夹继承%2B文件名。|4502258|想象一下这个结构|4502259|项目(xyz.project)|unordered-list-item|4502260|​|4502261|Folder1|4502262|4502263|4502264|4502265|4502266|Folder2|4502267|4502268|4502269|4502270|4502271|4502272|4502273|4502274|SomeFile.Txt|4502275|4502276|4502277|4502278|4502279|4502280|4502281|因此，该文件的命名空间为：|4502282|xyz.project.Folder1.Folder2.SomeFile.Txt|4502283|然后你就会读成这样|4502284|++++++++using+(var+stream+=+Assembly.GetExecutingAssembly().GetManifestResourceStream("xyz.project.Folder1.Folder2.SomeFile.Txt"))
++++++++{
++++++++++++TextReader+tr+=+new+StreamReader(stream);
++++++++++++string+fileContents+=+tr.ReadToEnd();
++++++++}|code-block|syntax|javascript|4502285|entityMap^0|0|0|0|0|0|1|1|0|0|1|2|2|1|0|0|0|1|2|3|3|2|1|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1N|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|1O|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|1P|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|1Q|8|@]|9|@]|A|$]]|$1|K|3|L|5|J|7|1R|8|@]|9|@]|A|$]]|$1|M|3|N|5|J|7|1S|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|1T|8|@]|9|@]|A|$]]|$1|P|3|L|5|6|7|1U|8|@]|9|@]|A|$]]|$1|Q|3|L|5|J|7|1V|8|@]|9|@]|A|$]]|$1|R|3|L|5|J|7|1W|8|@]|9|@]|A|$]]|$1|S|3|T|5|J|7|1X|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|V|3|-4|5|6|7|1Z|8|@]|9|@]|A|$]]|$1|W|3|L|5|6|7|20|8|@]|9|@]|A|$]]|$1|X|3|L|5|6|7|21|8|@]|9|@]|A|$]]|$1|Y|3|L|5|J|7|22|8|@]|9|@]|A|$]]|$1|Z|3|L|5|J|7|23|8|@]|9|@]|A|$]]|$1|10|3|L|5|J|7|24|8|@]|9|@]|A|$]]|$1|11|3|12|5|J|7|25|8|@]|9|@]|A|$]]|$1|13|3|-4|5|6|7|26|8|@]|9|@]|A|$]]|$1|14|3|-4|5|6|7|27|8|@]|9|@]|A|$]]|$1|15|3|-4|5|6|7|28|8|@]|9|@]|A|$]]|$1|16|3|L|5|6|7|29|8|@]|9|@]|A|$]]|$1|17|3|L|5|6|7|2A|8|@]|9|@]|A|$]]|$1|18|3|L|5|6|7|2B|8|@]|9|@]|A|$]]|$1|19|3|1A|5|6|7|2C|8|@]|9|@]|A|$]]|$1|1B|3|1C|5|6|7|2D|8|@]|9|@]|A|$]]|$1|1D|3|1E|5|6|7|2E|8|@]|9|@]|A|$]]|$1|1F|3|1G|5|1H|7|2F|8|@]|9|@]|A|$1I|1J]]|$1|1K|3|-4|5|6|7|2G|8|@]|9|@]|A|$]]]|1L|$]]

You can't read Resource Files with File.ReadAllText.

Instead you need to open a Resource Stream with Assembly.GetManifestResourceStream.

You don't pass it a path either, you pass it a namespace. The namespace of the file will be the Assemblies Default Namespace + The folder heieracy in the project the file is in + the name of the file.

Imagine this structure

<ul>
<li>Project (xyz.project)</li>
<li><ul>
<li>Folder1</li>
</ul></li>
<li><ul>
<li><ul>
<li>Folder2</li>
</ul></li>
</ul></li>
<li><ul>
<li><ul>
<li><ul>
<li>SomeFile.Txt</li>
</ul></li>
</ul></li>
</ul></li>
</ul>

So the namespace for the file will be:

xyz.project.Folder1.Folder2.SomeFile.Txt

Then you would read it like so

<pre><code> using (var stream = Assembly.GetExecutingAssembly().GetManifestResourceStream("xyz.project.Folder1.Folder2.SomeFile.Txt"))
 {
 TextReader tr = new StreamReader(stream);
 string fileContents = tr.ReadToEnd();
 }
</code></pre>

blocks|key|4590099|text|你好，建议的解决方案不起作用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4590100|此返回null|offset|length|style|CODE|4590101|Assembly.GetExecutingAssembly().GetManifestResourceStream("xyz.project.Folder1.Folder2.SomeFile.Txt")|code-block|syntax|javascript|4590102|另一种方法是使用来自MemoryStream的Ressource+Data|4590103|++byte[]+aa+=+Properties.Resources.YOURRESSOURCENAME;
++MemoryStream+MS+=new+MemoryStream(aa);
++StreamReader+sr+=+new+StreamReader(MS);|4590104|不太理想，但有效|4590105|entityMap^0|0|3|4|0|0|A|C|N|E|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@$D|W|E|X|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|Y|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|Z|8|@$D|10|E|11|F|G]|$D|12|E|13|F|G]]|9|@]|A|$]]|$1|O|3|P|5|J|7|14|8|@]|9|@]|A|$K|L]]|$1|Q|3|R|5|6|7|15|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|16|8|@]|9|@]|A|$]]]|T|$]]

Hello the proposed solution doesn't work

This return <code>null</code>:

<pre><code>Assembly.GetExecutingAssembly().GetManifestResourceStream("xyz.project.Folder1.Folder2.SomeFile.Txt")
</code></pre>

Another way is to use a <code>MemoryStream</code> from the <code>Ressource Data</code>:

<pre><code> byte[] aa = Properties.Resources.YOURRESSOURCENAME;
 MemoryStream MS =new MemoryStream(aa);
 StreamReader sr = new StreamReader(MS);
</code></pre>

Not ideal but it works

I have embed <code>sample.txt</code>(it contains just one line "aaaa" ) file into project's resources like in this <a href="https://stackoverflow.com/a/433182/1843300">answer</a>.
When I'm trying to read it like this:

<pre><code>string s = File.ReadAllText(global::ConsoleApplication.Properties.Resources.sample);
</code></pre>

I'm getting System.IO.FileNotFoundException' exception.
Additional information: Could not find file 'd:\Work\Projects\MyTests\ConsoleApplication\ConsoleApplication\bin\Debug\aaaa'.

So seemingly it's trying to take file name from my resource file instead of reading this file. Why is this happening? And how can I make it read sample.txt

Trying solution of @Ryios and getting Argument null exception

<pre><code> using (var stream = Assembly.GetExecutingAssembly().GetManifestResourceStream("ConsoleApplication.Resources.sample.txt"))
 {
 TextReader tr = new StreamReader(stream);
 string fileContents = tr.ReadToEnd();
 }
</code></pre>

The file is located in d:\Work\Projects\MyTests\ConsoleApplication\ConsoleApplication\Resources\sample.txt

p.s. Solved. I had to set Build Action - embed resource in sample.txt properties

Reading a file from resources

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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我已经将sample.txt(它只包含一行"aaaa“)文件嵌入到项目的资源中，就像在这个中一样。当我像这样读它的时候：string s = File.ReadAllText(global::ConsoleApplication.Properties.Resources.sample);我得到了System.IO.Fi...

问从资源中读取文件
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问从资源中读取文件EN