HTTP状态404描述请求的资源（）不可用。 Apache Tomcat / 7.0.27 [英] HTTP Status 404 description The requested resource () is not available. Apache Tomcat/7.0.27
问题描述
当我尝试通过 http：// localhost：7080 / first / 或< a href =http：// localhost：7080 / first / start-page.html =noreferrer> http：// localhost：7080 / first / start-page.html 我收到以下消息：
HTTP状态404 - 类型状态报告消息描述请求的资源（）不可用。 Apache Tomcat / 7.0.27
但是如果我要求localhost：7080 / tomcat的主页正确找到。
$ b $有人可以帮助我吗？
为什么我不能从eclipes得到我的start-page.html？
，但是当我从C：\Users\Mohsen\workspace.metadata.plugins\org.eclipse.wst.server.core\tmp0\wtpwebapps\first\WEB-INF\\ \\start-page.html，它运行正常。
<？xml version =1.0encoding =ISO-8859 -1？>
<！DOCTYPE html PUBLIC - // W3C // DTD XHTML 1.0 Strict // ENhttp://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd\">
< html xmlns =http://www.w3.org/1999/xhtml
xmlns：h =http://java.sun.com/jsf/html>
< h：head>
< meta http-equiv =Content-Typecontent =text / html; charset = ISO-8859-1/>
< title>第一页< / title>
< / h：head>
< h：body>
< fieldset>
< legend>随机结果页< / legend>
< h：form>
按按钮可以获得三个可能的结果页面之一。
< br />
< h：commandButton value =转到随机页面
/>
< / h：form>
< / fieldset>
...
< / h：body>< / html>
我想你创建了xhtml文件 文件和更改文件扩展名。这是错的您应该创建一个html文件并再次配置扩展名。然后点击下一个按钮。之后，您应该选择新的Facelet模板。然后完成
when i try to access my project home at http://localhost:7080/first/ or http://localhost:7080/first/start-page.html i get this message:
HTTP Status 404 - type Status report message description The requested resource () is not available. Apache Tomcat/7.0.27
but if i ask for localhost:7080/ the home page of tomcat is correctly found.
can someone help me?
why i can't get my start-page.html from eclipes? but when i run from C:\Users\Mohsen\workspace.metadata.plugins\org.eclipse.wst.server.core\tmp0\wtpwebapps\first\WEB-INF\start-page.html, it runs correctly.
<?xml version="1.0" encoding="ISO-8859-1" ?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml"
xmlns:h="http://java.sun.com/jsf/html">
<h:head>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1" />
<title>first page</title>
</h:head>
<h:body>
<fieldset>
<legend>Random Results Page</legend>
<h:form>
Press button to get one of three possible results pages.
<br/>
<h:commandButton value="Go to Random Page"
/>
</h:form>
</fieldset>
…
</h:body></html>
I think you created "xhtml file", creating a "file" and changing the file extension. This is wrong. You should create a html file and configure again extension. Then click the next button. After that, you should choose "New Facelet Template". Then finish.
这篇关于HTTP状态404描述请求的资源（）不可用。 Apache Tomcat / 7.0.27的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
