从servlet读取资源文件 [英] Reading a resource file from a servlet
本文介绍了从servlet读取资源文件的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我想从servlet中读取文本文件,该文件是使用Jersey制作的,并使用Gradle构建的.服务器是Tomcat 9.
I want to read a text file from a servlet which is made using Jersey and built using Gradle. The server is Tomcat 9.
我的文件结构:
project
+ src
+ package
+ App.java
+ res
+ file.txt
我的build.gradle
:
apply plugin: "java"
apply plugin: "war"
project.webAppDirName = "WebContent"
sourceSets {
main {
java {
srcDirs = ["src"]
}
resources {
srcDirs = ["res"]
}
}
}
repositories {
mavenCentral()
}
dependencies {
compile "org.glassfish.jersey.containers:jersey-container-servlet:2.25.1"
}
war {
archiveName "api.war"
}
Tomcat提取WAR后,file.txt
与源软件包位于同一目录中.
After Tomcat extracts the WAR, file.txt
is in the same directory as the source packages:
WEB-INF
+ classes
+ package
+ App.class
+ file.txt
我试图使用类加载器加载文件,但是类路径似乎是Tomcat的bin
目录.我也不想对路径进行硬编码.
在这里加载文件的最佳方法是什么?
I tried to load the file using the class loader but the classpath seems to be the bin
directory of Tomcat. I also don't want to hardcode the path.
What is the best way to load the file here?
推荐答案
InputStream in = package.App.class.getClassloader().getResourceAsStream("file.txt")
这篇关于从servlet读取资源文件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!
查看全文