我的播放应用程序是2.4.2.在开发人员模式中,我曾经使用Source.fromFile("./ public/files/abc.json")从后端控制器的公用文件夹中读取文件
当我将应用程序转换为生产模式时,我收到了File Not Found Exceptions.我发现公用文件夹在生产模式下打包在资产jar中.我能做什么才能在开发和生产模式下工作？
我在使用docker容器加载Java jar类路径中的JSON文件时遇到FileNotFoundException,它是一个Spring-Boot应用程序.此JSON文件在资源文件夹中可用.我能够在./target/classes/path下的docker中看到JSON文件.
Resource resource = resourceLoader.getResource("classpath:folderNm/file.json");
HashMap<String, String> headerMapping = (HashMap<String, String>) parser.parse(new FileReader(resource.getFile().getAbsolutePath()));
但是我得到了这个例外:
java.io.FileNotFoundException: class path resource [folderNm/file.json] cannot be resolved to absolute file path because it does not reside in the file system: jar:file:/app.jar!/folderNm/file.json
我试过了
- > resource.getFile().getPath();
- > resource.getFile().getCanonicalPath();
- >" ./target/classes/folderName/fileName"(硬编码的FilePath位置) - >" /app.jar!/folderNm/file.json"(硬编码的FilePath位置)
InputStream inputStream = getClass().getResourceAsStream("xyz.json");
BufferedReader br = new BufferedReader(new InputStreamReader(inputStream, "UTF-8"));
StringBuilder responseStrBuilder = new StringBuilder();
String inputStr;
while ((inputStr = br.readLine()) != null)
responseStrBuilder.append(inputStr);
以上都没有运行.请建议一种解决此问题的方法.
我有一个包含多个maven模块的项目
project/pom.xml
/external_services/pom.xml
/ifs/pom.xml
/src/test/java/
/MockIFSClient.java
/IFSClientTest.java
/src/test/java/resources/sample.json
/inventory/pom.xml
/business/pom.xml
/src/main/java/InventorySummary.java
/services/pom.xml
/src/main/java/InventorySummaryResource.java
/src/main/test/InventorySummaryResourceTest.java
MockIFSClient访问sample.json为
try {
inventoryPriceDetails = mapper.readValue(new File(getClass().getResource("/getInventoryAndPriceResponse.json").getPath()), new TypeReference<List<InventoryPriceDetail>>() {
});
} catch (final IOException e) {
throw new RuntimeException("could not read resource :" + e.getMessage());
}
所以IFSClientTest运行鳍,因为它们在同一个包中.
问题？尝试访问相同代码的
InventorySummaryResourceTest调用MockIFSClient,但现在它失败了
could not read resource :file:/Users/harith/IdeaProjects/inventory_api/external_services/ifs/target/ifs-1.0-SNAPSHOT-tests.jar!/sample.json (No such file or directory)
services/pom.xml具有依赖性
<dependency>
<groupId>com.org.project.external_services</groupId>
<artifactId>ifs</artifactId>
<version>${project.version}</version>
<type>test-jar</type>
<scope>test</scope>
</dependency>
题
我做了什么改变
new File(getClass().getResource("/getInventoryAndPriceResponse.json").getPath())
这样它也可以从不同的模块访问
当我尝试执行我的 aws lambda function
1) Error injecting constructor, java.lang.NullPointerException
at in.nikhilbhardwaj.path.route.resources.ServicesResource.<init>(ServicesResource.java:66)
at in.nikhilbhardwaj.path.route.resources.ServicesResource.class(ServicesResource.java:56)
while locating in.nikhilbhardwaj.path.route.resources.ServicesResource
for parameter 0 at in.nikhilbhardwaj.path.alexa.intent.HelloWorldIntentAction.<init>(HelloWorldIntentAction.java:44)
while locating in.nikhilbhardwaj.path.alexa.intent.HelloWorldIntentAction
while locating in.nikhilbhardwaj.path.alexa.intent.IntentAction annotated with @com.google.inject.multibindings.Element(setName=,uniqueId=10, type=MAPBINDER, keyType=java.lang.String)
at in.nikhilbhardwaj.path.alexa.intent.IntentModule.configure(IntentModule.java:17) (via modules: in.nikhilbhardwaj.path.alexa.AlexaStarterApplicationModule -> in.nikhilbhardwaj.path.alexa.intent.IntentModule -> com.google.inject.multibindings.MapBinder$RealMapBinder)
while locating java.util.Map<java.lang.String, in.nikhilbhardwaj.path.alexa.intent.IntentAction>
for parameter 0 at in.nikhilbhardwaj.path.alexa.intent.IntentHandlerServiceImpl.<init>(IntentHandlerServiceImpl.java:16)
while locating in.nikhilbhardwaj.path.alexa.intent.IntentHandlerServiceImpl
while locating in.nikhilbhardwaj.path.alexa.intent.IntentHandlerService
for parameter 0 at in.nikhilbhardwaj.path.alexa.AlexaStarterSpeechlet.<init>(AlexaStarterSpeechlet.java:26)
while locating in.nikhilbhardwaj.path.alexa.AlexaStarterSpeechlet
Caused by: java.lang.NullPointerException
at java.io.Reader.<init>(Reader.java:78)
at java.io.InputStreamReader.<init>(InputStreamReader.java:113)
at in.nikhilbhardwaj.path.route.resources.ServicesResource.initializeServiceIds(ServicesResource.java:88)
at in.nikhilbhardwaj.path.route.resources.ServicesResource.<init>(ServicesResource.java:68)
at in.nikhilbhardwaj.path.route.resources.ServicesResource$$FastClassByGuice$$3d6e91ec.newInstance(<generated>)
我已经classpath按照 …
我有一个非常简单的方法，它使用该getclass().getResourceAsStream()方法来读取文件。但是它总是返回 null，我无法弄清楚出了什么问题。这是我的一段代码。
InputStream sw = getClass().getResourceAsStream("/filename.txt");
BufferedReader bf = new BufferedReader( new InputStreamReader(sw));
sw始终保持为空。该文件filename.txt存在于我的项目的根目录中。
编辑：我找到了原因。我意识到我是从 Eclipse 运行我的项目，并且该项目不是我 PC 上类路径的一部分。但是，如果我将程序打包为 jar 文件然后运行它，则 jar 文件中的文件被视为资源，可以使用 getResourceAsStream() 方法读取。
您好我在Eclipse下编写了这样的函数:
public static ArrayList<String> getMails(){
ArrayList<String> mails = new ArrayList<String>();
try{
FileInputStream fstream = new FileInputStream("mails.txt");
DataInputStream in = new DataInputStream(fstream);
BufferedReader br = new BufferedReader(new InputStreamReader(in));
String strLine;
while ((strLine = br.readLine()) != null) {
mails.add(strLine.trim());
}
in.close();
}catch (Exception e){//Catch exception if any
System.err.println("Error: " + e.getMessage());
}
return mails;
}
mails.txt文件位于workspace/projectname下,我想将此项保存在workspace/projectname/bin /目录下,作为相对路径,因此每当我将workspace/projectname // bin目录复制到其他位置或计算机时,让它工作.但是,当我尝试这个时,我得到"FileNotFound"异常.我怎样才能解决这个问题 ？谢谢
我收到以下错误
java.io.FileNotFoundException: in.txt, (The system cannot find the file specified)
at java.io.FileInputStream.open(Native Method)
at java.io.FileInputStream.<init>(Unknown Source)
at java.io.FileInputStream.<init>(Unknown Source)
at java.io.FileReader.<init>(Unknown Source)
at FileTest.main(FileTest.java:50)
我确定我已经在src,bin和root目录下创建了一个in.txt文件.我也尝试在主参数中指定完整目录,但仍然无法正常工作.为什么Eclipse不接受它？
import java.io.*;
public class FileTest {
public static void main(String[] args) {
try {
String inFileName = args[0];
String outFileName = args[1];
BufferedReader ins = new BufferedReader(new FileReader(inFileName));
BufferedReader con = new BufferedReader(new InputStreamReader(System.in));
PrintWriter outs = new PrintWriter(new FileWriter(outFileName));
String first = ins.readLine(); // read from file
while (first != null) {
System.out.print("Type in …我想打开一个文件并返回其内容.虽然它与想要打开文件的类在同一目录中,但找不到该文件.如果你能帮助我解决这个问题会很酷.
这是代码:
@GET @Produces("text/html") @Path("/{partNO}/") @Consumes("text/html")
public String getPartNoResponseHTML(@PathParam("partNO") String parID) throws WebApplicationException {
PartNoTemplate partNo = getPartNoResponse(parID);
String result = "";
try {
result = readFile(PART_NO_TEMPLATE_FILE);
} catch (FileNotFoundException e) {
e.printStackTrace(System.out);
return e.getMessage() + e.toString();
// throw new WebApplicationException(Response.Status.NOT_FOUND);
} finally {
result = result.replace("{partNO}", parID);
result = result.replace("{inputFormat}", partNo.getFormat().toString());
}
return result;
}
我猜它找不到文件,因为它在tomcat上运行.我也在使用Jersey和JAX-RS.谢谢您的帮助,
马克西