我有2个项目.第一个只生成一个DLL.
第二个需要第一个DLL.但是当我从DLL调用一个方法时,我有一个FileNotFoundException,带有以下消息:
BDD,Version = 1.0.0.0,Culture = neutral,PublicKeyToken = null
我确定该文件存在.哪里我错了？
我尝试启动Spring Web应用程序时收到以下错误消息:
2012-04-12 13:53:20,491 ERROR [org.springframework.web.servlet.DispatcherServlet] -
Context initialization failed
org.springframework.beans.factory.BeanInitializationException: Could not load properties; nested exception is java.io.FileNotFoundException: class path resource [timex-servlet.properties] cannot be opened because it does not exist
java.io.FileNotFoundException: class path resource [timex-servlet.properties] cannot be opened because it does not exist
at org.springframework.core.io.ClassPathResource.getInputStream(ClassPathResource.java:137)
at org.springframework.core.io.support.PropertiesLoaderSupport.loadProperties(PropertiesLoaderSupport.java:172)
我正在通过eclipse运行Tomcat(版本6.x).我尝试将timex-servlet.properties放在以下目录中,但无济于事:
WebContent\WEB-INF
WebContent\WEB-INF\classes
WebContent\
以下是timex-servlet.xml中timex-servlet.properties的引用:
<bean id="placeholderConfig"
class="org.springframework.beans.factory.config.PropertyPlaceholderConfigurer">
<property name="location"
value="timex-servlet.properties" />
</bean>
有几个SO线程处理相同的消息,表示放置类路径:在属性文件引用前面.所以我尝试了以下,但也没有用:
<bean id="placeholderConfig"
class="org.springframework.beans.factory.config.PropertyPlaceholderConfigurer">
<property name="location"
value="classpath:timex-servlet.properties" />
</bean>
我想写一些文件,像这样:
FileWriter fw = new FileWriter("somefile.txt", true);
它由单个进程启动时正常工作.像这样:
java -jar XXXXXXX.jar
但是当由另一个进程调用它时,将抛出IOException.例:
java.io.FileNotFoundException: 'somefile.txt' (No such file or directory)
at java.io.FileOutputStream.openAppend(Native Method)
at java.io.FileOutputStream.<init>(FileOutputStream.java:192)
at java.io.FileOutputStream.<init>(FileOutputStream.java:116)
at java.io.FileWriter.<init>(FileWriter.java:61)
我写了一个Puppet来在运行CentOS 6.2的虚拟机上安装Play 2.1.1.
Puppet执行了以下两个命令:
/usr/bin/wget http://downloads.typesafe.com/play/2.1.1/play-2.1.1.zip
/usr/bin/unzip play-2.1.1.zip
当我play在包中运行命令时,它显示以下错误消息:
java.io.FileNotFoundException: ~/play-2.1.1/framework/sbt/boot/update.log (No such file or directory)
at java.io.FileOutputStream.open(Native Method)
at java.io.FileOutputStream.<init>(FileOutputStream.java:209)
at java.io.FileOutputStream.<init>(FileOutputStream.java:160)
at java.io.FileWriter.<init>(FileWriter.java:90)
at xsbt.boot.Update.<init>(Checks.java:51)
at xsbt.boot.Launch.update(Launch.scala:275)
at xsbt.boot.Launch$$anonfun$jnaLoader$1.apply(Launch.scala:120)
at scala.Option.getOrElse(Option.scala:108)
at xsbt.boot.Launch.jnaLoader$2f324eef(Launch.scala:115)
at xsbt.boot.Launch.<init>(Launch.scala:94)
at xsbt.boot.Launcher$.apply(Launch.scala:290)
at xsbt.boot.Launch$.apply(Launch.scala:16)
at xsbt.boot.Boot$.runImpl(Boot.scala:31)
at xsbt.boot.Boot$.main(Boot.scala:20)
at xsbt.boot.Boot.main(Boot.scala)
Error during sbt execution: java.io.FileNotFoundException: ~/play-2.1.1/framework/sbt/boot/update.log (No such file or directory)
我没有更改目录中的单行代码play-2.1.1.我搜索了错误消息,但没有找到任何有用的答案.任何人都可以指出问题是什么？
java filenotfoundexception sbt playframework playframework-2.0
我浪费了很多时间来解决这个问题.但不能.可能很简单.
应用程序的jar文件无法加载log4j.xml文件,该文件是log4j的配置文件.
这是我正在使用的代码
import org.apache.log4j.xml.DOMConfigurator;
public class LoggerConfig {
public void configLogger()
{
DOMConfigurator.configure("log4j.xml");
}
}
它是一个maven项目.我的目标结构是:
src
main
java
| com
| | my
| | | abc
| | | | test
| | | | | LoggerConfig.java
resource
| com
| | my
| | | abc
| | | | test
| | | | | log4j.xml
我的classpath文件如下所示:
<?xml version="1.0" encoding="UTF-8"?>
<classpath>
<classpathentry kind="src" output="target/classes" path="src/main/java"/>
<classpathentry kind="src" output="target/classes" path="src/main/resources"/>
<classpathentry kind="con" path="org.eclipse.jdt.launching.JRE_CONTAINER/org.eclipse.jdt.internal.debug.ui.launcher.StandardVMType/JavaSE-1.6"/>
<classpathentry kind="con" path="org.maven.ide.eclipse.MAVEN2_CLASSPATH_CONTAINER"/> …我正在开发一个依赖于某个数据文件的webapp,它在运行时会被混淆.没有数据文件,我似乎无法编译.为什么是这样？
这是我的core.clj
(def my-data (slurp "my-file.txt"))
然后当我尝试编译时:
$ lein ring war
我得到了这个例外
Exception in thread "main" java.io.FileNotFoundException: my-file.txt (No such file or directory), compiling:(core.clj:24:28)
我该怎么编译我的战争？我不需要在编译时篡改文件甚至检查是否存在.提前致谢!
[UPDATE]
这不是特定于war文件打包或响铃,例如:
(ns slurp-test.core
(:gen-class))
(def x (slurp "/tmp/foo.txt"))
(defn -main [& args]
(println x))
然后:
$ lein uberjar
Compiling slurp-test.core
(ns slurp-test.core
Exception in thread "main" java.io.FileNotFoundException: /tmp/foo.txt (No such file or directory), compiling:(core.clj:4:8)
我怎样才能解决这个问题？
我有一个文件夹，里面有不同的子文件夹。我必须遍历所有文件并检查 John 和 Jose 的出现并分别替换为 Mikel 和 Mourinho。
这是我用 Python 编写的脚本。它工作正常但是当我遇到一个.gif文件时它给了我一个错误并且它没有进一步迭代。
你能告诉我为什么吗？
错误是
Traceback (most recent call last):
File "C:\Users\sid\Desktop\script.py", line 33, in <module>
os.chmod(path ,stat.S_IWRITE)
FileNotFoundError: [WinError 2] The system cannot find the file specified:'C:\Users\sid\Desktop\test\\images/ds_dataobject.gif.bak'
我的代码：
import os,stat
import fileinput
import sys
rootdir ='C:\Users\spemmara\Desktop\test'
searchTerms={"John":"Mikel", "Jose":"Mourinho"}
def replaceAll(file,searchExp,replaceExp):
for line in fileinput.input(file, inplace=1):
if searchExp in line:
line = line.replace(searchExp,replaceExp)
sys.stdout.write(line)
for subdir, dirs, files in os.walk(rootdir):
for file in files:
path=subdir+'/'+file
print(path)
os.chmod(path ,stat.S_IWRITE)
for key,value …在我们的Android应用程序中,我们打开位于assets/config/notification.wav中的wav文件.要打开并播放声音,我们使用以下代码:
String soundClipPath = "config/notification.wav"
AssetFileDescriptor afd = mContext.getAssets().openFd(soundClipPath);
int soundID = mSoundPool.load(afd, 1);
这已经解决了大约一年的任何问题.突然间,这已停止工作了.我们还没有更新构建之间的任何依赖关系.我能找到的唯一区别是,构建工作是从去年开始.如果我们在netbeans中本地构建apk,但是当我们使用hudson构建apk时它不起作用.我试过以下没有运气:
使用上面的代码时,我们得到以下stacktrace:
java.io.FileNotFoundException: This file can not be opened as a file descriptor; it is probably compressed
at android.content.res.AssetManager.openAssetFd(Native Method)
at android.content.res.AssetManager.openFd(AssetManager.java:331)
...
尝试从res/raw文件夹加载时,以下内容:
android.content.res.Resources$NotFoundException: File res/raw/notification.wav from drawable resource ID #0x7f040000
at android.content.res.Resources.openRawResourceFd(Resources.java:981)
at android.media.SoundPool.load(SoundPool.java:191)
...
有人可以帮我们解决这个奇怪的问题....
最好的祝福,
亨里克
android filenotfoundexception android-assets android-maven-plugin
出于某种原因,我在fileNotFoundException阅读时都得到了一个.值得注意的是Toast打印"File exists!".我使用BufferedReader底部来测试文件的内容是否正确.
@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container,
Bundle savedInstanceState) {
View v = inflater.inflate(R.layout.fragment_calendar, container, false);
recipes = new ArrayMap<>();
filename = "calendar_recipes.txt";
bText= (EditText) v.findViewById(R.id.bEditText);
lText= (EditText) v.findViewById(R.id.lEditText);
dText= (EditText) v.findViewById(R.id.dEditText);
cal = (CalendarView) v.findViewById(R.id.calendarView);
date = cal.getDate();
File file = getActivity().getFileStreamPath(filename);
if(file.exists())
{
Toast.makeText(getActivity(), "File exists!", Toast.LENGTH_SHORT).show();
try
{
FileInputStream fileInputStream = new FileInputStream(getActivity().getFilesDir()+filename);
ObjectInputStream objectInputStream = new ObjectInputStream(fileInputStream);
Map recipes = (Map)objectInputStream.readObject();
}
catch(ClassNotFoundException | IOException | ClassCastException e) {
e.printStackTrace(); …我正在尝试使用Jexcel API创建一个Excel文件，并使用手机上的应用程序对其进行写入。当我运行该应用程序时，它会抛出一个FileNotFoundException。我什至尝试根据另一个问题的答案来创建文本文件，但是会引发相同的错误。我已在清单中授予了适当的权限，但我似乎仍然无法查明问题所在。请帮忙。
这是我的代码
public WritableWorkbook createWorkbook(String fileName){
//Saving file in external storage
File sdCard = Environment.getExternalStoragePublicDirectory(Environment.DIRECTORY_DOWNLOADS);
File directory = new File(sdCard.getAbsolutePath() + "/bills");
//create directory if not exist
if(!directory.isDirectory()){
directory.mkdirs();
}
//file path
file= new File(directory, fileName);
if (file.exists())
Log.e(taf,"file created");
else
Log.e(taf,"file not created");
WorkbookSettings wbSettings = new WorkbookSettings();
wbSettings.setLocale(new Locale("en", "EN"));
wbSettings.setUseTemporaryFileDuringWrite(true);
WritableWorkbook workbook;
workbook=null;
try {
workbook = Workbook.createWorkbook(file, wbSettings);
Log.i(taf,"workbook created");
//Excel sheet name. 0 represents first sheet
WritableSheet sheet = workbook.createSheet("MyShoppingList", 0);
try …