How to read zip file through getResourceAsStream with relative path?












0















I used to use this method to read text files in my maven's resources/ directory, so that I can use relative path:



public static BufferedReader fileReaderAsResource(String filePath) throws IOException {
InputStream is = Thread.currentThread().getContextClassLoader().getResourceAsStream(filePath);
if (is == null) {
throw new FileNotFoundException(" Not found: " + filePath);
}
return new BufferedReader(new InputStreamReader(is, DEFAULT_ENCODING));
}


Now I need to read zip file due to its size and I still want to use relative path to file in my "resources" directory. Is there a way to do this?



I have this method to read zip file, but it only reads in file through absolute path:



public static BufferedReader fileZipReader(String fileName) throws IOException, URISyntaxException {
URL zipUrl = IOUtils.class.getClassLoader().getResource(fileName);
File zipFile = new File(zipUrl.toURI());
ZipFile zip = new ZipFile(zipFile);
for (Enumeration e = zip.entries(); e.hasMoreElements(); ) {
ZipEntry zipEntry = (ZipEntry) e.nextElement();
if (!zipEntry.isDirectory()) {
return new BufferedReader(new InputStreamReader(zip.getInputStream(zipEntry)));
}
}
throw new FileNotFoundException("File not found: " + fileName);
}


How to read zip file through relative path to my standard maven's resources/ directory?










share|improve this question























  • File zipFile = new File(zipUrl.toURI()); A File cannot represent an URL pointing to a Zip entry. If code truly needs a file, it will be necessary to extract the information to a file, first.

    – Andrew Thompson
    Nov 20 '18 at 2:07


















0















I used to use this method to read text files in my maven's resources/ directory, so that I can use relative path:



public static BufferedReader fileReaderAsResource(String filePath) throws IOException {
InputStream is = Thread.currentThread().getContextClassLoader().getResourceAsStream(filePath);
if (is == null) {
throw new FileNotFoundException(" Not found: " + filePath);
}
return new BufferedReader(new InputStreamReader(is, DEFAULT_ENCODING));
}


Now I need to read zip file due to its size and I still want to use relative path to file in my "resources" directory. Is there a way to do this?



I have this method to read zip file, but it only reads in file through absolute path:



public static BufferedReader fileZipReader(String fileName) throws IOException, URISyntaxException {
URL zipUrl = IOUtils.class.getClassLoader().getResource(fileName);
File zipFile = new File(zipUrl.toURI());
ZipFile zip = new ZipFile(zipFile);
for (Enumeration e = zip.entries(); e.hasMoreElements(); ) {
ZipEntry zipEntry = (ZipEntry) e.nextElement();
if (!zipEntry.isDirectory()) {
return new BufferedReader(new InputStreamReader(zip.getInputStream(zipEntry)));
}
}
throw new FileNotFoundException("File not found: " + fileName);
}


How to read zip file through relative path to my standard maven's resources/ directory?










share|improve this question























  • File zipFile = new File(zipUrl.toURI()); A File cannot represent an URL pointing to a Zip entry. If code truly needs a file, it will be necessary to extract the information to a file, first.

    – Andrew Thompson
    Nov 20 '18 at 2:07
















0












0








0








I used to use this method to read text files in my maven's resources/ directory, so that I can use relative path:



public static BufferedReader fileReaderAsResource(String filePath) throws IOException {
InputStream is = Thread.currentThread().getContextClassLoader().getResourceAsStream(filePath);
if (is == null) {
throw new FileNotFoundException(" Not found: " + filePath);
}
return new BufferedReader(new InputStreamReader(is, DEFAULT_ENCODING));
}


Now I need to read zip file due to its size and I still want to use relative path to file in my "resources" directory. Is there a way to do this?



I have this method to read zip file, but it only reads in file through absolute path:



public static BufferedReader fileZipReader(String fileName) throws IOException, URISyntaxException {
URL zipUrl = IOUtils.class.getClassLoader().getResource(fileName);
File zipFile = new File(zipUrl.toURI());
ZipFile zip = new ZipFile(zipFile);
for (Enumeration e = zip.entries(); e.hasMoreElements(); ) {
ZipEntry zipEntry = (ZipEntry) e.nextElement();
if (!zipEntry.isDirectory()) {
return new BufferedReader(new InputStreamReader(zip.getInputStream(zipEntry)));
}
}
throw new FileNotFoundException("File not found: " + fileName);
}


How to read zip file through relative path to my standard maven's resources/ directory?










share|improve this question














I used to use this method to read text files in my maven's resources/ directory, so that I can use relative path:



public static BufferedReader fileReaderAsResource(String filePath) throws IOException {
InputStream is = Thread.currentThread().getContextClassLoader().getResourceAsStream(filePath);
if (is == null) {
throw new FileNotFoundException(" Not found: " + filePath);
}
return new BufferedReader(new InputStreamReader(is, DEFAULT_ENCODING));
}


Now I need to read zip file due to its size and I still want to use relative path to file in my "resources" directory. Is there a way to do this?



I have this method to read zip file, but it only reads in file through absolute path:



public static BufferedReader fileZipReader(String fileName) throws IOException, URISyntaxException {
URL zipUrl = IOUtils.class.getClassLoader().getResource(fileName);
File zipFile = new File(zipUrl.toURI());
ZipFile zip = new ZipFile(zipFile);
for (Enumeration e = zip.entries(); e.hasMoreElements(); ) {
ZipEntry zipEntry = (ZipEntry) e.nextElement();
if (!zipEntry.isDirectory()) {
return new BufferedReader(new InputStreamReader(zip.getInputStream(zipEntry)));
}
}
throw new FileNotFoundException("File not found: " + fileName);
}


How to read zip file through relative path to my standard maven's resources/ directory?







java maven zip inputstream






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 20 '18 at 1:51









user697911user697911

3,208113472




3,208113472













  • File zipFile = new File(zipUrl.toURI()); A File cannot represent an URL pointing to a Zip entry. If code truly needs a file, it will be necessary to extract the information to a file, first.

    – Andrew Thompson
    Nov 20 '18 at 2:07





















  • File zipFile = new File(zipUrl.toURI()); A File cannot represent an URL pointing to a Zip entry. If code truly needs a file, it will be necessary to extract the information to a file, first.

    – Andrew Thompson
    Nov 20 '18 at 2:07



















File zipFile = new File(zipUrl.toURI()); A File cannot represent an URL pointing to a Zip entry. If code truly needs a file, it will be necessary to extract the information to a file, first.

– Andrew Thompson
Nov 20 '18 at 2:07







File zipFile = new File(zipUrl.toURI()); A File cannot represent an URL pointing to a Zip entry. If code truly needs a file, it will be necessary to extract the information to a file, first.

– Andrew Thompson
Nov 20 '18 at 2:07














1 Answer
1






active

oldest

votes


















0














You can wrap an InputStream with a ZipInputStream, i.e. :



InputStream is = Thread.currentThread().getContextClassLoader().getResourceAsStream(filePath);
if (is == null) {
throw new FileNotFoundException(" Not found: " + filePath);
}
ZipInputStream zis = new ZipInputStream(is);


EDIT:



Use the method above named as "fileReaderZipAsResource", I read the file normally:



try {
BufferedReader br = fileReaderZipAsResource(qaFilePath);
String line;
while ((line = br.readLine()) != null) {
if (line.isEmpty()) {
throw new RuntimeException("Invalid entry ... 2");
}
line = line.trim();
textKGKB.add(line);
}
} catch (IOException ioe) {
ioe.printStackTrace();
}


But the debugging shows that the program doesn't enter the loop. It simply pass by the loop and continue the logic, without throwing an exception either. My text file is a 4-column text file delimited with tab key. I simply zip it and name it as xyy.zip, and pass it as parameter to the method above.



What's the problem? Does the wrapping of ZipInputStream really work?






share|improve this answer


























  • I tried this, and it didn't work. What I did is to simply zip the text file and use this method. Please see my edit.

    – user697911
    Nov 20 '18 at 4:11











  • Can you post your file?

    – John Camerin
    Nov 20 '18 at 13:22











Your Answer






StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53385124%2fhow-to-read-zip-file-through-getresourceasstream-with-relative-path%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0














You can wrap an InputStream with a ZipInputStream, i.e. :



InputStream is = Thread.currentThread().getContextClassLoader().getResourceAsStream(filePath);
if (is == null) {
throw new FileNotFoundException(" Not found: " + filePath);
}
ZipInputStream zis = new ZipInputStream(is);


EDIT:



Use the method above named as "fileReaderZipAsResource", I read the file normally:



try {
BufferedReader br = fileReaderZipAsResource(qaFilePath);
String line;
while ((line = br.readLine()) != null) {
if (line.isEmpty()) {
throw new RuntimeException("Invalid entry ... 2");
}
line = line.trim();
textKGKB.add(line);
}
} catch (IOException ioe) {
ioe.printStackTrace();
}


But the debugging shows that the program doesn't enter the loop. It simply pass by the loop and continue the logic, without throwing an exception either. My text file is a 4-column text file delimited with tab key. I simply zip it and name it as xyy.zip, and pass it as parameter to the method above.



What's the problem? Does the wrapping of ZipInputStream really work?






share|improve this answer


























  • I tried this, and it didn't work. What I did is to simply zip the text file and use this method. Please see my edit.

    – user697911
    Nov 20 '18 at 4:11











  • Can you post your file?

    – John Camerin
    Nov 20 '18 at 13:22
















0














You can wrap an InputStream with a ZipInputStream, i.e. :



InputStream is = Thread.currentThread().getContextClassLoader().getResourceAsStream(filePath);
if (is == null) {
throw new FileNotFoundException(" Not found: " + filePath);
}
ZipInputStream zis = new ZipInputStream(is);


EDIT:



Use the method above named as "fileReaderZipAsResource", I read the file normally:



try {
BufferedReader br = fileReaderZipAsResource(qaFilePath);
String line;
while ((line = br.readLine()) != null) {
if (line.isEmpty()) {
throw new RuntimeException("Invalid entry ... 2");
}
line = line.trim();
textKGKB.add(line);
}
} catch (IOException ioe) {
ioe.printStackTrace();
}


But the debugging shows that the program doesn't enter the loop. It simply pass by the loop and continue the logic, without throwing an exception either. My text file is a 4-column text file delimited with tab key. I simply zip it and name it as xyy.zip, and pass it as parameter to the method above.



What's the problem? Does the wrapping of ZipInputStream really work?






share|improve this answer


























  • I tried this, and it didn't work. What I did is to simply zip the text file and use this method. Please see my edit.

    – user697911
    Nov 20 '18 at 4:11











  • Can you post your file?

    – John Camerin
    Nov 20 '18 at 13:22














0












0








0







You can wrap an InputStream with a ZipInputStream, i.e. :



InputStream is = Thread.currentThread().getContextClassLoader().getResourceAsStream(filePath);
if (is == null) {
throw new FileNotFoundException(" Not found: " + filePath);
}
ZipInputStream zis = new ZipInputStream(is);


EDIT:



Use the method above named as "fileReaderZipAsResource", I read the file normally:



try {
BufferedReader br = fileReaderZipAsResource(qaFilePath);
String line;
while ((line = br.readLine()) != null) {
if (line.isEmpty()) {
throw new RuntimeException("Invalid entry ... 2");
}
line = line.trim();
textKGKB.add(line);
}
} catch (IOException ioe) {
ioe.printStackTrace();
}


But the debugging shows that the program doesn't enter the loop. It simply pass by the loop and continue the logic, without throwing an exception either. My text file is a 4-column text file delimited with tab key. I simply zip it and name it as xyy.zip, and pass it as parameter to the method above.



What's the problem? Does the wrapping of ZipInputStream really work?






share|improve this answer















You can wrap an InputStream with a ZipInputStream, i.e. :



InputStream is = Thread.currentThread().getContextClassLoader().getResourceAsStream(filePath);
if (is == null) {
throw new FileNotFoundException(" Not found: " + filePath);
}
ZipInputStream zis = new ZipInputStream(is);


EDIT:



Use the method above named as "fileReaderZipAsResource", I read the file normally:



try {
BufferedReader br = fileReaderZipAsResource(qaFilePath);
String line;
while ((line = br.readLine()) != null) {
if (line.isEmpty()) {
throw new RuntimeException("Invalid entry ... 2");
}
line = line.trim();
textKGKB.add(line);
}
} catch (IOException ioe) {
ioe.printStackTrace();
}


But the debugging shows that the program doesn't enter the loop. It simply pass by the loop and continue the logic, without throwing an exception either. My text file is a 4-column text file delimited with tab key. I simply zip it and name it as xyy.zip, and pass it as parameter to the method above.



What's the problem? Does the wrapping of ZipInputStream really work?







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 20 '18 at 4:11









user697911

3,208113472




3,208113472










answered Nov 20 '18 at 3:15









John CamerinJohn Camerin

436111




436111













  • I tried this, and it didn't work. What I did is to simply zip the text file and use this method. Please see my edit.

    – user697911
    Nov 20 '18 at 4:11











  • Can you post your file?

    – John Camerin
    Nov 20 '18 at 13:22



















  • I tried this, and it didn't work. What I did is to simply zip the text file and use this method. Please see my edit.

    – user697911
    Nov 20 '18 at 4:11











  • Can you post your file?

    – John Camerin
    Nov 20 '18 at 13:22

















I tried this, and it didn't work. What I did is to simply zip the text file and use this method. Please see my edit.

– user697911
Nov 20 '18 at 4:11





I tried this, and it didn't work. What I did is to simply zip the text file and use this method. Please see my edit.

– user697911
Nov 20 '18 at 4:11













Can you post your file?

– John Camerin
Nov 20 '18 at 13:22





Can you post your file?

– John Camerin
Nov 20 '18 at 13:22




















draft saved

draft discarded




















































Thanks for contributing an answer to Stack Overflow!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53385124%2fhow-to-read-zip-file-through-getresourceasstream-with-relative-path%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

鏡平學校

ꓛꓣだゔៀៅຸ໢ທຮ໕໒ ,ໂ'໥໓າ໼ឨឲ៵៭ៈゎゔit''䖳𥁄卿' ☨₤₨こゎもょの;ꜹꟚꞖꞵꟅꞛေၦေɯ,ɨɡ𛃵𛁹ޝ޳ޠ޾,ޤޒޯ޾𫝒𫠁သ𛅤チョ'サノބޘދ𛁐ᶿᶇᶀᶋᶠ㨑㽹⻮ꧬ꧹؍۩وَؠ㇕㇃㇪ ㇦㇋㇋ṜẰᵡᴠ 軌ᵕ搜۳ٰޗޮ޷ސޯ𫖾𫅀ल, ꙭ꙰ꚅꙁꚊꞻꝔ꟠Ꝭㄤﺟޱސꧨꧼ꧴ꧯꧽ꧲ꧯ'⽹⽭⾁⿞⼳⽋២៩ញណើꩯꩤ꩸ꩮᶻᶺᶧᶂ𫳲𫪭𬸄𫵰𬖩𬫣𬊉ၲ𛅬㕦䬺𫝌𫝼,,𫟖𫞽ហៅ஫㆔ాఆఅꙒꚞꙍ,Ꙟ꙱エ ,ポテ,フࢰࢯ𫟠𫞶 𫝤𫟠ﺕﹱﻜﻣ𪵕𪭸𪻆𪾩𫔷ġ,ŧآꞪ꟥,ꞔꝻ♚☹⛵𛀌ꬷꭞȄƁƪƬșƦǙǗdžƝǯǧⱦⱰꓕꓢႋ神 ဴ၀க௭எ௫ឫោ ' េㇷㇴㇼ神ㇸㇲㇽㇴㇼㇻㇸ'ㇸㇿㇸㇹㇰㆣꓚꓤ₡₧ ㄨㄟ㄂ㄖㄎ໗ツڒذ₶।ऩछएोञयूटक़कयँृी,冬'𛅢𛅥ㇱㇵㇶ𥄥𦒽𠣧𠊓𧢖𥞘𩔋цѰㄠſtʯʭɿʆʗʍʩɷɛ,əʏダヵㄐㄘR{gỚṖḺờṠṫảḙḭᴮᵏᴘᵀᵷᵕᴜᴏᵾq﮲ﲿﴽﭙ軌ﰬﶚﶧ﫲Ҝжюїкӈㇴffצּ﬘﭅﬈軌'ffistfflſtffतभफɳɰʊɲʎ𛁱𛁖𛁮𛀉 𛂯𛀞నఋŀŲ 𫟲𫠖𫞺ຆຆ ໹້໕໗ๆทԊꧢꧠ꧰ꓱ⿝⼑ŎḬẃẖỐẅ ,ờỰỈỗﮊDžȩꭏꭎꬻ꭮ꬿꭖꭥꭅ㇭神 ⾈ꓵꓑ⺄㄄ㄪㄙㄅㄇstA۵䞽ॶ𫞑𫝄㇉㇇゜軌𩜛𩳠Jﻺ‚Üမ႕ႌႊၐၸဓၞၞၡ៸wyvtᶎᶪᶹစဎ꣡꣰꣢꣤ٗ؋لㇳㇾㇻㇱ㆐㆔,,㆟Ⱶヤマފ޼ޝަݿݞݠݷݐ',ݘ,ݪݙݵ𬝉𬜁𫝨𫞘くせぉて¼óû×ó£…𛅑הㄙくԗԀ5606神45,神796'𪤻𫞧ꓐ㄁ㄘɥɺꓵꓲ3''7034׉ⱦⱠˆ“𫝋ȍ,ꩲ軌꩷ꩶꩧꩫఞ۔فڱێظペサ神ナᴦᵑ47 9238їﻂ䐊䔉㠸﬎ffiﬣ,לּᴷᴦᵛᵽ,ᴨᵤ ᵸᵥᴗᵈꚏꚉꚟ⻆rtǟƴ𬎎

Why https connections are so slow when debugging (stepping over) in Java?