Openjpa Error In Opening Zip File
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Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community java.util.zip.zipexception error in opening zip file linux of 6.2 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up java.util.zip.ZipException: error in opening zip file up vote 37 down vote favorite 6 I have a Jar java.util.zip.zipexception: zip file is empty file, which contains other nested Jars. When I invoke the new JarFile() constructor on this file, I get an exception which says: java.util.zip.ZipException: error in opening zip file When I manually unzip the contents of this Jar file and zip it up again, it works fine. Also, please note that this exception is seen only on WebSphere 6.1.0.7 and higher versions. The same thing works fine on tomcat and WebLogic. Also, when
Java.util.zip.zipexception Error In Opening Zip File Eclipse
I use JarInputStream instead of JarFile, I am able to read the contents of the Jar file without any exceptions. Please let me know if you have any ideas on how this can be fixed. Thanks, Sandhya java share|improve this question edited Aug 23 at 21:02 Petro 1,2591623 asked Nov 28 '08 at 7:12 user41536 239156 2 Thanks for the hint about rezipping the file -- that fixed for me. –Bryan Larsen May 4 '10 at 22:24 I've had this problem on Mac when Windows and Linux worked just fine. Using JarInputStream fixed the problem for me. –Boris van Schooten Dec 20 '11 at 9:47 I have faced the same problem on Tomcat Start UP [catalina.properties]: org.apache.catalina.startup.TldConfig tldScanJar WARNING: Failed to process JAR [jar:../opensaml.jar!/] for TLD files ZipException to resolve this problem add opensaml.~.jar into Application lib folder. –Yash Apr 11 at 12:19 add a comment| 8 Answers 8 active oldest votes up vote 11 down vote Make sure your jar file is not corrupted. If it's corrupted or not able to unzip, this error will occur. share|improve this answer edited Jul 28 at 18:56 FrankerZ 5,13141447 answered Sep 27 '10 at 7:14 arulraj.net 1,26711628 +1 More often than not this is the reason. –JamesC Oct
due to the jar not having he agent. Was using openjpa-2.3.0.jar and should have used openjpa-all-2.3.0.jar - Not the path to the jar being wrong. java.lang.runtimeexception java.util.zip.zipexception error in opening zip file - Although its hard to tell from the error message. java.util.zip.zipexception: error in opening zip file at java.util.zip.zipfile.open(native method) VM Optoin in intellij was -javaagent:/full/linux/path/openjpa-2.3.0.jar Here is the full path vm optoins with openjpa-all-2.3.0.jar: -javaagent:/full/linux/path/MyProject/target/MyProject-0.0.1-SNAPSHOT/WEB-INF/lib/openjpa-all-2.3.0.jar Posted by Douglas
Java.util.zip.zipexception Error In Opening Zip File Weblogic
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This Site Careers Other all forums Forum: Object Relational Mapping JPA file accessed from https://coderanch.com/t/562272/ORM/databases/JPA-file-accessed-ear-files different .ear files deployed on weblogic server chandr prakash Ranch Hand Posts: 32 posted 4 years ago I am writing a application in which following JPA file,will be accessed from different application. To do this i created .jar file of following java file zip file and added in classpath of server. package entity.library; import java.io.Serializable; import javax.persistence.*; @Entity @Table(name="STUDENT_DETAILS") @NamedQueries({ @NamedQuery(name="findById", query="select t from TestJPA t where t.studentid = ?1"), @NamedQuery(name="findAll", query="select t from TestJPA t"), }) public class TestJPA implements Serializable { private static final long serialVersionUID = 1L; error in opening @Id @Column(name="STUDENTID") private String studentid; @Column(name="STUDENTNAME") private String studentname; public TestJPA() { } public String getStudentid() { return this.studentid; } public void setStudentid(String studentid) { this.studentid = studentid; } public String getStudentname() { return this.studentname; } public void setStudentname(String studentname) { this.studentname = studentname; } } I created testclient.jar(entity.library.TestJPA.class) and added this jar in classpath of weblogic server.(I can see entry of jar in classpath of weblogic server.) I am accessing this JPA file from .ear module. My persistence.xml