Using Java 6 or later, the classpath option supports wildcards. Note the following:
- Use straight quotes (
"
)
- Use
*
, not *.jar
Windows
java -cp "Test.jar;lib/*" my.package.MainClass
Unix
java -cp "Test.jar:lib/*" my.package.MainClass
This is similar to Windows, but uses :
instead of ;
. If you cannot use wildcards, bash
allows the following syntax (where lib
is the directory containing all the Java archive files):
java -cp "$(printf %s: lib/*.jar)"
(Note that using a classpath is incompatible with the -jar
option. See also: Execute jar file with multiple classpath libraries from command prompt)
Understanding Wildcards
From the Classpath document:
Class path entries can contain the basename wildcard character *
, which is considered equivalent to specifying a list of all the files
in the directory with the extension .jar
or .JAR
. For example, the
class path entry foo/*
specifies all JAR files in the directory named
foo. A classpath entry consisting simply of *
expands to a list of all
the jar files in the current directory.
A class path entry that contains *
will not match class files. To
match both classes and JAR files in a single directory foo, use either
foo;foo/*
or foo/*;foo
. The order chosen determines whether the
classes and resources in foo
are loaded before JAR files in foo
, or
vice versa.
Subdirectories are not searched recursively. For example, foo/*
looks
for JAR files only in foo
, not in foo/bar
, foo/baz
, etc.
The order in which the JAR files in a directory are enumerated in the
expanded class path is not specified and may vary from platform to
platform and even from moment to moment on the same machine. A
well-constructed application should not depend upon any particular
order. If a specific order is required then the JAR files can be
enumerated explicitly in the class path.
Expansion of wildcards is done early, prior to the invocation of a
program's main method, rather than late, during the class-loading
process itself. Each element of the input class path containing a
wildcard is replaced by the (possibly empty) sequence of elements
generated by enumerating the JAR files in the named directory. For
example, if the directory foo
contains a.jar
, b.jar
, and c.jar
, then
the class path foo/*
is expanded into foo/a.jar;foo/b.jar;foo/c.jar
,
and that string would be the value of the system property
java.class.path
.
The CLASSPATH
environment variable is not treated any differently from
the -classpath
(or -cp
) command-line option. That is, wildcards are
honored in all these cases. However, class path wildcards are not
honored in the Class-Path jar-manifest
header.
Note: due to a known bug in java 8, the windows examples must use a backslash preceding entries with a trailing asterisk: https://bugs.openjdk.java.net/browse/JDK-8131329
There's another solution that also works.
- Go to the Build Path settings in the project properties.
- Remove the JRE System Library
- Add it back; Select "Add Library" and select the JRE System Library. The default worked for me.
This works because you have multiple classes in different jar files. Removing and re-adding the JRE lib will make the right classes be first.
If you want a fundamental solution make sure you exclude the jar files with the same classes.
For me I have: javax.xml.soap.SOAPPart
in three different jars: axis-saaj-1.4.jar
, saaj-api-1.3.jar
and the rt.jar
Best Answer
This answer is same as Laura's answer , however, in new eclipse versions you will not be able to see a "create project from existing source" option.
Hence you can do this instead:
Goto File > New > Project
Select the type of project, click Next
Uncheck Use default location
Click on Browse to navigate to your source folder, or type in the path to your source
Click Finish
Taken from this discussion forum in eclipse.org