1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
|
/*******************************************************************************
* Copyright (c) 2007 IBM Corporation and others. All rights reserved. This
* program and the accompanying materials are made available under the terms of
* the Eclipse Public License v1.0 which accompanies this distribution, and is
* available at http://www.eclipse.org/legal/epl-v10.html
*
* Contributors: IBM Corporation - initial API and implementation
******************************************************************************/
package org.eclipse.equinox.prov.query;
import java.util.ArrayList;
import java.util.Iterator;
import org.eclipse.core.runtime.IProgressMonitor;
import org.eclipse.core.runtime.NullProgressMonitor;
import org.eclipse.equinox.internal.prov.metadata.Messages;
import org.eclipse.equinox.prov.metadata.*;
import org.eclipse.osgi.service.resolver.VersionRange;
public class Query {
public static boolean match(IInstallableUnit source, String id, VersionRange range, RequiredCapability[] requirements, boolean and) {
if (id != null && !source.getId().equals(id))
return false;
if (range != null && !range.isIncluded(source.getVersion()))
return false;
if (requirements == null)
return true;
// if any of the requirements are met by any of the capabilities of the source...
for (int k = 0; k < requirements.length; k++) {
boolean valid = false;
ProvidedCapability[] capabilities = source.getProvidedCapabilities();
for (int j = 0; j < capabilities.length; j++)
//TODO Need to deal with option and multiplicity flag
if ((requirements[k].getName().equals(capabilities[j].getName()) && requirements[k].getNamespace().equals(capabilities[j].getNamespace()) && (requirements[k].getRange() == null ? true : requirements[k].getRange().isIncluded(capabilities[j].getVersion())))) {
valid = true;
}
// if we are OR'ing then the first time we find a requirement that is met, return success
if (valid && !and)
return true;
// if we are AND'ing then the first time we find a requirement that is NOT met, return failure
if (!valid && and)
return false;
}
// if we get past the requirements check and we are AND'ing then return true
// since all requirements must have been met. If we are OR'ing then return false
// since none of the requirements were met.
return and;
}
public static IInstallableUnit[] query(IQueryable[] sources, String id, VersionRange range, RequiredCapability[] requirements, boolean and, IProgressMonitor progress) {
if (progress == null)
progress = new NullProgressMonitor();
progress.beginTask(Messages.QUERY_PROGRESS, sources.length);
ArrayList result = new ArrayList();
for (int i = 0; i < sources.length; i++) {
IQueryable source = sources[i];
IInstallableUnit[] list = source.query(id, range, requirements, and, progress);
for (int j = 0; j < list.length; j++)
result.add(list[j]);
progress.worked(1);
}
progress.done();
return (IInstallableUnit[]) result.toArray(new IInstallableUnit[result.size()]);
}
public static Iterator getIterator(IQueryable[] sources, String id, VersionRange range, RequiredCapability[] requirements, boolean and) {
Iterator[] result = new Iterator[sources.length];
for (int i = 0; i < sources.length; i++)
result[i] = sources[i].getIterator(id, range, requirements, and);
return new CompoundIterator(result);
}
}
|
