better project name for temp project + pass dummy status

This commit is contained in:
Maxim.Mossienko
2014-01-24 14:27:37 +01:00
parent 10f6a0f9eb
commit 4aca40ba36
@@ -97,6 +97,9 @@ public class PlatformProjectOpenProcessor extends ProjectOpenProcessor {
@Nullable ProjectOpenedCallback callback,
final boolean isReopen) {
VirtualFile baseDir = virtualFile;
boolean dummyProject = false;
String dummyProjectName = null;
if (!baseDir.isDirectory()) {
baseDir = virtualFile.getParent();
while (baseDir != null) {
@@ -108,8 +111,10 @@ public class PlatformProjectOpenProcessor extends ProjectOpenProcessor {
if (baseDir == null) { // no reasonable directory -> create new temp one or use parent
if (Registry.is("ide.open.file.in.temp.project.dir")) {
try {
File directory = FileUtil.createTempDirectory(virtualFile.getName(), null, true);
dummyProjectName = virtualFile.getPath();
File directory = FileUtil.createTempDirectory(dummyProjectName, null, true);
baseDir = LocalFileSystem.getInstance().refreshAndFindFileByIoFile(directory);
dummyProject = true;
} catch (IOException ex) {
LOG.error(ex);
}
@@ -178,7 +183,8 @@ public class PlatformProjectOpenProcessor extends ProjectOpenProcessor {
}
if (project == null) {
project = projectManager.newProject(projectDir.getParentFile().getName(), projectDir.getParent(), true, false);
String projectName = dummyProject ? dummyProjectName : projectDir.getParentFile().getName();
project = projectManager.newProject(projectName, projectDir.getParent(), true, dummyProject);
}
if (project == null) return null;