repository hg clone not found,command returned code 255
Hello everyone.
I'm new to hgLab.I just installed the server side site.Created a Project and a repository.
But when I try to do a CLONE the repository following the url.
An error happen:
repository hg clone not found... command
returned code 255
Here is my clone url:
hg clone http://[email blocked]:85/hg/slug1/r1
I found that there is a prefix with an @ signal before my server url.
Is that prefix correct?
If not how to remove that from my clone url?
Thanks in advance.
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Error.png 4.42 KB
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CloneUrl.png 6.96 KB
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Support Staff 1 Posted by Anton Gogolev on 07 Nov, 2013 02:55 PM
This:
is not a Clone URL, but a full command that you're supposed to run from the command prompt. Your actual Clone URL ishttp://[email blocked]:85/hg/slug1/r1
.2 Posted by gqy117 on 07 Nov, 2013 03:10 PM
Hello Anton.
Thanks for the quick replay :)
I have run the command in the server by using cmd.exe
It seems it works in some degree.
And run the following command
cd R1
(echo. && echo [ui]) >> ..hg\hgrc (echo username = gqy117 ^<[email blocked]^> && echo.) >> ..hg\hgrc echo [auth] >> ..hg\hgrc
(echo hglab.prefix = http://gqy117.cloudapp.net:85/hg && echo hglab.username = gqy117) >> ..hg\hgrc
After that it seems the Clone Instructions still display the same thing as before.
I'm just not sure what should I do next?
Thanks.
Support Staff 3 Posted by Anton Gogolev on 07 Nov, 2013 03:16 PM
The second set of commands is entirely optional just writes a bunch of
stuff to .hg/hgrc, which is a configuration file for your local Mercurial
repository. It sets up your username, authentication prefix and other
settings. These commands do not affect Cloning Instructions page in any
way.
Now that you've successfully cloned a repository, you can start working
with it as usual: committing, viewing logs, diffing, pulling, pushing, etc.
If you're new to Mercurial, here's an excellent tutorial to get you
started: http://hginit.com/ .
4 Posted by gqy117 on 07 Nov, 2013 03:20 PM
Oh ,I've figured it out.
I should run this command in the client side,using cmd.exe
Thank you for your kindly help!
Support Staff 5 Posted by Anton Gogolev on 07 Nov, 2013 03:24 PM
Glad I could help! If you need further assistance, feel free to contact me.