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remote: actually check if remote exits

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When converting the git remote command to a builtin in 211c89 ("Make
git-remote a builtin"), a few calls to check if a remote exists were
converted from:
       if (!exists $remote->{$name}) {
       	  [...]
to:
       remote = remote_get(argv[1]);
       if (!remote)
          [...]

The new check is not quite correct, because remote_get() never returns
NULL if a name is given.  This leaves us with the somewhat cryptic error
message "error: Could not remove config section 'remote.test'", if we
are trying to remove a remote that does not exist, or a similar error if
we try to rename a remote.

Use the remote_is_configured() function to check whether the remote
actually exists, and die with a more sensible error message ("No such
remote: $remotename") instead if it doesn't.

Signed-off-by: Thomas Gummerer <t.gummerer@gmail.com>
Reviewed-by: Jeff King <peff@peff.net>
Signed-off-by: Junio C Hamano <gitster@pobox.com>
This commit is contained in:
Thomas Gummerer
2016-02-16 10:47:51 +01:00
committed by Junio C Hamano
parent 674468b364
commit cc8e538d45
2 changed files with 20 additions and 2 deletions
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4
builtin/remote.c
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@ -634,7 +634,7 @@ static int mv(int argc, const char **argv)
rename.remote_branches = &remote_branches;
oldremote = remote_get(rename.old);
if (!oldremote)
if (!remote_is_configured(oldremote))
die(_("No such remote: %s"), rename.old);
if (!strcmp(rename.old, rename.new) && oldremote->origin != REMOTE_CONFIG)
@ -773,7 +773,7 @@ static int rm(int argc, const char **argv)
usage_with_options(builtin_remote_rm_usage, options);
remote = remote_get(argv[1]);
if (!remote)
if (!remote_is_configured(remote))
die(_("No such remote: %s"), argv[1]);
known_remotes.to_delete = remote;