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Don't force a bad permission message

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This commit is contained in:
RogueException
2015-12-28 16:21:55 -04:00
parent 5d25f98e9a
commit 8f4bbe66ea
1 changed files with 2 additions and 4 deletions
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6
src/Discord.Net.Commands/CommandService.cs
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@@ -9,8 +9,6 @@ namespace Discord.Commands
/// <summary> A Discord.Net client with extensions for handling common bot operations like text commands. </summary> /// <summary> A Discord.Net client with extensions for handling common bot operations like text commands. </summary>
public sealed partial class CommandService : IService public sealed partial class CommandService : IService
{ {
private const string DefaultPermissionError = "You do not have permission to access this command.";
private readonly CommandServiceConfig _config; private readonly CommandServiceConfig _config;
private readonly CommandGroupBuilder _root; private readonly CommandGroupBuilder _root;
private DiscordClient _client; private DiscordClient _client;
@@ -117,7 +115,7 @@ namespace Discord.Commands
string errorText; string errorText;
if (!command.CanRun(eventArgs.User, eventArgs.Channel, out errorText)) if (!command.CanRun(eventArgs.User, eventArgs.Channel, out errorText))
{ {
RaiseCommandError(CommandErrorType.BadPermissions, eventArgs, new Exception(errorText ?? DefaultPermissionError)); RaiseCommandError(CommandErrorType.BadPermissions, eventArgs, errorText != null ? new Exception(errorText) : null);
return; return;
} }
@@ -262,7 +260,7 @@ namespace Discord.Commands
StringBuilder output = new StringBuilder(); StringBuilder output = new StringBuilder();
string error; string error;
if (!command.CanRun(user, channel, out error)) if (!command.CanRun(user, channel, out error))
output.AppendLine(error ?? DefaultPermissionError); output.AppendLine(error ?? "You do not have permission to access this command.");
else else
ShowCommandHelpInternal(command, user, channel, output); ShowCommandHelpInternal(command, user, channel, output);
return (replyChannel ?? channel).SendMessage(output.ToString()); return (replyChannel ?? channel).SendMessage(output.ToString());