大约有 10,000 项符合查询结果(耗时:0.0237秒) [XML]

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Git: “please tell me who you are” error

...s that I bootstrap together using Chef + some ad-hoc bash scripts. The problem is, when I want to run an update on one of these app servers, I get: ...
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Multiple dex files define Landroid/support/v4/accessibilityservice/AccessibilityServiceInfoCompat

...a module from a module dependency: You have to turn your compile project(':foo') into compile(project(':foo')) { exclude module: 'support-v4' }. Note the parenthesis. – espinchi Nov 22 '14 at 17:28 ...
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What is the best Battleship AI?

Battleship! 25 Answers 25 ...
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GIT commit as different user without email / or only email

... name family Set a Git email: $ git config --global user.email email@foo.com Confirm that you have set the Git email correctly: $ git config --global user.email email@foo.com share | imp...
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What's the difference between __PRETTY_FUNCTION__, __FUNCTION__, __func__?

... undecorated, but still adorned, version of the name. For a method named "foo", gcc will give you "foo", VC will give "my_namespace::my_class::foo". – Adrian McCarthy Jul 1 '15 at 15:56 ...
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Is there a macro recorder for Eclipse? [closed]

Is there a good Eclipse plugin for recording and playing back macros? 9 Answers 9 ...
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Get name of property as a string

...eClass { public static string SomeProperty { get { return "Foo"; } } } public class RemoteMgr { public static void ExposeProperty<T>(Expression<Func<T>> property) { var expression = GetMemberInfo(property); string path = string.Concat(ex...
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git command to move a folder inside another

...ails with error fatal: bad source, source=oldFolderName/somepath/somefile.foo, destination=newFolderName/somepath/somefile.foo if there are any unadded files, so I just found out. share | improve...
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RVM is not a function, selecting rubies with 'rvm use …' will not work

List the ruby versions 11 Answers 11 ...
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Pass a data.frame column name to a function

...ock here is that a natural (but incorrect) attempt often looks like this: foo <- function(df,col_name,col1,col2){ df$col_name <- df$col1 + df$col2 df } #Call foo() like this: foo(dat,z,x,y) The problem here is that df$col1 doesn't evaluate the expression col1. It simply loo...