小编Gil*_*Gil的帖子

我刚刚成功创建了一个本地克隆(使用我的凭据),但是当我尝试"拉"时,我收到一条错误消息:(手动将repos地址更改为抽象地址)

http:_ // username @ url:port/git/proj/team_repos.git:未经授权

org.eclipse.jgit.api.errors.TransportException

这从日志中更详细:

java.version = 1.7.0_10 java.vendor = Oracle Corporation BootLoader常量:OS = linux,ARCH = x86_64,WS = gtk,NL = en_SG框架参数:-vm /usr/java/jdk1.7.0_10/jre/bin命令-line参数:-os linux -ws gtk -arch x86_64 -vm /usr/java/jdk1.7.0_10/jre/bin

错误Sun Apr 14 19:00:11 IDT 2013 http_:// username @ url:port/git/proj/team_repos.git:未授权

org.eclipse.jgit.api.errors.TransportException:http:// username @ url:port/git/proj/team_repos.git:org.eclipse.jgit.api.FetchCommand.call未授权(FetchCommand.java:137 )org.eclipse.jgit.api.PullCommand.call(PullCommand.java:245)org.eclipse.egit.core.op.PullOperation $ 1.run(PullOperation.java:90)org.eclipse.core.internal .resources.Workspace.run(Workspace.java:2344)org.eclipse.core.internal.resources.Workspace.run(Workspace.java:2326)org.eclipse.egit.core.op.PullOperation.execute(PullOperation) .java:121)org.eclipse.egit.ui.internal.pull.PullOperationUI.execute(PullOperationUI.java:115)at org.eclipse.egit.ui.internal.pull.PullOperationUI $ 1.run(PullOperationUI.java: 90)在org.eclipse.core.internal.jobs.Worker.run(Worker.java:53)引起:org.eclipse.jgit.errors.TransportException:http_:// username @ url:port/git/proj/team_repos.git:未在org.eclipse.jgit.transport.Transp的org.eclipse.jgit.transport.TransportHttp.connect(TransportHttp.java:476)上获得授权 在org.eclipse.jgit.transport.FetchProcess.executeImp(FetchProcess.java:136)的org.eclipse.jgit.transport.FetchProcess.execute(FetchProcess.java:122)的org.eope.openFetch(TransportHttp.java:305) org.eclipse.jgit.api.FetchCommand.call上的.eclipse.jgit.transport.Transport.fetch(Transport.java:1104)(FetchCommand.java:128)... 8更多

没有找到任何解决方案..任何人？

我正在寻找一个linux命令,它搜索文本文件中的字符串,并在文件中的每个出现时突出显示(颜色),没有发出文本行(如grep那样).

我正在尝试从列表中过滤项目,并且我收到语法错误: SyntaxError:语法无效

代码:

a['name'] = 'Dan'
b['name'] = 'Joe'

ppl = [a,b]
inputName = raw_input('Enter name:').strip()
person = [p in ppl if p['name']==inputName].pop()

任何的想法？

def fibonacci(num):
    a=0
    b=1
    for i in range(num):
        a, b=b, a+b
    print a

循环内的线如何工作？某种程度上a&b的价值观发生了变化,似乎可以理解......

编辑:由于某些原因我感到困惑,认为b = b的中间exp是新的东西...没有读得很好..它真的是(a,b)=(b,a + b)这是python中swap的基本形式(: