x/exp/cmd/gorelease: 测试发布失败

c86crjj0  于 4个月前  发布在  Go
关注(0)|答案(3)|浏览(41)
#!watchflakes
post <- pkg == "golang.org/x/exp/cmd/gorelease" && test == "TestRelease"

自动创建的问题以收集这些故障。
示例( log ):

--- FAIL: TestRelease (0.47s)
    --- FAIL: TestRelease/basic/v1_autobase_verify (1.01s)
        gorelease_test.go:405: unexpected error: error downloading module example.com/basic@v1.0.1: CreateFile C:\Users\gopher\AppData\Local\Temp\1\gorelease_test-gocache1462627586\pkg\mod\cache\download\example.com\basic\@v\v1.0.1.partial: Access is denied.

watchflakes

m4pnthwp

m4pnthwp1#

找到新的 Jmeter 板测试碎片:

#!watchflakes
post <- pkg == "golang.org/x/exp/cmd/gorelease" && test == "TestRelease"

2023-02-13 19:21 windows-amd64-longtest exp@5e25df02 go@4a1829b6 x/exp/cmd/gorelease.TestRelease ( log )

--- FAIL: TestRelease (0.47s)
    --- FAIL: TestRelease/basic/v1_autobase_verify (1.01s)
        gorelease_test.go:405: unexpected error: error downloading module example.com/basic@v1.0.1: CreateFile C:\Users\gopher\AppData\Local\Temp\1\gorelease_test-gocache1462627586\pkg\mod\cache\download\example.com\basic\@v\v1.0.1.partial: Access is denied.

watchflakes

hec6srdp

hec6srdp2#

找到新的 Jmeter 板测试碎片:

#!watchflakes
post <- pkg == "golang.org/x/exp/cmd/gorelease" && test == "TestRelease"

2023-09-05 20:02 windows-amd64-longtest exp@92128663 go@972cc3e7 x/exp/cmd/gorelease.TestRelease ( log )

--- FAIL: TestRelease (0.57s)
    --- FAIL: TestRelease/basic/v1_querybase_verify (1.06s)
        gorelease_test.go:405: unexpected error: error downloading module example.com/basic@v1.0.1: CreateFile C:\Users\gopher\AppData\Local\Temp\1\gorelease_test-gocache531837263\pkg\mod\cache\download\example.com\basic\@v\v1.0.1.partial: Access is denied.
    --- FAIL: TestRelease/basic/v1_autobase_verify (1.11s)
        gorelease_test.go:405: unexpected error: error downloading module example.com/basic@v1.0.1: CreateFile C:\Users\gopher\AppData\Local\Temp\1\gorelease_test-gocache531837263\pkg\mod\cache\download\example.com\basic\@v\v1.0.1.partial: Access is denied.

watchflakes

dy1byipe

dy1byipe3#

解:根据题意,我们可以得到以下结论:

  1. 第一行和第二行的规律是:后一个数比前一个数大2;

  2. 第三行和第四行的规律是:后一个数比前一个数小3。

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