sed -r在廚師中無法正常工作

Question

我正在廚師中編寫食譜，在其中我已將腳本資源與bash解釋器一起使用來執行sed命令：

    script 'find the lastest version of the major release available' do
        interpreter "bash"
        code <<-EOH
        SUBMGR_REPO=$(ls -1d /net/gutstools.am.lilly.com/guts_dev/distr/redhat/* | egrep `echo as6-u7_64 | sed -r 's/([aw]s[[:digit:]]+).*$/\1-u[[:digit:]]+(_64)?$/'` 2>/dev/null | sort -nr -t/ -k 7.6,7 | awk -F/ '{ print $7 }' | head -1)
        echo $SUBMGR_REPO
        EOH
end

如果我從終端運行上述命令，則可以正常運行，如下所示：

[c244728_lx@brainiac-ia-0008 redhat]$ SUBMGR_REPO=$(ls -1d /net/gutstools.am.lilly.com/guts_dev/distr/redhat/* | egrep `echo as6-u7_64 | sed -r 's/([aw]s[[:digit:]]+).*$/\1-u[[:digit:]]+(_64)?$/'` 2>/dev/null | sort -nr -t/ -k 7.6,7 | awk -F/ '{ print $7 }' | head -1)
[c244728_lx@brainiac-ia-0008 redhat]$ echo $SUBMGR_REPO
as6-u8_64
[c244728_lx@brainiac-ia-0008 redhat]$ ls
as5-u10_64  as5-u8_64  as6-u4_64               as6-u5_64.tar   as7-u1_64                 ws5-u10     ws5-u6_64  ws5-u9     ws6-u2_64  ws6-u5     ws6-u7_64
as5-u11_64  as5-u9_64  as6-u5_64               as6-u6_64       as7-u2_64                 ws5-u10_64  ws5-u7     ws5-u9_64  ws6-u3     ws6-u5_64  ws6-u8_64
as5-u5_64   as6-u1_64  as6-u5_64_1             as6-u7_64       krb5-patch                ws5-u11     ws5-u7_64  ws6-u1     ws6-u3_64  ws6-u6     ws7-u1_64
as5-u6_64   as6-u2_64  as6-u5_64-GBIP          as6-u7_64-GBIP  openssl.tar               ws5-u11_64  ws5-u8     ws6-u1_64  ws6-u4     ws6-u6_64  ws7-u2_64
as5-u7_64   as6-u3_64  as6-u5_64-kernel-patch  as6-u8_64       ruby22_rail41_CHG0053697  ws5-u6      ws5-u8_64  ws6-u2     ws6-u4_64  ws6-u7
[c244728_lx@brainiac-ia-0008 redhat]$

但是廚師配方失敗，它在$ SUBMGR_REPO中給出空值。 請幫我解決這個問題。

謝謝。

Answer 1

您應該真正考慮一下重構，但是我敢打賭，問題在於您有一些未轉義的反斜杠需要修復（例如\\\\1 ）