如何查找 fstab 文件中的冲突

如何查找 fstab 文件中的冲突

有时我们注意到 /etc/fstab 文件中存在冲突,如下例所示

/dev/sdg 出现两次!

/data/sdb 出现两次!

#
# /etc/fstab
# Created by anaconda on Wed Nov  9 13:26:03 2016
#
# Accessible filesystems, by reference, are maintained under '/dev/disk'
# See man pages fstab(5), findfs(8), mount(8) and/or blkid(8) for more info
#
/dev/mapper/vg00-OS-linux_root /                       xfs     defaults        
0 0
UUID=cc749f07-ad72-49e8-ab19-ec6532f5e9 /boot                   xfs     
defaults        0 0
/dev/mapper/vg00-OS-linux_var /var                    xfs     defaults        
0 0
/dev/mapper/vg00-OS-linux_swap swap                    swap    defaults        
0 0


/dev/sdc /data/sdc ext4 defaults,noatime 0 0
/dev/sdb /data/sdb ext4 defaults,noatime 0 0
/dev/sde /data/sde ext4 defaults,noatime 0 0
/dev/sdf /data/sdf ext4 defaults,noatime 0 0
/dev/sdd /data/sdd ext4 defaults,noatime 0 0
/dev/sdg /data/sdb ext4 defaults,noatime 0 0
/dev/sdg /data/sdg ext4 defaults,noatime 0 0
/dev/sdh /data/sdh ext4 defaults,noatime 0 0
/dev/sdi /data/sdi ext4 defaults,noatime 0 0
/dev/sdj /data/sdj ext4 defaults,noatime 0 0
/dev/sdk /data/sdk ext4 defaults,noatime 0 0
/dev/sdl /data/sdl ext4 defaults,noatime 0 0

我们想要创建简单的验证来查找 fstab 文件中第一个字段或第二个字段的冲突

为此目的最好的语法是什么?

验证应在第一个字段或第二个字段中找到重复的单词,(语法行应尽可能短)

预期产出 -失败/正常

&(如果失败,应该打印第一个字段/第二个字段中的所有重复单词)

答案1

awk ' !/^#/ { if (seendev[$1]++) { print; ++rc; } if (seenmnt[$2]++) { print; ++rc; } } 
     END { exit rc }' < /etc/fstab

上面的 awk 单行代码将打印与第 1 列(设备)或第 2 列(安装点)重复的任何行,并且如果发生上述情况,还将以非零返回代码退出。

答案2

#!/bin/sh
# Filter out comments and blank lines from /etc/fstab, pick 1st/2nd field,
# then sort and find duplicates.
DUP_DEVS="$(grep -v -e '^#' -e '^$' /etc/fstab | awk '{ print $1;}' | sort | uniq -d)"
DUP_MOUNTS="$(grep -v -e '^#' -e '^$' /etc/fstab | awk '{ print $2;}' | sort | uniq -d)"

if [ -z "$DUP_DEVS" ] && [ -z "$DUP_MOUNTS" ]; then
    echo "ok"
    exit 0
else
    echo "fail"
    if [ ! -z "$DUP_DEVS" ]; then
        echo "Duplicate devices:"
        echo "$DUP_DEVS"
    else
        echo "Duplicate mount points:"
        echo "$DUP_MOUNTS"
    fi
    exit 1
fi

答案3

完整的解决方案awk:

awk '/^\/dev\//{ 
         if (devs[$1]++) dups[$1];
         if (data[$2]++) dups[$2] 
     }
     END{ 
         res="Ok"; 
         for (d in dups) { res="Fail\nDuplicates: "; break } 
         print "Result: "res; 
         for (d in dups) print d 
     }' /etc/fstab

输出:

Result: Fail
Duplicates: 
/dev/sdg
/data/sdb

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