shell Exercises

shell Exercise: Record a question every day

1, demand: - Batch change the file name

         All files with a .txt extension found under / 123 directory

1. Bulk edit .txt is .txt.bak
2. All .bak files are packaged compressed 123.tar.gz
3. The name of the batch file to restore, that is to increase the .bak delete 

#! / bin / bash 
## txt file to find 
the Find / 123 The -type f -name " * .txt " > / tmp / txt.list 
## batch modify the file name 
for f in ` CAT / tmp / txt.list`
 do 
    Music Videos $ F $ f.bak
 DONE 
## to create a directory, the directory already exists in order to avoid, so to add a complex extension 
D = ` DATE +% Y% m% D% H% m% S '
 mkdir / tmp / $ D 123_ 
## .bak copy files to / tmp / 123_ $ D
 for F in ` CAT / tmp / txt.list`
 do 
    CPf.bak $ / tmp / 123_ $ D
 DONE 
## packing compression 
CD / tmp /
 the tar CZF 123 . the tar .gz 123_ $ D / 
## reduction 
for F in ` CAT / tmp / txt.list`
 do 
    Music Videos $ F. f $ BAK
 DONE

 2, demands: - the statistical average user

 写个shell，看看你的Linux系统中是否有自定义用户（普通用户），若是有，一共有几个？并输出姓名！

Assuming all users are common to more than 1000 uid 

# ! / Bin / the bash 
## find the row with the ordinary user, and count the number of 
n- = ` awk -F ' : '  ' $. 3> = 1000 ' / etc / the passwd | WC - l` 
## determines the number of general users 
IF [$ n--gt 0 ]
 the then 
   ## outputs a normal user and how many 
    echo  " There are $ n-Common users. " 

   the first line ## to find the average user's line and print ( line with the user's name) 
    the uname = ` awk -F ' : '  ' $. 3> = 1000 ' / etc / the passwd | awk -F ':' '{print $1}'`
    echo "the user were $uname ."
else
    echo "No common users."
fi

3. Requirements: - detect file changes

    There are two Linux servers A and B, if A can ssh directly to B, no password. A and B have a directory called / data / web / file down here there are many, of course, we do not know the specific subdirectory of several layers, before the file in that directory if A and B are exactly the same. But now uncertain whether the agreement. Solid we need to write a script to achieve this function, the machine detects similarities and differences between A and B machines / data / web / directory file, a file on our A machine as standard. For example, if one less machine B a.txt file, then we should be able to detect it, or b.txt files on the B machine has changes, we should be able to detect it (more files on machine B we do not consider) .

 提示： 使用核心命令 md5sum a.txt 算出md5值，去和B机器上的比较。

#! / bin / the bash 
# machine A to B is assumed that the machine has done no password set 
the dir = / Data / Web 
## is assumed as an IP machine B 192. 168.0 . 100 
B_ip = 192.168 . 0.100 
Find $ the dir -type F | xargs md5sum> / tmp / md5.txt
 SSH $ B_ip " the Find $ dir The -type f | xargs md5sum> /tmp/md5_b.txt " 
scp $ B_ip: /tmp/md5_b.txt / tmp 

for f in ` awk  ' { $ 2} Print ' / tmp / md5.txt`
 do 
    IF  grep -q"$f" /tmp/md5_b.txt
    then
        md5_a=`grep $f /tmp/md5.txt|awk '{print $1}'`
        md5_b=`grep $f /tmp/md5_b.txt|awk '{print $1}'`
        if [ $md5_a != $md5_b ]
        then
             echo "$f changed."
        fi
    else
        echo "$f deleted. "
    fi
done