vi 中按位置替换。[已解决]

chen13993

1. printf "%-10s %-10s %-10s\n" No Name Mark
   
2. printf "%-10s %-10s %-10.2f\n" 1 Sarath 80.310106
   
3. printf "%-10s %-10s %-10.2f\n" 2 James 90.9989
   
4. printf "%-10s %-10s %-10.2f\n" 3 Jeff 77.10610

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在vi中编辑这么几行文字，我想要替换每行第一个10为5，第二个10为6.第三个10为7，改怎么做？

jason680

回复 1# chen13993

$ perl -pe '$n=5;s/-10/-$n++/ge;' FILE
printf "%-5s %-6s %-7s\n" No Name Mark
printf "%-5s %-6s %-7.2f\n" 1 Sarath 80.310106
printf "%-5s %-6s %-7.2f\n" 2 James 90.9989
printf "%-5s %-6s %-7.2f\n" 3 Jeff 77.10610

# to add -i argument to replace the FILE when everything was fine
$ perl -i -pe '$n=5;s/-10/-$n++/ge;' FILE

$ cat FILE
printf "%-5s %-6s %-7s\n" No Name Mark
printf "%-5s %-6s %-7.2f\n" 1 Sarath 80.310106
printf "%-5s %-6s %-7.2f\n" 2 James 90.9989
printf "%-5s %-6s %-7.2f\n" 3 Jeff 77.10610

   

chen13993

回复 2# jason680


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