bash - 用零填充十六进制数字的awk

Question

Input:

[0x000] : {foo   ,bar    ,0x0000000000a00000  ,abcd             ,143     ,65535   }
[0x001] : {foo   ,blah   ,0x0000000001e00000  ,abcd             ,142     ,65535   }
[0x002] : {foo   ,bar    ,0x0000000003000000  ,abcd             ,141     ,155     }
[0x003] : {foo   ,bar    ,0x0000000003000000  ,abcd             ,144     ,156     }

Desired output:

0x0000 N/A 143
0x0002 155 141
0x0003 156 144

Currently using: sed -e 's/65535/N\\/A/g' -e 's/[][},]//g' | awk '/foo/ {print $1,$NF,$(NF-1)}' sed -e 's/65535/N\\/A/g' -e 's/[][},]//g' | awk '/foo/ {print $1,$NF,$(NF-1)}'

I'd like to pad the first field to a 4 char hex number? How should I do it?

The input itself can already be a 4 char hex. In that case it should leave it as is.

Also, is there a way to merge the sed and the awk into a single command?

Thanks!

Answer 1

This awk line should do it alone:

awk 'BEGIN{FS="[ \t,:{}\\[\\]x]+"} $10==65535{$10="N/A"} {printf "0x%.4x %s %s\n", $3, $10, $9}'

Works with mawk and gawk .

Explanation:

• BEGIN{FS=...} : sets the field separator to <space> , <tab> , , , [ , ] , : , { , } and x .
• $10==65535{$8="N/A"} If the 10th variable equals to 65535 is is set to N/A
• printf ... print the needed values in the desired output

Answer 2

您可以使用mawk ，我刚刚修改了您的脚本，但是您可以使用awk （在本例中为mawk ）本身完成所有这些操作：

(sed -e 's/65535/N\/A/g' -e 's/[][},]//g' | mawk '/foo/ {printf ("0x%04x %s %s\n", $1,$NF,$(NF-1))}') < File

Answer 3

Here is another awk

awk -F, '$6+0=="65535" {split($1,a,"[][]");$0=(length(a[2])==6?a[2]:"0x0"substr(a[2],3,4)) " N/A " $5}1' file
0x0000 N/A 143
0x2001 N/A 142
0x0002 N/A 141

If line contains 65535 change it, if not just print it.
Test if length of hex is 6 character, if yes, just use it.
If not, pad an extra 0 to it.

Example data:

cat file
[0x000] : {foo   ,bar    ,0x0000000000a00000  ,abcd             ,143     ,65535   }
[0x2001] : {foo   ,bar    ,0x0000000001e00000  ,abcd             ,142     ,65535   }
[0x002] : {foo   ,bar    ,0x0000000003000000  ,abcd             ,141     ,65535   }
This line should not be modified
[0x002] : {foo   ,bar    ,0x0000000003000000  ,abcd             ,141     ,65533   }

awk -F, '$6+0=="65535" {split($1,a,"[][]");$0=(length(a[2])==6?a[2]:"0x0"substr(a[2],3,4)) " N/A " $5}1' file
0x0000 N/A 143
0x2001 N/A 142
0x0002 N/A 141
This line should not be modified
[0x002] : {foo   ,bar    ,0x0000000003000000  ,abcd             ,141     ,65533   }

Answer 4

 sed 's#\[0x\([[:xdigit:]]*\)].*,\([0-9]\{1,\}\)[[:space:]]*,[^,]*#0x0\1 N/A \2#;s/0x0\([[:xdigit:]]\{4\}\)/0x\1/' YourFile

• take 2 group, first the hexa at start and the last before last , (using sed pattern behavior) and reformat with content of the 2 group
• could be shorter using instead of [[:space:]] if space char are used instead of tab, and use of [0-9a-f] instead of [[:xdigit:]]
• added the 4 hexa correction (thanks @jotne)

Posix version so --posix for GNU sed

Answer 5

If you are Ok with perl use this below command:

perl -lane '$F[6]=~s/,/N\/A /g;
            if($F[7]=~/65535/){printf "0x%04s $F[6]\n", $1 if(/^\[0x([^\]]*)\]/)}
            else{print}' your_file

Answer 6

You can use printf :

printf "0x%.4x %s %s\n" $(sed -e 's/65535/N\/A/g' -e 's/[][},]//g' input  | awk '/foo/ {print $1,$NF,$(NF-1)}')
0x0000 N/A 143
0x0001 N/A 142
0x0002 N/A 141

Note: I am reusing your previous work here, only sent its output to printf

Answer 7

With GNU awk for handling non-decimal input data:

$ awk --non-decimal-data -F'[][ ,]+' '/foo/{printf "0x%04x %s %s\n", $2, ($9==65535?"N/A":$9), $8}' file
0x0000 N/A 143
0x0001 N/A 142
0x0002 N/A 141