bash - 减少 awk 遍数

Question

我在 Teradata 数据库上工作，有一个空间检查脚本。脚本应该根据在脚本开始时定义的空间使用值将标志提升为 CRITICAL 或 WARNING。

我的示例 SQL 文件输出是 (DatabaseOutput.log) 文件，如下所示，该文件用作 awk block 的输入。

### Some multiline Database query, resulting in Database Space usage

 *** Query completed. 11 rows found. 4 columns returned.
 *** Total elapsed time was 11 seconds.

## Output of the query, I am interested in DatabaseName, Perc2, MaxPerm

DatabaseName                                          Perc                    Perc2      MaxPerm
----------------------------------------------------  ----------------------  -------  -----------
AAA                                                   9.21899768137583E-001    92.19     10102320
BBB                                                   9.19923819717036E-001    91.99       524160
CCC                                                   9.17517791271651E-001    91.75      1687440
DDD                                                   9.15820363471060E-001    91.58       816720
EEE                                                   9.09293748338489E-001    90.93       149760
FFF                                                   9.07840905921109E-001    90.78      6934080
GGG                                                   9.04946085680591E-001    90.49      7273440
HHH                                                   8.54498111733230E-001    85.45      2538960
III                                                   8.22783559253080E-001    82.28      7598400
JJJ                                                   8.02181524253446E-001    80.22      8077680

+---------+---------+---------+---------+---------+---------+---+---------+-

要求的输出是

WARNING                           AAA     92.19  10102320
WARNING                           BBB     91.99    524160
WARNING                           CCC     91.75   1687440
WARNING                           DDD     91.58    816720
WARNING                           EEE     90.93    149760
WARNING                           FFF     90.78   6934080
WARNING                           GGG     90.49   7273440

我的工作 awk 代码需要三遍。它可以减少到一次 awk 吗？

工作 awk 代码:

cLvlCRIT=95
cLvlWARN=90

cat DatabaseOutput.log |
    awk '/-------------------/,/^$/' | # captures output block; it excludes query, logon, logoff information and header line but keeps separator's line.
        awk '{if (NR >= 2) {print}}' | # removes separator line, prints all lines from line 2 to EOF
            awk -v lLvlCRIT=$cLvlCRIT -v lLvlWARN=$cLvlWARN ' {
            if ( $1 != "StartCapture" && $3 >= lLvlCRIT ) {
                printf("%11s%30s%10s%10s\n", "CRITICAL",$1,$3,$4)
               }
            if ( $1 != "StartCapture" && $3 >= lLvlWARN && $3 < lLvlCRIT ) {
               printf("%11s%30s%10s%10s\n", "WARNING",$1,$3,$4)
              }
} '

提前致谢!

Answer 1

您可以使用标志在 awk 中获取 block :

awk -v lLvlCRIT="$cLvlCRIT" -v lLvlWARN="$cLvlWARN" '
/^----------------------------------------------------/ {block=1; next}
/^$/ && block {exit}    # if there is only one data block pattern - exit
                        # otherwise just reset block to 0 to find next block
block { your code on the block }'

因此要重现您的示例:

awk -v lLvlCRIT="$cLvlCRIT" -v lLvlWARN="$cLvlWARN" '
/^----------------------------------------------------/ {block=1; next}
/^$/ && block {exit}
block {if ( $3 >= lLvlCRIT )
     printf("%11s%30s%10s%10s\n", "CRITICAL",$1,$3,$4)       
else if ( $3 >= lLvlWARN )
     printf("%11s%30s%10s%10s\n", "WARNING",$1,$3,$4)  }' file
WARNING                           AAA     92.19  10102320
WARNING                           BBB     91.99    524160
WARNING                           CCC     91.75   1687440
WARNING                           DDD     91.58    816720
WARNING                           EEE     90.93    149760
WARNING                           FFF     90.78   6934080
WARNING                           GGG     90.49   7273440