Question

我在Teradata数据库上工作，并有一个空格检查脚本。脚本应该将标志提升为CRITICAL或WARNING，具体取决于脚本开头定义的空间使用值。

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

### 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

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

必需的输出是

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

我有工作的awk代码需要三次传递..可以减少到一个awk传递吗？

使用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

您可以使用标志来获取awk中的块：

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 }'

重现你的例子：

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

Answer 2

Google UUOC并且永远不会使用范围表达式，因为它们使得琐碎的任务变得非常简单，但是需要完全重写或复制条件，甚至更有趣：

awk -v lLvlCRIT="$cLvlCRIT" -v lLvlWARN="$cLvlWARN" '
inBlock {
    if      ( $3 >= lLvlCRIT ) { level = "CRITICAL" }
    else if ( $3 >= lLvlWARN ) { level = "WARNING" }
    else if (NF)               { next }
    else                       { exit }
    printf "%11s%30s%10s%10s\n", level, $1, $3, $4
}
/-------------------/ { inBlock=1 }
' DatabaseOutput.log

Answer 3

您的awk可能如下所示：

awk -v lLvlCRIT="$cLvlCRIT" -v lLvlWARN="$cLvlWARN" '
/^---/,/^$/ {
   if ( $0 ~ "^---" || $0 ~ "^$" ) next
   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)               
}' DatabaseOutput.lo

在awk中指定模式范围可能很棘手，并且标记是首选方法。有关详细信息，请参阅Specifying Record Ranges with Patterns。

Answer 4

这将根据您的示例输入进行。

 #!/bin/bash

    cLvlCRIT=95
    cLvlWARN=90

   grep -E '^[a-zA-Z]+[ ]+[0-9.]+' DatabaseOutput.log |
         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)
                  }
    } '

问候！

Answer 5

关注awk也可以帮助您。

awk -v cLvlCRIT="$cLvlCRIT" -v cLvlWARN="$cLvlWARN" -v space="                           " '
/^$/||/^+/{
  flag="";
  next
}
/^----------/{
  flag=1;
  next
}
flag && $3>=cLvlWARN{
  val=$1 OFS $3 OFS $4;
  printf("%s"space"%s\n",$3>=cLvlCRIT?"CRITICAL":($3>=cLvlWARN && $3<cLvlCRIT?"WARNING":""),val)
}
'   Input_file

Answer 6

这里主要关注的是awk的有效输入。下面使用标志的解决方案是一种方法。这也考虑了您输入的具体模式。

crit=95
warn=90
awk -v crit=$crit -v warn=$warn '
/^DatabaseName/{flag=1;next}
{$2=""}!flag{next}
$3>crit{printf "Critical\t\t%s%s",$0,ORS;next}
$3>warn{printf "Warning \t\t%s%s",$0,ORS}' DatabaseOutput.lo

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

^{Sidenote ：非常确定，awk方法对于TB级文件来说会很慢。}

减少awk通行证的数量

6 个答案: