awk不按預期打印變量

Question

嘗試真的很難讓awk打印出以下變量。但無論我怎樣努力，

awk -F, -v x=$CLIENT_ID -v y=$BRANCH -v z=$UUID -v b=$HERMES_GROUP_CSV_ID 'BEGIN { 
    OFS = ","; ORS = "\n" 
    } { 
     if (length($3) == 0) { 
      printf "\nCLIENT $x at $y Linux System Time: $z Pacific Time: $b #####: Column 3, Row "; printf NR; printf " data missing in the Client $x group input csv. Please check\n" 
     } 
    }' ${INPUT_FILE} 

它總是打印出

CLIENT $x at $y Linux System Time: $z Pacific Time: $b #####: Column 3, Row 249 data missing in the Client $x group input csv. Please check 

可以在任何大師指教？謝謝。

Answer 1

您正在使用$x作爲變量引用，但awk中的$用於引用輸入中的字段。變量在沒有裝飾的情況下使用，如x。所以：

awk -F, -v x=$CLIENT_ID -v y=$BRANCH -v z=$UUID -v b=$HERMES_GROUP_CSV_ID 'BEGIN { 
    OFS = ","; ORS = "\n" 
    } { 
     if (length($3) == 0) { 
      print "\nCLIENT "x" at "y" Linux System Time: "z" Pacific Time: "b" #####: Column 3, Row "; printf NR; printf " data missing in the Client "x" group input csv. Please check\n" 
     } 
    }' ${INPUT_FILE} 

看起來x這裏引用，但它不是：關鍵是要有x出現不帶引號的字符串中，以便它可以作爲變量進行擴展。