[英]find a pattern and print line based on finding the first pattern sed, awk grep
[英]Grep a line then print awk until a certain substring
我的代碼是
var=$(cat $FILE | grep "$alineimlookingfor" | awk '{print $1, $2, $4, $7, $9... all the way to $20}'
echo "$var"
但是,我希望在達到(0)或(1)之類的值時停止$ 9- $ 20。 這將使我的輸出格式看起來更好,因為(0)或(1)之后的任何東西都是垃圾。
有沒有人對實現它的方法有所了解?
輸入:
2013-02-21 00:12:03,374 [Thread] IN ProcedureTask - Finished Sales Summary 22 This (0) - 21-JUNE-10
輸出:
2013-02-21 00:12:03,374 IN ProcedureTask - Finished Sales Summary 22 This (0)
編輯:感謝所有對這條線特別注意的人和特別感謝
更新:
awk -v pattern="$alineimlookingfor" '
$0 ~ pattern {
rec = $1 OFS $2 OFS $4 OFS $7
for (i=9; i<=NF; i++) {
rec = rec OFS $i
if ($i ~ /\([01]\)/) {
break
}
}
print rec
}
' "$FILE"
應該完全符合您的要求
筆記:
NF
是含F ields的N-棕土在當前記錄的AWK變量。 rec
,由頻輸出˚Field 小號 eparator分離。 rec
變量
for
循環 rec
字符串。 首先,請注意awk
可以執行cat
和grep
所做的事情,因此我們可以立即簡化管道
awk -v pattern="$alineimlookingfor" '$0 ~ pattern {print $1, $2, $4, $7, $9... all the way to $20}' "$FILE"
接下來,聽起來你想要
awk -v pattern="$alineimlookingfor" '
$0 ~ pattern {
for (i=9; i<NF; i++) {
if ($i == "(0)" || $i == "(1)") {
NF = i
break
}
}
print
}
' "$FILE"
這會更改“此記錄中的字段數”變量,以便忽略后續字段。
測試
alineimlookingfor=ProcedureTask
awk -v pattern="$alineimlookingfor" '
$0 ~ pattern {
for (i=9; i<NF; i++) {
if ($i == "(0)" || $i == "(1)") {
NF = i
break
}
}
print
}
' <<'END'
foo
2013-02-21 00:12:03,374 [Thread] IN ProcedureTask - Finished Sales Summary 22 This (0) - 21-JUNE-10
bar
2013-02-21 00:12:03,374 [Thread] IN ProcedureTask - Finished Sales Summary 22 This (1) - 21-JUNE-10
baz
2013-02-21 00:12:03,374 [Thread] IN ProcedureTask - Finished Sales Summary 22 This (2) - 21-JUNE-10
END
輸出
2013-02-21 00:12:03,374 [Thread] IN ProcedureTask - Finished Sales Summary 22 This (0)
2013-02-21 00:12:03,374 [Thread] IN ProcedureTask - Finished Sales Summary 22 This (1)
2013-02-21 00:12:03,374 [Thread] IN ProcedureTask - Finished Sales Summary 22 This (2) - 21-JUNE-10
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