我有一个文件,我想在其中打印两个图案之间的所有线条。 Pattern1 =# Begin TRACE A Data
和 Pattern 2 = # Done Data $capture
,我想打印pattern1和pattern2之间的每一行。
文件一:
# Lower Limit
LIMIT_FLAG=0
LIMIT_POINT0=2884982910000.000000 -102800 -1
LIMIT_POINT1=2892982910000.000000 -102800 -1
# Limit Done
# Begin SPA Emission Mask
MASK SEGMENTS=0
MASK REFERENCE MODE=0
MASK REFERENCE LEVEL=0
MASK CENTER FREQUENCY=0
**
# SPA Emission Mask Done
# Begin SPA Data
<AP P_DATA>
** # Begin TRACE A Data **
P_0=-103.976000 , 2884.982910 MHz
P_1=-103.580000 , 2884.997456 MHz
P_2=-103.748000 , 2885.012001 MHz
P_3=-104.020000 , 2885.026547 MHz
P_4=-103.472000 , 2885.041092 MHz
P_5=-103.720000 , 2885.055638 MHz
P_6=-103.752000 , 2885.070183 MHz
P_7=-103.512000 , 2885.084729 MHz
P_8=-103.664000 , 2885.099274 MHz
P_9=-103.948000 , 2885.113820 MHz
P_10=-103.720000 , 2885.128365 MHz
P_11=-103.480000 , 2885.142911
# Done Data $capture
# Begin SPA Emission Mask
MASK SEGMENTS=0
MASK REFERENCE MODE=0
MASK REFERENCE LEVEL=0
MASK CENTER FREQUENCY=0
# End SPA Data
<APP_DATA_END>
# End SPA Data
<APP_DATA_END>
预期输出:
-103.976000 2884.982910
-103.580000 2884.997456
-103.748000 2885.012001
-104.020000 2885.026547
-103.472000 2885.041092
....
....
不应打印多余的行或空行,而应仅打印行数据。
答案1
sed -n '/# Begin TRACE A Data/,/# Done Data $capture/{s/ MHz//;s/,/ /;s/.*=//p;}' filename
/pattern1/,/pattern2/
仅选择从第一个到第二个模式的行,因此内部的所有内容{}
仅针对该范围执行s/ MHz//
删除尾随单元s/,/ /
将逗号替换为空格s/.*=//p
删除所有内容=
并打印模式,因此只有范围内的行=
才会被打印(选项-n
抑制默认输出)
实际上,对于您的示例数据,您也可以这样做
sed -n 's/ MHz//;s/.*=//;s/,/ /p'
因为只有您想要的行包含逗号。
答案2
Awk
解决方案:
awk '/# Begin TRACE A Data/{ f = 1; next }
/# Done Data \$capture/{ f = 0 }
NF && f{ gsub(/^P.+=|,| MHz/, ""); print }' file
输出:
-103.976000 2884.982910
-103.580000 2884.997456
-103.748000 2885.012001
-104.020000 2885.026547
-103.472000 2885.041092
-103.720000 2885.055638
-103.752000 2885.070183
-103.512000 2885.084729
-103.664000 2885.099274
-103.948000 2885.113820
-103.720000 2885.128365
-103.480000 2885.142911