Question

我有一个脚本问题：

将正确的变量传递给awk
awk不喜欢用于指定begninnging值的特定命令以及在指定模式之间打印的结束值。

这是states.txt的内容：

Alabama

Area: 52,423 sq.mi (135,775 sq.km.), 30th
Land: 50,750 sq.mi. (131,442 sq.km.), 28th
Water: 1,673 sq.mi. (4,333 sq.km.), 23rd
Coastline: 53 mi. (85 km.), 17th
Shoreline: 607 mi. (977 km.), 19th

Alaska

Area: 656,425 sq.mi (1,700,134 sq.km.), 1st
Land: 570,374 sq.mi. (1,477,263 sq.km.), 1st
Water: 86,051 sq.mi. (222,871 sq.km.), 1st
Coastline: 6,640 mi. (10,686 km.), 1st
Shoreline: 33,904 mi. (54,563 km.), 1st

Arizona

Area: 114,006 sq.mi (295,274 sq.km.), 6th
Land: 113,642 sq.mi. (294,332 sq.km.), 6th
Water: 364 sq.mi. (943 sq.km.), 48th

Arkansas

Area: 53,182 sq.mi (137,741 sq.km.), 29th
Land: 52,075 sq.mi. (134,874 sq.km.), 27th
Water: 1,107 sq.mi. (2,867 sq.km.), 31st

California

Area: 163,707 sq.mi (423,999 sq.km.), 3rd
Land: 155,973 sq.mi. (403,969 sq.km.), 3rd
Water: 7,734 sq.mi. (20,031 sq.km.), 6th
Coastline: 840 mi. (1,352 km.), 3rd
Shoreline: 3,427 mi. (5,515 km.), 5th

Colorado

Area: 104,100 sq.mi (269,618 sq.km.), 8th
Land: 103,730 sq.mi. (268,660 sq.km.), 8th
Water: 371 sq.mi. (961 sq.km.), 46th'

等等等等

我要做的是开发一个脚本，在解析它时单独为每个状态提取信息。

所以脚本看起来像这样：

for state in $(cat states.txt | egrep -v 'Area|Land|Water' | grep [A-Z]) ; do 

echo $state >> ./statelist.txt ; 

done ;

for statesnip in $(cat ./statelist.txt | awk 'NR>1{print p "_" $0 ORS} {p=$0}' | grep [A-Z]) ; do 

    state1=$(echo $statesnip | awk -F _ '{print $1}') ; 
    state2=$(echo $statesnip | awk -F _ '{print $2}') ; 

    cat ./states.txt | awk '/$state1/{f=1}; /$state2/{f=0}' >> $state1.tmp.txt ; 

done;

rm -f ./statelist.txt

所以这就是破坏：

第一个是传入awk的变量：

，如

awk -v state1=$state1 -v state2=$state2 '/state1/{f=1} f; /state2/{f=0}';

或

awk -v state1=${state1} state2=${state2} '/state1/{f=1} f; /state2/{f=0}';

我收到错误

第二个问题是，当我将变量调整为-v格式时，awk并不喜欢它（它只是整个文件的猫，很多次）。

 awk -v state1=${state1} -v state2=${state2} 'state1{f=1} f; state2{f=0}'

我只是反复得到整个文件的完整标记。

预期输出应如下所示：

cat ./statelist.txt

Alabama
Alaska
Arizona
Arkansas
California
Colorado

cat ./statelist.txt | awk 'NR>1{print p "_" $0 ORS} {p=$0}' | grep [A-Z]

Alabama_Alaska
Alaska_Arizona
Arizona_Arkansas
Arkansas_California
California_Colorado

cat ./Alabama.txt:

Alabama

Area: 52,423 sq.mi (135,775 sq.km.), 30th
Land: 50,750 sq.mi. (131,442 sq.km.), 28th
Water: 1,673 sq.mi. (4,333 sq.km.), 23rd
Coastline: 53 mi. (85 km.), 17th
Shoreline: 607 mi. (977 km.), 19th

cat ./Alaska.txt

Alaska

Area: 656,425 sq.mi (1,700,134 sq.km.), 1st
Land: 570,374 sq.mi. (1,477,263 sq.km.), 1st
Water: 86,051 sq.mi. (222,871 sq.km.), 1st
Coastline: 6,640 mi. (10,686 km.), 1st
Shoreline: 33,904 mi. (54,563 km.), 1st

cat ./Arizona.txt

Arizona

Area: 114,006 sq.mi (295,274 sq.km.), 6th
Land: 113,642 sq.mi. (294,332 sq.km.), 6th
Water: 364 sq.mi. (943 sq.km.), 48th

cat ./Arkansas.txt

Arkansas

Area: 53,182 sq.mi (137,741 sq.km.), 29th
Land: 52,075 sq.mi. (134,874 sq.km.), 27th
Water: 1,107 sq.mi. (2,867 sq.km.), 31st

cat ./California.txt

California

Area: 163,707 sq.mi (423,999 sq.km.), 3rd
Land: 155,973 sq.mi. (403,969 sq.km.), 3rd
Water: 7,734 sq.mi. (20,031 sq.km.), 6th
Coastline: 840 mi. (1,352 km.), 3rd
Shoreline: 3,427 mi. (5,515 km.), 5th

cat ./Colorado.txt

Colorado

Area: 104,100 sq.mi (269,618 sq.km.), 8th
Land: 103,730 sq.mi. (268,660 sq.km.), 8th
Water: 371 sq.mi. (961 sq.km.), 46th'

Answer 1

任何时候你在shell中编写循环只是为了操作文本你都有错误的方法。

在这种情况下，它看起来就像你真正需要的一切是：

awk 'NF==1{out=$1".txt"} {print > out}' states.txt

如果不是，请澄清。哦，非gawk，您可能需要在close(out)之前添加out=...。

Answer 2

虽然问题暗示awk正被用于解析文件，但给定的脚本使用的命令多于使用awk的其他命令。 Awk可以用来完成整个工作。

awk \
  ' \
    BEGIN \
    { FS = ":" }
    NF == 1 && /^[A-Z]/ \
    { FILE = $0 ".txt"; printf "\n%s\n\n", $0 >FILE }
    NF > 1 \
    { print >FILE }
  ' states.txt

虽然一个较小的脚本可以完成这项工作，但这个有一点额外的。使用冒号作为字段分隔符可以快速区分数据和标题行。空行被忽略，printf（）用于在输出文件中生成标题行。这意味着输入文件中不需要空格，这意味着额外的空格或空白行不会弄乱输出。这可能是也可能不是你想要的。

使用awk解析文本文件的各个部分

2 个答案: