在Linux中使用AWK合并两个文件

我有一个1.txt文件：

beto[email protected]||o||0174686211||o||7880291304ca0404f4dac3dc205f1adf||o||Mario||o||Mario||o||Kawati
[email protected]||o||174732943.0174732943||o||e10adc3949ba59abbe56e057f20f883e||o||Tiziano||o||Tiziano||o||D'Intino
[email protected]||o||0174844404||o||8d496ce08a7ecef4721973cb9f777307||o||Melanie||o||Melanie||o||Kiesel
[email protected]||o||0174847613||o||536c1287d2dc086030497d1b8ea7a175||o||Sihem||o||Sihem||o||Sousou
[email protected]||o||174902297.0174902297||o||9893ac33a018e8d37e68c66cae23040e||o||Nabile||o||Nabile||o||Nassime
[email protected]||o||174912161.0174912161||o||0c770713436695c18a7939ad82bc8351||o||Donald||o||Donald||o||Duck
[email protected]||o||0174991962||o||d161dc716be5daf1649472ddf9e343e6||o||Dagmar||o||Dagmar||o||Cernakova
[email protected]||o||0175099675||o||d26005df3e5b416d6a39cc5bcfdef42b||o||Esmeralda||o||Esmeralda||o||Trogu
catherines[email protected]||o||0175128896||o||2e9ce84389c3e2c003fd42bae3c49d12||o||Cat||o||Cat||o||Sou
[email protected]||o||0175228687||o||a7766a502e4f598c9ddb3a821bc02159||o||Anna||o||Anna||o||Beratsja
[email protected]||o||0175306898||o||297642a68e4e0b79fca312ac072a9d41||o||Celine||o||Celine||o||Jacinto
[email protected]||o||0175410459||o||a6565ca2bc8887cde5e0a9819d9a8ee9||o||Adem||o||Adem||o||Bulut

2.txt文件：

9893ac33a018e8d37e68c66cae23040e:134:@a1
536c1287d2dc086030497d1b8ea7a175:~~@!:/92\
8d496ce08a7ecef4721973cb9f777307:demodemo

1.txt的FS为“ || o ||”对于2.txt是“：”
我想基于1.txt的第三列必须与2.txt文件的第一列匹配并且应由2.txt文件的第二列替换的情况下,将两个文件合并到单个文件result.txt中.

预期的输出将包含所有匹配的行：
我向您展示其中之一：

[email protected]||o||174902297.0174902297||o||134:@a1||o||Nabile||o||Nabile||o||Nassime

我尝试了脚本：

awk -F"||o||"  'NR==FNR{s=$0; sub(/:[^:]*$/, "", s); a[s]=$NF;next} {s = $5; for (i=6; i<=NF; ++i) s = s "," $i; if (s in a) { NF = 5; $5=a[s]; print } }' FS=: <(tr -d '\r' < 2.txt) FS="||o||" OFS="||o||" <(tr -d '\r' < 1.txt) > result.txt

但是结果是一个空文件.任何帮助将不胜感激.

解决方法:

如果您的实际Input_file与所示示例相同,则执行awk可能会帮助您.

awk -v s1="||o||" '
FNR==NR{
  a[$9]=$1 s1 $5;
  b[$9]=$13 s1 $17 s1 $21;
  next
}
($1 in a){
  print a[$1] s1 $2 FS $3 s1 b[$1]
}
' FS="|" 1.txt FS=":" 2.txt

编辑：由于OP稍微改变了要求,因此根据新的要求提供代码,它将在哪里创建2个文件,而1个文件的ID在1.txt中存在,而不是在2.txt中存在,而其他则相反.

awk -v s1="||o||" '
FNR==NR{
  a[$9]=$1 s1 $5;
  b[$9]=$13 s1 $17 s1 $21;
  c[$9]=$0;
  next
}
($1 in a){
  val=$1;
  $1="";
  sub(/:/,"");
  print a[val] s1 $0 s1 b[val];
  d[val]=$0;
  next
}
{
  print > "NOT_present_in_2.txt"
}
END{
for(i in d){
  delete c[i]
};
for(j in c){
  print j,c[j] > "NOT_present_in_1.txt"
}}
' FS="|" 1.txt FS=":" OFS=":" 2.txt