[英]Replace a column value using AWK
I have a file named as source-data.lst (created using ls -l) which contains the following information:我有一个名为source-data.lst的文件(使用 ls -l 创建),其中包含以下信息:
-rwxrwxrwx 1 soumyadipd soumyadipd 379 Apr 2 17:07 filegen.awk
-rw-rw-r-- 1 soumyadipd soumyadipd 129 Mar 31 13:35 file_name_list.txt
-rw-rw-r-- 1 soumyadipd soumyadipd 624 Apr 2 16:32 log
-rw-r--r-- 1 soumyadipd root 0 Apr 3 12:38 source-data.lst
-rw-rw-r-- 1 soumyadipd soumyadipd 676 Apr 2 16:32 temp
-rw-rw-r-- 1 soumyadipd soumyadipd 157 Mar 31 15:10 Type1_LP_OUT.txt
I need to update the file size by 1 for every row.我需要将每一行的文件大小更新 1。
by executing awk '{ $5=$5+1; print $0 }' source-data.lst
通过执行
awk '{ $5=$5+1; print $0 }' source-data.lst
awk '{ $5=$5+1; print $0 }' source-data.lst
from terminal the output will be as follows:从终端
awk '{ $5=$5+1; print $0 }' source-data.lst
输出如下:
[soumyadipd@linuxpc awkscripts]$ awk '{ $5=$5+1; print $0 }' source-data.lst
-rwxrwxrwx 1 soumyadipd soumyadipd 380 Apr 2 17:07 filegen.awk
-rw-rw-r-- 1 soumyadipd soumyadipd 130 Mar 31 13:35 file_name_list.txt
-rw-rw-r-- 1 soumyadipd soumyadipd 625 Apr 2 16:32 log
-rw-r--r-- 1 soumyadipd root 1 Apr 3 12:38 source-data.lst
-rw-rw-r-- 1 soumyadipd soumyadipd 677 Apr 2 16:32 temp
-rw-rw-r-- 1 soumyadipd soumyadipd 158 Mar 31 15:10 Type1_LP_OUT.txt
[soumyadipd@linuxpc awkscripts]$
But the problem is that I have an AWK script named as filegen.awk as:但问题是我有一个名为filegen.awk的 AWK 脚本:
BEGIN{
noOfSourceData=0;
while ((getline data < "source-data.lst") > 0) {
noOfSourceData++;
sourceRecordsList[noOfSourceData] = data;
}
}{
}END{
for (sd=1; sd<= noOfSourceData; sd++) {
#print sourceRecordsList[sd]
print sourceRecordsList[sd] > "temp"
awk '{ $5=$5+1; print $0 }' temp
}
}
Using this filegen.awk I have to increase the file size of source-data.lst .使用这个filegen.awk我必须增加source-data.lst的文件大小。 So executing the script as follows I found the following error:
所以按如下执行脚本我发现了以下错误:
[soumyadipd@linuxpc awkscripts]$ awk -f filegen.awk source-data.lst > log
awk: filegen.awk:16: awk '{ $5=$5+1; print $0 }' temp
awk: filegen.awk:16: ^ invalid char ''' in expression
[soumyadipd@linuxpc awkscripts]$
Kindly help me to make a solution... thanks in advance.请帮助我制定解决方案......提前致谢。
Instead of using getline
into a variable getline data < "source-data.lst"
you could use just getline
from a file:您可以只使用文件中的
getline
,而不是将getline
用于变量getline data < "source-data.lst"
:
awk '
BEGIN{
noOfSourceData=0;
while ((getline < "source-data.lst") > 0) {
noOfSourceData++;
$5=$5+1;
sourceRecordsList[noOfSourceData] = $0;
}
}
{
}
END{
for (sd=1; sd<= noOfSourceData; sd++)
print sourceRecordsList[sd]
}' dummy.txt
I solved my problem, and the code of filegen.awk is as follows:我的问题解决了, filegen.awk的代码如下:
BEGIN{
noOfSourceData=0;
while ((getline data < "source-data.lst") > 0) {
noOfSourceData++;
sourceRecordsList[noOfSourceData] = data;
}
}{
}END{
for (sd=1; sd<= noOfSourceData; sd++) {
system("echo \"" sourceRecordsList[sd] "\" | awk '{ $5=$5+1; print $0 }'");
}
}
And I get little idea from here while searching for the solution.在寻找解决方案时,我从这里得到的想法很少。
Thanks everybody for replying.谢谢大家的回复。
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