Bash:打印string值导致错误

fatal: not enough arguments to satisfy format string `%s SPT=80' ^ ran out for this one 

这是我的代码

 for ((h = 1 ; h < 4 ; h++ )); do x=$(awk -vi=h -vj=17 'FNR == 2 {printf "%s " $j}' newiptables.log) echo $x 

这是我的文件

 Dec 26 09:17:51 localhost kernel: IN=eth0 OUT= MAC=00:10:c6:a8:da:68:00:90:7f:9c:50:5a:08:00 SRC=198.252.206.16 DST=10.128.1.225 LEN=313 TOS=0x00 PREC=0x00 TTL=64 ID=59334 PROTO=TCP SPT=80 DPT=56506 WINDOW=46535 RES=0x00 ACK PSH URGP=0 Dec 26 09:17:52 localhost kernel: IN=eth0 OUT= MAC=00:10:c6:a8:da:68:00:90:7f:9c:50:5a:08:00 SRC=198.252.206.16 DST=10.128.1.225 LEN=1440 TOS=0x00 PREC=0x00 TTL=64 ID=47303 PROTO=TCP SPT=80 DPT=56506 WINDOW=46535 RES=0x00 ACK URGP=0 Dec 26 09:17:52 localhost kernel: IN=eth0 OUT= MAC=00:10:c6:a8:da:68:00:90:7f:9c:50:5a:08:00 SRC=198.252.206.16 DST=10.128.1.225 LEN=1440 TOS=0x00 PREC=0x00 TTL=64 ID=47559 PROTO=TCP SPT=80 DPT=56506 WINDOW=46535 RES=0x00 ACK URGP=0 

Solutions Collecting From Web of "Bash:打印string值导致错误"

问题是在awkprintf命令中缺少一个逗号:

 awk -vi=h -vj=17 'FNR == 2 {printf "%s ", $j}' newiptables.log ^ |== This is needed 

从手册引用:

一个简单的printf语句如下所示:

  printf format, item1, item2, ...