使用sed

Question

我需要从中删除我的后缀日志行：

Jun 12 06:19:20 cm postfix: 123123: to=<xxxxx@xxx.xxx>, relay=x.x.x.x, conn_use=3, delay=2, delays=1.6/0/0.01/0.32, dsn=2.0.0, status=sent (250 2.0.0 Ok: queued as XXXXXXX)

到此：

Jun 12 06:19:20 cm postfix: 123123: to=<xxxxx@xxx.xxx>, relay=x.x.x.x, delay=2, delays=1.6/0/0.01/0.32, dsn=2.0.0, status=sent (250 2.0.0 Ok: queued as XXXXXXX)

我只需要删除“conn_use = 3”。该值可以是变量（例如conn_use = 12.5） 我尝试过： sed's / conn_use =。* \ //' 但它删除任何东西直到最后（基本上，在“as”之后）：XXXXXXX）

Answer 1

sed 方法：

sed 's/conn_use=[^,]*, //' file

输出：

Jun 12 06:19:20 cm postfix: 123123: to=<xxxxx@xxx.xxx>, relay=x.x.x.x, delay=2, delays=1.6/0/0.01/0.32, dsn=2.0.0, status=sent (250 2.0.0 Ok: queued as XXXXXXX)

• [^,]* - 匹配除,
之外的任何字符

Answer 2

瞧瞧：

sed 's/conn_use=[^,]*, //'<<< 'Jun 12 06:19:20 cm postfix: 123123: to=<xxxxx@xxx.xxx>, relay=x.x.x.x, conn_use=3, delay=2, delays=1.6/0/0.01/0.32, dsn=2.0.0, status=sent (250 2.0.0 Ok: queued as XXXXXXX)'

输出：

Jun 12 06:19:20 cm postfix: 123123: to=<xxxxx@xxx.xxx>, relay=x.x.x.x, delay=2, delays=1.6/0/0.01/0.32, dsn=2.0.0, status=sent (250 2.0.0 Ok: queued as XXXXXXX)