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先给出几个正确运行的例子，最后一个有语法错误，可是我不懂错误的含义，请教诸位

sed version: GNU sed 4.1.5
test文件共6行
$ cat test
1 this is the first line
2 this is the second line
3 this is the third line
i this is the fourth line
1 this is the fifth line
2 this is the sixth line
在每行下append一行‘---’
$ sed 'a\---' test
1 this is the first line
---
2 this is the second line
---
3 this is the third line
---
i this is the fourth line
---
1 this is the fifth line
---
2 this is the sixth line
---
找到含有1的行，移到下一行，将this换为THAT
$ sed '/1/{n;s/this/THAT;}' test
1 this is the first line
2 THAT is the second line
3 this is the third line
i this is the fourth line
1 this is the fifth line
2 THAT is the sixth line
找到含有1的行，跳过，其余的每行下append一行‘---’
$ sed '/1/n;a\---' test
1 this is the first line
2 this is the second line
---
3 this is the third line
---
i this is the fourth line
---
1 this is the fifth line
2 this is the sixth line
---
出语法错误的一行，原目的是在含有1的一行的下一行（这里是含有2的那两行）下append一行‘---’
$ sed '/1/{n;a\---;}' test
sed：-e 表达式 #1，字符 0：未匹配的“{”

请问诸位这一句的错误在哪？正确的语法应该如何实现呢。谢谢。

看花了。你为什么不这样写。看得清楚些
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{}的批量操作，记得是针对一个地址寻址的。换地址，要重新{}的。

呵呵，加上那些字是为了作替换(s)实验用的。
我在重新看一下{}的用法

eexpress 写了：看花了。你为什么不这样写。看得清楚些
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{}的批量操作，记得是针对一个地址寻址的。换地址，要重新{}的。

可是这一句是正确的啊
sed '/1/{n;s/this/THAT;}' test

百度了一下wu终于找到了

$ sed -e '/1/{n;a\---' -e '}' test

我试一下先，哈哈