Awk 为什么'；这个字段分隔符regex是否允许我提取所需的值？

我的输入文件

sumacomando

如下所示：

"firstName": "gdrgo",   "xxxxx": "John", "xxxxx": "John", "xxxxx": "John", "xxxxx": "John", "xxxxx": "John",   "lastName": "222",dfg
"xxxxx": "John",    "firstName": "beto",   "xxxxx": "John", "xxxxx": "John", "xxxxx": "John",   "lastName": "111","xxxxx": "John",
"xxxxx": "John",    "firstName": "beto",   "xxxxx": "John", "xxxxx": "John", "xxxxx": "John",   "lastName": "111","xxxxx": "John",
"xxxxx": "John",   "xxxxx": "John",    "firstName": "beto2", "xxxxx": "John","lastName": "555", "xxxxx": "John","xxxxx": "John",
"xxxxx": "John",   "xxxxx": "John",    "firstName": "beto2", "xxxxx": "John","lastName": "444", "xxxxx": "John","xxxxx": "John",
"firstName": "gdrgo",   "xxxxx": "John", "xxxxx": "John", "xxxxx": "John", "xxxxx": "John", "xxxxx": "John",   "lastName": "222",dfg
"xxxxx": "John",   "xxxxx": "John",    "firstName": "beto2", "xxxxx": "John","lastName": "444", "xxxxx": "John","xxxxx": "John",

我使用以下命令：

awk -v RS="\n" \
    -v FS='firstName": "|",[^+]*lastName": "|",' \
    '{sum[$1]=$2;} {print sum[$1]}' sumacomando

哪些产出：

gdrgo
111
111
555
444
gdrgo
444

但我预料到：

222
111
111
555
444
222
444

---

我做错了什么？

输入有点不规则，不清楚数组

求和的目的是什么，但要准确地给出您想要的：

awk -F'^.*"lastName": "|",' '{ print $2 }' sumacomando

字段分隔符regex
“^.*“lastName”：“|”和“
匹配从行的开头到
“lastName”：“
，然后是
”，
，因此，第二个字段-
$2
-有效地成为
lastName
字段关联值的内容。
数据文件是否真的像您显示的那样在逗号后加空格？你的字段分隔符正则表达式很难阅读-你不认为你应该更加尊重Awk吗？你认为中期选举能处理什么？您有
，[^+]*lastName:“
，在我看来，它像一个双引号、一个逗号、零个或多个不是加号的字符，然后是
lastName
、一个双引号、冒号、空格、双引号。您是否简单地打印出字段（明确地），以便您可以看到Awk如何解释您的正则表达式？（
for（i=1；i