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我希望根据每个文件中的所有值为unique-ID/hash-ID我的文件中的每一行创建一个JSONsJSON object

我从这里开始,但不确定是否必须明确列名,或者是否有办法在不明确列名的情况下包含所有列。

mlr --json put -S '$hash_id=$f_name."".$l_name."".$title' then reorder -e -f job file.json

file.json输入:

{"f_nams":"Hana","title":"Mrs","l_name":"Smith"}
{"f_nams":"Mike","title":"Mr","l_name":"Larry"}
{"f_nams":"Jhon","title":"Mr","l_name":"Doe"}

期望的输出:

{"f_nams":"Hana","title":"Mrs","l_name":"Smith","hash_id":"hash_value_based_on_all_columns"}
{"f_nams":"Mike","title":"Mr","l_name":"Larry","hash_id":"hash_value_based_on_all_columns"}
{"f_nams":"Jhon","title":"Mr","l_name":"Doe","hash_id":"hash_value_based_on_all_columns"}
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1 回答 1

1

假设输入 file.json 的格式如下所示:

cat file.json
{"f_nams":"Hana","title":"Mrs","l_name":"Smith"}
{"f_nams":"Mike","title":"Mr","l_name":"Larry"}
{"f_nams":"Jhon","title":"Mr","l_name":"Doe"}

然后一种方法是使用以下 perl 脚本来生成所需的输出:

perl -MMIME::Base64 -ne '
/"f_nams":"(\w+)","title":"(\w+)","l_name":"(\w+)"/ && do {
($fn,$tt,$ln)=($1,$2,$3);
$x=$fn . $tt . $ln;
chomp($hashvalue = encode_base64($x));
s/\}/,"hash_id":"$hashvalue"\}/;print}' file.json

产生:

{"f_nams":"Hana","title":"Mrs","l_name":"Smith","hash_id":"SGFuYU1yc1NtaXRo"}
{"f_nams":"Mike","title":"Mr","l_name":"Larry","hash_id":"TWlrZU1yTGFycnk="}
{"f_nams":"Jhon","title":"Mr","l_name":"Doe","hash_id":"Smhvbk1yRG9l"}
于 2020-08-02T18:40:58.220 回答