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a1 = ["re1": "111", "re2": "2222", "re1": "111", "re2": "2222"]
a2 = ["re3": "333", "re4": "444","re8": "888", "re9": "999"]

效果：
a3 = ["re1": "111", "re2": "2222", "re3": "333", "re4": "444","re1": "111", "re2": "2222","re8": "888", "re9": "999"]

如何使用jq将两个文件中的数组合并到一个数组中？

我想合并两个包含JSON的文件。它们每个都包含一组JSON对象。

registration.json

[    { "name": "User1", "registration": "2009-04-18T21:55:40Z" },    { "name": "User2", "registration": "2010-11-17T15:09:43Z" }]

useredits.json

[    { "name": "User1", "editcount": 164 },    { "name": "User2", "editcount": 150 },    { "name": "User3", "editcount": 10 }]

在理想的情况下，我希望合并操作的结果如下：

[    { "name": "User1", "editcount": 164, "registration": "2009-04-18T21:55:40Z" },    { "name": "User2", "editcount": 150, "registration": "2010-11-17T15:09:43Z" }]

我找到了https://github.com/stedolan/jq/issues/1247#issuecomment-348817802，但我得到了

jq: error: module not found: jq

答案

jq解决方案：

jq -s "[ .[0] + .[1] | group_by(.name)[]           | select(length > 1) | add ]" registration.json useredits.json

输出：

[  {    "name": "User1",    "registration": "2009-04-18T21:55:40Z",    "editcount": 164  },  {    "name": "User2",    "registration": "2010-11-17T15:09:43Z",    "editcount": 150  }]

另一答案

以下假设你有jq 1.5或更高版本，并且：

• joins.jq如下所示位于目录〜/ .jq /或目录〜/ .jq / join /
• pwd中没有名为joins.jq的文件
• registration.json已被修复，使其成为有效的JSON（顺便说一句，这可以由jq本身完成）。

然后使用的调用是：

jq -s "include "joins"; joins(.name)" registration.json useredits.json

joins.jq

# joins.jq Version 1 (12-12-2017)def distinct(s):  reduce s as $x ({}; .[$x | (type[0:1] + tostring)] = $x)  |.[];# Relational Join# joins/6 provides similar functionality to the SQL INNER JOIN statement:#   SELECT (Table1|p1), (Table2|p2)#     FROM Table1#     INNER JOIN Table2 ON (Table1|filter1) = (Table2|filter2)# where filter1, filter2, p1 and p2 are filters.# joins(s1; s2; filter1; filter2; p1; p2)# s1 and s2 are streams of objects corresponding to rows in Table1 and Table2;# filter1 and filter2 determine the join criteria;# p1 and p2 are filters determining the final results.# Input: ignored# Output: a stream of distinct pairs [p1, p2]# Note: items in s1 for which filter1 == null are ignored, otherwise all rows are considered.#def joins(s1; s2; filter1; filter2; p1; p2):  def it: type[0:1] + tostring;  def ix(s;f):    reduce s as $x ({};  ($x|f) as $y | if $y == null then . else .[$y|it] += [$x] end);  # combine two dictionaries using the cartesian product of distinct elements  def merge:    .[0] as $d1 | .[1] as $d2    | ($d1|keys_unsorted[]) as $k    | if $d2[$k] then distinct($d1[$k][]|p1) as $a | distinct($d2[$k][]|p2) as $b | [$a,$b]      else empty end;   [ix(s1; filter1), ix(s2; filter2)] | merge;def joins(s1; s2; filter1; filter2):  joins(s1; s2; filter1; filter2; .; .) | add ;# Input: an array of two arrays of objects# Output: a stream of the joined objectsdef joins(filter1; filter2):  joins(.[0][]; .[1][]; filter1; filter2);# Input: an array of arrays of objects.# Output: a stream of the joined objects where f defines the join criterion.def joins(f):  # j/0 is defined so TCO is applicable  def j:    if length < 2 then .[][]    else [[ joins(.[0][]; .[1][]; f; f)]] + .[2:] | j    end;   j ;

另一答案

虽然没有严格回答问题，但命令如下

jq -s "flatten | group_by(.name) | map(reduce .[] as $x ({}; . * $x))"      registration.json useredits.json

生成此输出：

[    { "name": "User1", "editcount": 164, "registration": "2009-04-18T21:55:40Z" },    { "name": "User2", "editcount": 150, "registration": "2010-11-17T15:09:43Z" },    { "name": "User3", "editcount": 10 }]

资料来源：jq - error when merging two JSON files "cannot be multiplied"

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