bash 合并 JSON 数组
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Merge Arrays of JSON
提问by Waffy
So my objective is to merge json files obtain this format:
所以我的目标是合并 json 文件以获得这种格式：
{
"title": "NamesBook",
"list": [
{
"name": "Ajay"
},
{
"name": "Al"
}
]
}
And I have files that look like this format:
我有看起来像这种格式的文件：
blahblah.json
blahblah.json
{
"title": "NamesBook",
"list": [
{
"name": "Ajay"
}
]
}
blueblue.json
蓝蓝.json
{
"title": "NamesBook",
"list": [
{
"name": "Al"
}
]
}
I can store the list array of all my names in a variable with the following:
我可以将我所有名字的列表数组存储在一个变量中，如下所示：
x = jq -s '.[].list' *.json
And then I was planning on appending the variable to an empty array in a file I created, out.json, which looks like this:
然后我计划将变量附加到我创建的文件 out.json 中的空数组中，如下所示：
{
"type": "NamesBook",
"list": []
}
However, when my script runs over the line
但是，当我的脚本越过该行时
jq '.list[] += "$x"' out.json'
It brings up a jq error:
它带来了一个 jq 错误：
Cannot iterate over null.
无法迭代 null。
Even when I add a random element, the same error shows up. Tips on how I should proceed? Are there other tools in jq to help achieve merging arrays?
即使我添加了一个随机元素，也会出现同样的错误。关于我应该如何进行的提示？jq 中是否还有其他工具可以帮助实现合并数组？
回答by zeppelin
You can merge your files with add(jq 1.3+):
您可以使用add(jq 1.3+)合并文件：
jq -s '.[0].list=[.[].list|add]|.[0]' *.json
or flatten(jq 1.5+):
或展平（jq 1.5+）：
jq -s '.[0].list=([.[].list]|flatten)|.[0]' *.json
[.[].list]- creates an array of all "list" arrays
[.[].list]- 创建一个包含所有“列表”数组的数组
[
[
{
"name": "Ajay"
}
],
[
{
"name": "Al"
}
]
]
[.[].list]|flatten- flattenit (or .[].list|add- addall the arrays together)
[.[].list]|flatten-展平它（或.[].list|add-将所有数组加在一起）
[
{
"name": "Ajay"
},
{
"name": "Al"
}
]
.[0].list=([.[].list]|flatten)|.[0]- replace the first "list" with the merged one, output it.
.[0].list=([.[].list]|flatten)|.[0]- 用合并的列表替换第一个“列表”，输出它。
{
"title": "NamesBook",
"list": [
{
"name": "Ajay"
},
{
"name": "Al"
}
]
}
回答by Jeff Mercado
Assuming every file will have the same titleand you're simply combining the listcontents, you could do this:
假设每个文件都相同，title并且您只是将list内容组合在一起，您可以这样做：
$ jq 'reduce inputs as $i (.; .list += $i.list)' blahblah.json blueblue.json
This just takes the first item and adds to its list, the list of all the other inputs.
这只是获取第一项并将所有其他输入的列表添加到其列表中。
回答by peak
The OP did not specify what should happen if there are objects for which .title is not "NamesBook". If the intent is to select objects with .title equal to "NamesBook", one could write:
如果存在 .title 不是“NamesBook”的对象，OP 没有指定应该发生什么。如果目的是选择 .title 等于“NamesBook”的对象，可以这样写：
map(select(.title == "NamesBook"))
| {title: .[0].title, list: map( .list ) | add}
This assumes that jq is invoked with the -s option.
这假设 jq 是使用 -s 选项调用的。
Incidentally, addis the way to go here: simple and fast.
顺便说一句，add这里的方法是：简单快捷。
回答by Jannis Ioannou
Let me also provide just what the title asks for, because I'm sure a lot of people that stepped on this question look for something simpler:
让我也提供标题所要求的内容，因为我相信很多遇到这个问题的人都在寻找更简单的东西：
echo -e '["a","b"]\n["c","d"]' | jq '.[]' | jq -s
