你要找的东西是这样的：

$ string=abc,def
$ jq --arg cities "$string" '[., ($cities / ",")] | transpose | map(.[0] + {city: .[1]})' file.json
[
  {
    "loc": "51.12345, 12.12345",
    "city": "abc"
  },
  {
    "loc": "65.12345, 15.12345",
    "city": "def"
  }
]
$

使用nodejs时：

> var json =   [
...     {
...       "loc": "51.12345, 12.12345",
...       "city": "CityName1"
...     },
...     {
...       "loc": "65.12345, 15.12345",
...       "city": "CityName2"
...     }
...   ]
> var c = 0
> ["abc", "def"].forEach(n => json[c++].city = n)
> console.log(JSON.stringify(json, null, 4))
[
    {
        "loc": "51.12345, 12.12345",
        "city": "abc"
    },
    {
        "loc": "65.12345, 15.12345",
        "city": "def"
    }
]

从脚本：

#!/bin/bash

node<<EOF | sponge file.json
var json = $(< file.json)
var c = 0;
["abc", "def"].forEach(n => json[c++].city = n)
console.log(JSON.stringify(json, null, 4))
EOF

输出文件：

[
    {
        "loc": "51.12345, 12.12345",
        "city": "abc"
    },
    {
        "loc": "65.12345, 15.12345",
        "city": "def"
    }
]

使用reduce是另一种方法，提供相同的长度，它允许迭代条目及其索引，然后可以使用索引访问输入数组。

string="abc,def"
jq --arg var "$string" '
  reduce ($var / "," | to_entries)[] as {$key, $value} (.;
    .[$key].city = $value
  )
'

我得到一个jq版本1.4的错误
为了使这种方法与jq1.4兼容，用一个普通变量替换变量destructuring，并在以后访问它的部分，例如。

string="abc,def"
jq --arg var "$string" '
  reduce ($var / "," | to_entries)[] as $item (.;
    .[$item.key].city = $item.value
  )
'

输出：

[
  {
    "loc": "51.12345, 12.12345",
    "city": "abc"
  },
  {
    "loc": "65.12345, 15.12345",
    "city": "def"
  }
]