arrays - 尝试访问元素时路径表达式无效

Question

当尝试更改数组中的单个元素时，我得到尝试访问元素时无效的路径表达式 - 但仅当从 --rawInput 捕获数组时.

示例:

# input: [ 1, 0 ]
. as $list | $list[0] = 30
# output: [ 30, 0 ]

但这不起作用:

# input: 1,0
split(",") | map(tonumber) as $list | $list[0] = 30
# Invalid path expression near attempt to access element 0 of [1,0]

• Working example
• Not working example

有什么想法吗？

Answer 1

由于以下原因，您的尝试失败:

Note that the LHS of assignment operators refers to a value in .. Thus $var.foo = 1 won’t work as expected ($var.foo is not a valid or useful path expression in .); use $var | .foo = 1 instead.

来自Assignment section of the jq manual .

它可能只在您的第一个 jq 命令中有效，因为 $list 和 . 是相等的。

接下来您可以使用以下内容:

split(",") | map(tonumber) as $list | $list | .[0] = 30

或更简单地针对您的情况:

split(",") | map(tonumber) | .[0]=30