尝试访问元素附近的无效路径表达式
Invalid path expression near attempt to access element
尝试更改数组中的单个元素时，我得到 Invalid path expression near attempt to access element - 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]
有什么想法吗？
您的尝试失败，原因如下：
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 命令中起作用，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
或者更简单地说你的情况：
split(",") | map(tonumber) | .[0]=30
尝试更改数组中的单个元素时，我得到 Invalid path expression near attempt to access element - 但仅当从 --rawInput.
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 | $list[0] = 30
# Invalid path expression near attempt to access element 0 of [1,0]
有什么想法吗？
您的尝试失败，原因如下：
Note that the LHS of assignment operators refers to a value in
.. Thus$var.foo = 1won’t work as expected ($var.foois not a valid or useful path expression in.); use$var | .foo = 1instead.
来自Assignment section of the jq manual.
它可能只在您的第一个 jq 命令中起作用，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
或者更简单地说你的情况：
split(",") | map(tonumber) | .[0]=30