bash 在 Shell 脚本中将值存储到变量
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Store value to variable in Shell script
提问by Sivailango
"$emails" has the array of values, i want to parse the values from it, To do it, i am using the jq. if i do below command
“$emails”具有值数组，我想从中解析值，为此，我正在使用jq。如果我执行以下命令
echo "$emails" | ./jq '.total_rows'
i could get the value i.e 4, i want to store the returned results into some variable,
我可以得到值，即 4，我想将返回的结果存储到某个变量中，
total_rows="$emails" | ./jq '.total_rows'
but total_rows has no value.
但 total_rows 没有价值。
echo $total_rows
How do store the returned result into variable?
如何将返回的结果存储到变量中？
回答by Nidhoegger
You have to use the right quotation, like this:
你必须使用正确的引用，像这样：
total_rows=`echo "$emails" | ./jq '.total_rows'`
The `` will execute the command and give total_rowsthe value of it, so whatever would be the output of
`` 将执行命令并给出total_rows它的值，所以无论输出是什么
echo "$emails" | ./jq '.total_rows'
will so be stored in total_rows.
将因此存储在total_rows.
As mentioned in the comments by Tom Fenech, it is better to use $()for command substitution. It provides a better readability. So what you can do is:
正如 Tom Fenech 在评论中提到的，最好使用$()命令替换。它提供了更好的可读性。所以你可以做的是：
total_rows=$(echo "$emails" | ./jq '.total_rows')
