bash 检查数组的长度是否等于bash中的变量
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Checking if length of array is equal to a variable in bash
提问by Kr?lleb?lle
I want to check if the length of a bash array is equal to a bash variable (int) or not. My current code looks like this:
我想检查 bash 数组的长度是否等于 bash 变量（int）。我当前的代码如下所示：
if [ "${#selected_columns}" -eq "${number_of_columns}" ]; then
echo "They are equal!"
fi
This returns false since the echo statement is never run. However, doing this produces 4 for both of them:
这将返回 false，因为从不运行 echo 语句。但是，这样做会为它们生成 4 个：
echo "${#selected_columns[@]}"
echo "${number_of_columns}"
What's wrong here? Has it something to do with string versus int?
这里有什么问题？它与字符串与整数有关吗？
采纳答案by sampson-chen
In your:
在你的：
if [ "${#selected_columns}" -eq "${number_of_columns}" ]; then
echo "They are equal!"
fi
${#selected_columns}is missing [@].
${#selected_columns}不见了[@]。
Fixed:
固定的：
if [ "${#selected_columns[@]}" -eq "${number_of_columns}" ]; then
echo "They are equal!"
fi
回答by user3226688
According to bash's man page:
根据 bash 的手册页：
${#name[subscript]} expands to the length of ${name[subscript]}. If subscript is * or @, the expansion is the number of elements in the array. Referencing an array variable without a subscript is equivalent to referencing the array with a subscript of 0.
${#name[subscript]} 扩展到 ${name[subscript]} 的长度。如果下标为 * 或 @，则扩展为数组中的元素数。引用一个没有下标的数组变量相当于引用一个下标为 0 的数组。
Using ${name}on index arrays will result to ${name[0]}, then you got the length of ${name[0]}, not counting elements of whole array. So that's not problem about comparing string with integer. AFAIK, comparing "integer number in string" with "integer assigned by let" is never a problem in bash scripting.
${name}在索引数组上使用将导致${name[0]}，然后您得到 的长度${name[0]}，而不是计算整个数组的元素。因此，将字符串与整数进行比较不是问题。AFAIK，将“字符串中的整数”与“由 let 分配的整数”进行比较在 bash 脚本中从来都不是问题。
