Bash 从变量中获取最后一行
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Bash get last line from a variable
提问by Matt Backslash
If I have a variable with multiple lines (text) in it, how can I get the last line out of it?
如果我有一个包含多行（文本）的变量，我怎样才能从中取出最后一行？
I already figured out how to get the first line:
我已经想出了如何获得第一行：
STRING="This is a
multiple line
variable test"
FIRST_LINE=(${STRING[@]})
echo "$FIRST_LINE"
# output:
"This is a"
Probably there should be an operator for the last line. Or at least I assume that because with @the first line comes out.
可能最后一行应该有一个运算符。或者至少我认为这是因为@第一行出来了。
回答by redneb
An easy way to do this is to use tail:
一个简单的方法是使用tail：
echo "$STRING" | tail -n1
回答by anubhava
Using bash string manipulations:
使用 bash 字符串操作：
$> str="This is a
multiple line
variable test"
$> echo "${str##*$'\n'}"
variable test
${str##*$'\n'}will remove the longest match till \nfrom start of the string thus leaving only the last line in input.
${str##*$'\n'}将删除最长的匹配直到\n从字符串的开头，从而只留下输入中的最后一行。
回答by chepner
If you want an array with one element per line from STRING, use
如果您想要一个每行一个元素的数组 from STRING，请使用
readarray -t lines <<< "$STRING"
Then, the first line would be ${lines[0]}, and the last line would be ${lines[-1]}. In older versions of bash, negative indices aren't allowed and you'll have to compute the last index manually: ${lines[${#lines[@]}-1]}.
然后，第一行是${lines[0]}，最后一行是${lines[-1]}。在旧版本的 中bash，不允许使用负索引，您必须手动计算最后一个索引：${lines[${#lines[@]}-1]}。
