在Bash中,测试数组是否包含某个值的最简单方法是什么?
当前回答
The answer with most votes is very concise and clean, but it can have false positives when a space is part of one of the array elements. This can be overcome when changing IFS and using "${array[*]}" instead of "${array[@]}". The method is identical, but it looks less clean. By using "${array[*]}", we print all elements of $array, separated by the first character in IFS. So by choosing a correct IFS, you can overcome this particular issue. In this particular case, we decide to set IFS to an uncommon character $'\001' which stands for Start of Heading (SOH)
$ array=("foo bar" "baz" "qux")
$ IFS=$'\001'
$ [[ "$IFS${array[*]}$IFS" =~ "${IFS}foo${IFS}" ]] && echo yes || echo no
no
$ [[ "$IFS${array[*]}$IFS" =~ "${IFS}foo bar${IFS}" ]] && echo yes || echo no
yes
$ unset IFS
这解决了大多数假阳性问题,但需要一个好的IFS选择。
注意:如果之前设置了IFS,最好保存并重新设置,而不是使用未设置的IFS
相关:
访问bash命令行参数$@ vs $*
其他回答
for i in "${array[@]}"
do
if [ "$i" -eq "$yourValue" ] ; then
echo "Found"
fi
done
字符串:
for i in "${array[@]}"
do
if [ "$i" == "$yourValue" ] ; then
echo "Found"
fi
done
扩展上面来自Sean DiSanti的答案,我认为下面是一个简单而优雅的解决方案,它避免了对数组进行循环,并且不会由于部分匹配而给出假阳性
function is_in_array {
local ELEMENT="${1}"
local DELIM=","
printf "${DELIM}%s${DELIM}" "${@:2}" | grep -q "${DELIM}${ELEMENT}${DELIM}"
}
可以这样称呼:
$ haystack=("needle1" "needle2" "aneedle" "spaced needle")
$ is_in_array "needle" "${haystack[@]}"
$ echo $?
1
$ is_in_array "needle1" "${haystack[@]}"
$ echo $?
0
如果您想做一个快速而复杂的测试,看看是否值得遍历整个数组以获得精确匹配,Bash可以像对待标量一样对待数组。测试标量中的匹配项,如果没有,则跳过循环节省时间。显然你会得到假阳性。
array=(word "two words" words)
if [[ ${array[@]} =~ words ]]
then
echo "Checking"
for element in "${array[@]}"
do
if [[ $element == "words" ]]
then
echo "Match"
fi
done
fi
这将输出“Checking”和“Match”。使用array=(word "two words" something),它只会输出"Checking"。使用array=(单词“two widgets”什么的)将没有输出。
如果你不想重复,这是值得研究的:
#!/bin/bash
myarray=("one" "two" "three");
wanted="two"
if `echo ${myarray[@]/"$wanted"/"WAS_FOUND"} | grep -q "WAS_FOUND" ` ; then
echo "Value was found"
fi
exit
片段改编自:http://www.thegeekstuff.com/2010/06/bash-array-tutorial/ 我认为这很聪明。
编辑: 你可以这样做:
if `echo ${myarray[@]} | grep -q "$wanted"` ; then
echo "Value was found"
fi
但后者仅在数组包含唯一值时有效。我认为,在143中寻找1只会给出假阳性。
containsElement () { for e in "${@:2}"; do [[ "$e" = "$1" ]] && return 0; done; return 1; }
现在正确处理空数组。