我在Bash中有一个字符串:
string="My string"
如何测试它是否包含另一个字符串?
if [ $string ?? 'foo' ]; then
echo "It's there!"
fi
哪里是我不认识的接线员。我使用echo和grep吗?
if echo "$string" | grep 'foo'; then
echo "It's there!"
fi
这看起来有点笨拙。
我在Bash中有一个字符串:
string="My string"
如何测试它是否包含另一个字符串?
if [ $string ?? 'foo' ]; then
echo "It's there!"
fi
哪里是我不认识的接线员。我使用echo和grep吗?
if echo "$string" | grep 'foo'; then
echo "It's there!"
fi
这看起来有点笨拙。
当前回答
因此,这个问题有很多有用的解决方案——但哪一个最快/使用的资源最少?
使用此框架重复测试:
/usr/bin/time bash -c 'a=two;b=onetwothree; x=100000; while [ $x -gt 0 ]; do TEST ; x=$(($x-1)); done'
每次更换测试:
[[ $b =~ $a ]] 2.92 user 0.06 system 0:02.99 elapsed 99% CPU
[ "${b/$a//}" = "$b" ] 3.16 user 0.07 system 0:03.25 elapsed 99% CPU
[[ $b == *$a* ]] 1.85 user 0.04 system 0:01.90 elapsed 99% CPU
case $b in *$a):;;esac 1.80 user 0.02 system 0:01.83 elapsed 99% CPU
doContain $a $b 4.27 user 0.11 system 0:04.41 elapsed 99%CPU
(doContain在F.Houri的回答中)
对于傻笑:
echo $b|grep -q $a 12.68 user 30.86 system 3:42.40 elapsed 19% CPU !ouch!
因此,无论是在扩展测试还是案例中,简单的替代选项都可以预测地获胜。这个箱子是便携式的。
输出到100000 greps是可想而知的痛苦!关于无需使用外部实用程序的旧规则是正确的。
其他回答
我不确定是否使用if语句,但您可以使用case语句获得类似的效果:
case "$string" in
*foo*)
# Do stuff
;;
esac
此堆栈溢出答案是唯一一个陷阱空格和破折号字符的答案:
# For null cmd arguments checking
to_check=' -t'
space_n_dash_chars=' -'
[[ $to_check == *"$space_n_dash_chars"* ]] && echo found
我喜欢sed。
substr="foo"
nonsub="$(echo "$string" | sed "s/$substr//")"
hassub=0 ; [ "$string" != "$nonsub" ] && hassub=1
编辑,逻辑:
使用sed从字符串中删除子字符串的实例如果新字符串与旧字符串不同,则存在子字符串
使用jq:
string='My long string'
echo $string | jq -Rr 'select(contains("long"))|"It is there"'
jq中最困难的事情是打印单个引用:
echo $string | jq --arg quote "'" -Rr 'select(contains("long"))|"It\($quote)s there"'
仅使用jq检查条件:
if jq -Re 'select(contains("long"))|halt' <<< $string; then
echo "It's there!"
fi
这里回答的问题的扩展如何判断POSIX sh中的字符串是否包含另一个字符串?:
此解决方案适用于特殊字符:
# contains(string, substring)
#
# Returns 0 if the specified string contains the specified substring,
# otherwise returns 1.
contains() {
string="$1"
substring="$2"
if echo "$string" | $(type -p ggrep grep | head -1) -F -- "$substring" >/dev/null; then
return 0 # $substring is in $string
else
return 1 # $substring is not in $string
fi
}
contains "abcd" "e" || echo "abcd does not contain e"
contains "abcd" "ab" && echo "abcd contains ab"
contains "abcd" "bc" && echo "abcd contains bc"
contains "abcd" "cd" && echo "abcd contains cd"
contains "abcd" "abcd" && echo "abcd contains abcd"
contains "" "" && echo "empty string contains empty string"
contains "a" "" && echo "a contains empty string"
contains "" "a" || echo "empty string does not contain a"
contains "abcd efgh" "cd ef" && echo "abcd efgh contains cd ef"
contains "abcd efgh" " " && echo "abcd efgh contains a space"
contains "abcd [efg] hij" "[efg]" && echo "abcd [efg] hij contains [efg]"
contains "abcd [efg] hij" "[effg]" || echo "abcd [efg] hij does not contain [effg]"
contains "abcd *efg* hij" "*efg*" && echo "abcd *efg* hij contains *efg*"
contains "abcd *efg* hij" "d *efg* h" && echo "abcd *efg* hij contains d *efg* h"
contains "abcd *efg* hij" "*effg*" || echo "abcd *efg* hij does not contain *effg*"