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

我发现我经常需要这个功能,所以我在.bashrc中使用了一个自制的shell函数,这样我就可以随时重用它,并使用一个容易记住的名字:

function stringinstring()
{
    case "$2" in
       *"$1"*)
          return 0
       ;;
    esac
    return 1
}

要测试$string1(例如,abc)是否包含在$string2(例如,123abcABC)中,我只需要运行striginstring“$string1”“$string2”并检查返回值,例如

stringinstring "$str1" "$str2"  &&  echo YES  ||  echo NO

其他回答

这个怎么样:

text="   <tag>bmnmn</tag>  "
if [[ "$text" =~ "<tag>" ]]; then
   echo "matched"
else
   echo "not matched"
fi

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

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*"

通用的大海捞针示例下面是变量

#!/bin/bash

needle="a_needle"
haystack="a_needle another_needle a_third_needle"
if [[ $haystack == *"$needle"* ]]; then
    echo "needle found"
else
    echo "needle NOT found"
fi