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

我使用这个函数(一个不包括但很明显的依赖项)。它通过了以下测试。如果函数返回值>0,则找到字符串。你也可以很容易地返回1或0。

function str_instr {
   # Return position of ```str``` within ```string```.
   # >>> str_instr "str" "string"
   # str: String to search for.
   # string: String to search.
   typeset str string x
   # Behavior here is not the same in bash vs ksh unless we escape special characters.
   str="$(str_escape_special_characters "${1}")"
   string="${2}"
   x="${string%%$str*}"
   if [[ "${x}" != "${string}" ]]; then
      echo "${#x} + 1" | bc -l
   else
      echo 0
   fi
}

function test_str_instr {
   str_instr "(" "'foo@host (dev,web)'" | assert_eq 11
   str_instr ")" "'foo@host (dev,web)'" | assert_eq 19
   str_instr "[" "'foo@host [dev,web]'" | assert_eq 11
   str_instr "]" "'foo@host [dev,web]'" | assert_eq 19
   str_instr "a" "abc" | assert_eq 1
   str_instr "z" "abc" | assert_eq 0
   str_instr "Eggs" "Green Eggs And Ham" | assert_eq 7
   str_instr "a" "" | assert_eq 0
   str_instr "" "" | assert_eq 0
   str_instr " " "Green Eggs" | assert_eq 6
   str_instr " " " Green "  | assert_eq 1
}

其他回答

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*"

正如Paul在绩效比较中提到的:

if echo "abcdefg" | grep -q "bcdef"; then
    echo "String contains is true."
else
    echo "String contains is not true."
fi

这是符合POSIX的,就像Marcus提供的答案中的“case“$string”一样,但它比case语句答案更容易阅读。还要注意,这将比使用case语句慢得多。正如保罗所指出的,不要在循环中使用它。

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

一个是:

[ $(expr $mystring : ".*${search}.*") -ne 0 ] && echo 'yes' ||  echo 'no'

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