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

grep-q对于这个目的很有用。

同样使用awk:

string="unix-bash 2389"
character="@"
printf '%s' "$string" | awk -vc="$character" '{ if (gsub(c, "")) { print "Found" } else { print "Not Found" } }'

输出:

未找到

string="unix-bash 2389"
character="-"
printf '%s' "$string" | awk -vc="$character" '{ if (gsub(c, "")) { print "Found" } else { print "Not Found" } }'

输出:

建立

原始来源:http://unstableme.blogspot.com/2008/06/bash-search-letter-in-string-awk.html

这个怎么样:

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

您可以使用逻辑&&以更紧凑

#!/bin/bash

# NO MATCH EXAMPLE
string="test"
[[ "$string" == *"foo"* ]] && {
        echo "YES"
}

# MATCH EXAMPLE
string="tefoost"
[[ "$string" == *"foo"* ]] && {
        echo "YES"
}

由于POSIX/BusyBox问题在没有提供正确答案(IMHO)的情况下结束,我将在这里发布答案。

最简单的答案是:

[ ${_string_##*$_substring_*} ] || echo Substring found!

or

[ "${_string_##*$_substring_*}" ] || echo 'Substring found!'

注意,对于某些shell(ash),双哈希是必须的。当未找到子字符串时,上面将计算[stringvalue]。它不会返回错误。找到子字符串时,结果为空,其结果为[]。这将抛出错误代码1,因为字符串被完全替换(由于*)。

最短最常见的语法:

[ -z "${_string_##*$_substring_*}" ] && echo 'Substring found!'

or

[ -n "${_string_##*$_substring_*}" ] || echo 'Substring found!'

另一个:

[ "${_string_##$_substring_}" != "$_string_" ] && echo 'Substring found!'

or

[ "${_string_##$_substring_}" = "$_string_" ] || echo 'Substring found!'

注意单个等号!

公认的答案是最好的,但由于有不止一种方法可以做到这一点,这里有另一种解决方案:

if [ "$string" != "${string/foo/}" ]; then
    echo "It's there!"
fi

${var/search/replace}是$var,如果找到了第一个搜索实例,则将其替换为replace(它不会更改$var)。如果您试图将foo替换为空,并且字符串发生了更改,那么很明显找到了foo。