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

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。

试试oobash。

它是Bash 4的OO风格字符串库。它支持德语元音变音。它是用巴什语写的。

有许多函数可用:-base64Decode、-base64Encode、-acapital、-center、-charAt、concat、-concontains、-count、-endsWith、-equals、-equalsIgnoreCase、-reverse、-hashCode、-indexOf、-isAlnum、-isAlpha、-isAscii、-isDigit、-isEmpty、-isHexDigit、-isLowerCase、-isSpace、-isPrintable、-isUpperCase、-isVisible、-lastIndexOf、-length、-matches、-replaceAll、-replaceFirst、-startsWith,-substring、-swapCase、-toLowerCase、-toString、-toUpperCase、-trim和-zfill。

查看包含的示例:

[Desktop]$ String a testXccc
[Desktop]$ a.contains tX
true
[Desktop]$ a.contains XtX
false

oobash可在Sourceforge.net上获得。

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
}

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。

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