[[ $string == *foo* ]] && echo "It's there" || echo "Couldn't find"

一个是：

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

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

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

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
}

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