string 如何检查字符串在 Bash shell 中是否有空格
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How to check if a string has spaces in Bash shell
提问by derrdji
Say a string might be like "a b '' c '' d". How can I check that there is single/double quote and space contained in the string?
说一个字符串可能像"a b '' c '' d". 如何检查字符串中是否包含单/双引号和空格？
采纳答案by Steve B.
case "$var" in
*\ * )
echo "match"
;;
*)
echo "no match"
;;
esac
回答by Paused until further notice.
You can use regular expressions in bash:
您可以在 bash 中使用正则表达式：
string="a b '' c '' d"
if [[ "$string" =~ \ |\' ]] # slightly more readable: if [[ "$string" =~ ( |\') ]]
then
echo "Matches"
else
echo "No matches"
fi
Edit:
编辑：
For reasons obvious above, it's better to put the regex in a variable:
由于上面显而易见的原因，最好将正则表达式放在变量中：
pattern=" |'"
if [[ $string =~ $pattern ]]
And quotes aren't necessary inside double square brackets. They can't be used on the right or the regex is changed to a literal string.
并且双方括号内不需要引号。它们不能在右侧使用，或者正则表达式更改为文字字符串。
回答by codeforester
You could do this, without the need for any backslashes or external commands:
您可以这样做，而无需任何反斜杠或外部命令：
# string matching
if [[ $string = *" "* ]]; then
echo "string contains one more spaces"
else
echo "string doesn't contain spaces"
fi
# regex matching
re="[[:space:]]+"
if [[ $string =~ $re ]]; then
echo "string contains one or more spaces"
else
echo "string doesn't contain spaces"
fi
Related:
有关的：
回答by glenn Hymanman
[[ "$str" = "${str%[[:space:]]*}" ]] && echo "no spaces" || echo "has spaces"
回答by Paul
string="a b '' c '' d"
if [ "$string" == "${string//[\' ]/}" ]
then
echo did not contain space or single quote
else
echo did contain space or single quote
fi
回答by alex tingle
The portable way to do this is with grep:
这样做的便携式方法是grep：
S="a b '' c '' d"
if echo $S | grep -E '[ "]' >/dev/null
then
echo "It's a match"
fi
...a bit ugly, but guaranteed to work everywhere.
...有点难看，但保证可以在任何地方工作。
回答by Grzegorz Oledzki
How about an approach similar to:
类似的方法如何：
$ A="some string"; echo $A | grep \ | wc -l
1
$ A="somestring"; echo $A | grep \ | wc -l
0
?
?
回答by ezpz
function foo() {
echo "String: $*"
SPACES=$(($#-1))
echo "Spaces: $SPACES"
QUOTES=0
for i in $*; do
if [ "$i" == "'" ]; then
QUOTES=$((QUOTES+1))
fi
done
echo "Quotes: $QUOTES"
echo
}
S="string with spaces"
foo $S
S="single' 'quotes"
foo $S
S="single '' quotes"
foo $S
S="single ' ' quotes"
foo $S
yields:
产量：
String: string with spaces
Spaces: 2
Quotes: 0
String: single' 'quotes
Spaces: 1
Quotes: 0
String: single '' quotes
Spaces: 2
Quotes: 0
String: single ' ' quotes
Spaces: 3
Quotes: 2
回答by rgiannico
What about this:
那这个呢：
[[ $var == ${var//[ \"]/_} ]] && echo "quotes or spaces not found"
or if you like this:
或者如果你喜欢这个：
if [[ $var == ${var//[ \"]/_} ]] ; then
echo "quotes or spaces not found"
else
echo "found quotes or spaces"
fi
Explanation:I'm evaluating a comparison between the variable ${var} and the variable ${var} itself after a on-the-fly non-destructive string substitution of all the quotes and spaces with an underscore.
说明：我正在评估变量 ${var} 和变量 ${var} 本身之间的比较，在对所有引号和空格进行动态非破坏性字符串替换后，带有下划线。
Examples:
例子：
${var// /_} # Substitute all spaces with underscores
The following code substitute all characters between the squared brackets (space and quotes) with an underscore. Note that quotes has to be protected with backslash:
以下代码用下划线替换方括号（空格和引号）之间的所有字符。请注意，引号必须用反斜杠保护：
${var//[ \"]/_}
回答by dz.
I do wonder why nobody mentioned the [:space:] set. Usually your not only interested in detecting the space character. I often need to detect any white space, e.g. TAB. The "grep" example would look like this:
我确实想知道为什么没有人提到 [:space:] 集。通常您不仅对检测空格字符感兴趣。我经常需要检测任何空白，例如 TAB。“grep”示例如下所示：
$ echo " " | egrep -q "[:space:]" && echo "Has no Whitespace" || echo "Has Whitespace"
Has Whitespace
$ echo "a" | egrep -q "[:space:]" && echo "Has no Whitespace" || echo "Has Whitespace"
Has no Whitespace
