bash 检查字符串是否包含非数字字符
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Check if string contains non digit characters
提问by saeed hardan
How can I check if a given string contains non numeric characters, examples :
如何检查给定字符串是否包含非数字字符，示例：
x11z returns > 0
x also returns > 0
1111~ also returns > 0
By character I mean everything not between 0-9. I saw similar threads but non of them talks about "non 0-9" except they show if its a-zor A-Z.
我所说的性格是指不在 之间的一切0-9。我看到了类似的线程，但他们都没有谈论“非 0-9”，除非他们显示它是a-z或A-Z。
回答by tarrsalah
Just by using bashpattern matching:
只需使用bash模式匹配：
[[ "$MY_VAR" =~ ^[^0-9]+$ ]] && echo "no digit in $MY_VAR"
回答by squiguy
Just use a negated character class:
只需使用否定字符类：
grep [^0-9]
This will match any non-numeric character, and not strings composed of only digits.
这将匹配任何非数字字符，而不是仅由数字组成的字符串。
回答by Rod Klinger
Most of these suggestions return true if the first character is a digit, but don't find errors within the string. The function below returns true if the entire string is digits, false if any non-digits are detected in the string.
如果第一个字符是数字，这些建议中的大多数都会返回 true，但不会在字符串中发现错误。如果整个字符串是数字，则下面的函数返回 true，如果在字符串中检测到任何非数字，则返回 false。
function isdigit () {
[ $# -eq 1 ] || return 1;
[[ = *[^0-9]* ]] && return 1
return 0
}
回答by jaypal singh
Something like this:
像这样的东西：
if [[ "xf44wd" =~ [0-9]+ ]]; then
echo "contains $?"
else
echo "does no contains $?"
fi
or
或者
if [[ ! "xf44wd" =~ [0-9]+ ]]; then
echo "does not contains $?"
else
echo "contains $?"
fi
回答by iruvar
Another option, a bash "containment" check
另一种选择，bash“遏制”检查
[[ "xf4fgh" = *[^0-9]* ]]
echo $?
0
[[ "1234" = *[^0-9]* ]]
echo $?
1
