bash 如何验证十进制数？
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How do I validate decimal numbers?
提问by user2003069
Hi I'm working on an assignment and got stuck on this part, how do I validate decimal numbers/numbers in shell?
嗨，我正在做一项作业并在这部分卡住了，我如何在 shell 中验证十进制数字/数字？
It can accept numbers but not decimal numbers. I want it to be able to accept both.
它可以接受数字，但不能接受十进制数字。我希望它能够同时接受两者。
This is what I have so far
这是我到目前为止
if echo $value | egrep '^[0-9]+$' >/dev/null 2>&1 ; then
echo "OK"
else
echo "There Is An Error"
echo "Please Try Again"
fi
回答by Guru
Instead of using grep, you can use the bash to check expression:
您可以使用 bash 来检查表达式，而不是使用 grep：
#!/bin/bash
value=98.23
if [[ "$value" =~ ^[0-9]+(\.[0-9]+)?$ ]]
then
echo good
else
echo bad
fi
回答by Cris
use this regex instead ^[0-9]*(\.[0-9]+)?$
改用这个正则表达式 ^[0-9]*(\.[0-9]+)?$
回答by glenn Hymanman
Using bash's pattern matching:
使用 bash 的模式匹配：
shopt -s extglob
while read line; do
if [[ $line == ?([-+])+([0-9])?(.*([0-9])) ]] ||
[[ $line == ?(?([-+])*([0-9])).+([0-9]) ]]
then
echo "$line is a number"
else
echo "$line NOT a number"
fi
done << END
1
-1
a
1a
1.0
1.
.0
.
-.0
+
+0
+.0
END
outputs
输出
1 is a number
-1 is a number
a NOT a number
1a NOT a number
1.0 is a number
1. is a number
.0 is a number
. NOT a number
-.0 is a number
+ NOT a number
+0 is a number
+.0 is a number
The patterns:
图案：
- optional sign, followed by one or more digits, followed optionally by a dot and zero or more digits
- optional sign, followed by zero or more digits, followed by a mandatory dot, followed by one or more digits.
- 可选符号，后跟一位或多位数字，后跟可选的点和零位或多位数字
- 可选符号，后跟零个或多个数字，后跟一个强制点，后跟一个或多个数字。
回答by Jigar
Try this: It checks negative and decimal number also.
试试这个：它也检查负数和十进制数。
echo $value | egrep '^-[0-9]+$|^[0-9]+$|^[0-9].[0-9]+$|^-[0-9].[0-9]+$' > /dev/null
回答by MetalGodwin
Works in both Bash 3.0 and 4.0.
适用于 Bash 3.0 和 4.0。
isInteger() {
[[ =~ ^[0-9]+$ ]];
}
isDecimal() {
[[ =~ ^[0-9]+\.[0-9]+$ ]] && ! isInteger ;
}
computer:~ # isDecimal 123 && echo true || echo false
false
computer:~ # isDecimal 12.34 && echo true || echo false
true
computer:~ # isDecimal 12.34a && echo true || echo false
false
computer:~ # isDecimal 0.0000001 && echo true || echo false
true
To check if number, simply test against both functions.
要检查数字，只需对这两个函数进行测试。
回答by J. Katzwinkel
How about this:
这个怎么样：
if [ ! -z $(echo "$value" | grep -o "^[1-9][0-9]*\.\?[0-9]*$") ]; then echo ok; fi
-ztests for an empty string. So the negation [ ! -z "" ]will be fulfilled if the given string starts with a matching pattern.
-z测试空字符串。因此，[ ! -z "" ]如果给定的字符串以匹配模式开头，则否定将得到满足。
回答by William Pursell
In standard shell ([[is non-standard) testwill do the validation for you.
在标准外壳（[[非标准）中test将为您进行验证。
if test "$value" -eq 0 -o "$value" -ne 0 2> /dev/null; then
: # $value is an integer
else
: # $value is not an integer
fi
