Linux 如何在 Bash 中测试变量是否为数字？

How do I test if a variable is a number in Bash?

提问by Flávio Amieiro

I just can't figure out how do I make sure an argument passed to my script is a number or not.

我只是不知道如何确保传递给我的脚本的参数是否为数字。

All I want to do is something like this:

我想要做的就是这样：

test *isnumber*  && VAR= || echo "need a number"

Any help?

有什么帮助吗？

采纳答案by Charles Duffy

One approach is to use a regular expression, like so:

一种方法是使用正则表达式，如下所示：

re='^[0-9]+$'
if ! [[ $yournumber =~ $re ]] ; then
   echo "error: Not a number" >&2; exit 1
fi

If the value is not necessarily an integer, consider amending the regex appropriately; for instance:

如果该值不一定是整数，请考虑适当修改正则表达式；例如：

^[0-9]+([.][0-9]+)?$

...or, to handle numbers with a sign:

...或者，处理带符号的数字：

^[+-]?[0-9]+([.][0-9]+)?$

回答by Charles Duffy

This is a little rough around the edges but a little more novice friendly.

这有点粗糙，但对新手更友好。

if [ $number -ge 0 ]
then
echo "Continue with code block"
else
echo "We matched 0 or $number is not a number"
fi

This will cause an error and print "Illegal number:" if $number is not a number but it will not break out of the script. Oddly there is not a test option I could find to just test for an integer. The logic here will match any number that is greater than or equal to 0.

这将导致错误并打印“非法数字：”如果 $number 不是数字但它不会脱离脚本。奇怪的是，我找不到一个测试选项来测试一个整数。这里的逻辑将匹配任何大于或等于 0 的数字。

回答by Alberto Zaccagni

The following solution can also be used in basic shells such as Bourne without the need for regular expressions. Basically any numeric value evaluation operations using non-numbers will result in an error which will be implicitly considered as false in shell:

以下解决方案也可用于基本 shell，例如 Bourne，而无需使用正则表达式。基本上任何使用非数字的数值计算操作都会导致错误，在 shell 中将被隐式视为 false：

"$var" -eq "$var"

as in:

如：

#!/bin/bash

var=a

if [ -n "$var" ] && [ "$var" -eq "$var" ] 2>/dev/null; then
  echo number
else
  echo not a number
fi

You can can also test for $? the return code of the operation which is more explicit:

您也可以测试 $? 更明确的操作返回码：

[ -n "$var" ] && [ "$var" -eq "$var" ] 2>/dev/null
if [ $? -ne 0 ]; then
   echo $var is not number
fi

Redirection of standard error is there to hide the "integer expression expected" message that bash prints out in case we do not have a number.

标准错误的重定向是为了隐藏 bash 打印出的“预期整数表达式”消息，以防我们没有数字。

CAVEATS(thanks to the comments below):

警告（感谢以下评论）：

• Numbers with decimal points are notidentified as valid "numbers"
• Using [[ ]]instead of [ ]will always evaluate to true
• Most non-Bash shells will always evaluate this expression as true
• The behavior in Bash is undocumented and may therefore change without warning
• If the value includes spaces after the number (e.g. "1 a") produces error, like bash: [[: 1 a: syntax error in expression (error token is "a")
• If the value is the same as var-name (e.g. i="i"), produces error, like bash: [[: i: expression recursion level exceeded (error token is "i")

• 带小数点的数字不被识别为有效的“数字”
• 使用[[ ]]而不是[ ]将始终评估为true
• 大多数非 Bash shell 将始终将此表达式评估为 true
• Bash 中的行为没有记录，因此可能会在没有警告的情况下更改
• 如果该值在数字后包含空格（例如“1 a”）会产生错误，例如 bash: [[: 1 a: syntax error in expression (error token is "a")
• 如果该值与 var-name 相同（例如 i="i"），则会产生错误，例如 bash: [[: i: expression recursion level exceeded (error token is "i")

回答by Marnix

I use the following (for integers):

我使用以下（对于整数）：

## ##### constants
##
## __TRUE - true (0)
## __FALSE - false (1)
##
typeset -r __TRUE=0
typeset -r __FALSE=1

## --------------------------------------
## isNumber
## check if a value is an integer 
## usage: isNumber testValue 
## returns: ${__TRUE} - testValue is a number else not
##
function isNumber {
  typeset TESTVAR="$(echo "" | sed 's/[0-9]*//g' )"
  [ "${TESTVAR}"x = ""x ] && return ${__TRUE} || return ${__FALSE}
}

isNumber  
if [ $? -eq ${__TRUE} ] ; then
  print "is a number"
fi

回答by mrucci

This tests if a number is a non negative integer and is both shell independent (i.e. without bashisms) and uses only shell built-ins:

这将测试一个数字是否为非负整数并且与 shell 无关（即没有 bashisms）并且仅使用 shell 内置函数：

INCORRECT.

不正确。

As this first answer (below) allows for integers with characters in them as long as the first are not first in the variable.

由于第一个答案（如下）允许包含字符的整数，只要第一个不是变量中的第一个。

[ -z "${num##[0-9]*}" ] && echo "is a number" || echo "is not a number";

CORRECT.

正确。

As jillescommented and suggested in his answerthis is the correct way to do it using shell-patterns.

正如jilles在他的回答中评论和建议的那样，这是使用 shell-patterns 的正确方法。

[ ! -z "${num##*[!0-9]*}" ] && echo "is a number" || echo "is not a number";

回答by ultrasawblade

http://tldp.org/LDP/Bash-Beginners-Guide/html/sect_04_03.html

http://tldp.org/LDP/Bash-Beginners-Guide/html/sect_04_03.html

You can also use bash's character classes.

您还可以使用 bash 的字符类。

if [[ $VAR = *[[:digit:]]* ]]; then
 echo "$VAR is numeric"
else
 echo "$VAR is not numeric"
fi

Numerics will include space, the decimal point, and "e" or "E" for floating point.

数字将包括空格、小数点和浮点数的“e”或“E”。

But, if you specify a C-style hex number, i.e. "0xffff" or "0XFFFF", [[:digit:]] returns true. A bit of a trap here, bash allows you do to something like "0xAZ00" and still count it as a digit (isn't this from some weird quirk of GCC compilers that let you use 0x notation for bases other than 16???)

但是，如果您指定 C 风格的十六进制数，即“0xffff”或“0XFFFF”，[[:digit:]] 返回真。这里有点陷阱，bash 允许你对“0xAZ00”之类的东西做一些事情，并且仍然把它算作一个数字（这不是来自 GCC 编译器的一些奇怪的怪癖，它让你对 16 以外的基数使用 0x 符号？ )

You might want to test for "0x" or "0X" before testing if it's a numeric if your input is completely untrusted, unless you want to accept hex numbers. That would be accomplished by:

如果您的输入完全不受信任，您可能希望在测试它是否是数字之前测试“0x”或“0X”，除非您想接受十六进制数字。这将通过以下方式实现：

if [[ ${VARIABLE:1:2} = "0x" ]] || [[ ${VARIABLE:1:2} = "0X" ]]; then echo "$VAR is not numeric"; fi

回答by ata

I tried ultrasawblade's recipe as it seemed the most practical to me, and couldn't make it work. In the end i devised another way though, based as others in parameter substitution, this time with regex replacement:

我尝试了 Ultrasawblade 的配方，因为它对我来说似乎是最实用的，但无法实现。最后，我设计了另一种方法，基于参数替换中的其他方法，这次使用正则表达式替换：

[[ "${var//*([[:digit:]])}" ]]; && echo "$var is not numeric" || echo "$var is numeric"

It removes every :digit: class character in $var and checks if we are left with an empty string, meaning that the original was only numbers.

它删除 $var 中的每个 :digit: 类字符并检查我们是否留下空字符串，这意味着原始字符串只是数字。

What i like about this one is its small footprint and flexibility. In this form it only works for non-delimited, base 10 integers, though surely you can use pattern matching to suit it to other needs.

我喜欢这个的地方在于它的占地面积小和灵活性。在这种形式中，它仅适用于非分隔的、以 10 为基数的整数，但您当然可以使用模式匹配来满足其他需求。

回答by jilles

Without bashisms (works even in the System V sh),

没有 bashisms（即使在 System V sh 中也能工作），

case $string in
    ''|*[!0-9]*) echo bad ;;
    *) echo good ;;
esac

This rejects empty strings and strings containing non-digits, accepting everything else.

这拒绝空字符串和包含非数字的字符串，接受其他所有内容。

Negative or floating-point numbers need some additional work. An idea is to exclude -/ .in the first "bad" pattern and add more "bad" patterns containing the inappropriate uses of them (?*-*/ *.*.*)

负数或浮点数需要一些额外的工作。一个想法是在第一个“坏”模式中排除-/.并添加更多包含不适当使用它们的“坏”模式（?*-*/ *.*.*）

回答by WWWIZARDS

Quick & Dirty: I know it's not the most elegant way, but I usually just added a zero to it and test the result. like so:

Quick & Dirty：我知道这不是最优雅的方式，但我通常只是给它添加一个零并测试结果。像这样：

function isInteger {
  [ $((+0)) != 0 ] && echo " is a number" || echo " is not a number"
 }

x=1;      isInteger $x
x="1";    isInteger $x
x="joe";  isInteger $x
x=0x16 ;  isInteger $x
x=-32674; isInteger $x   

$(($1+0)) will return 0 or bomb if $1 is NOT an integer. for Example:

如果 $1 不是整数，则 $(($1+0)) 将返回 0 或炸弹。例如：

function zipIt  { # quick zip - unless the 1st parameter is a number
  ERROR="not a valid number. " 
  if [ $((+0)) != 0 ] ; then  # isInteger() 
      echo " backing up files changed in the last  days."
      OUT="zipIt--day.tgz" 
      find . -mtime - -type f -print0 | xargs -0 tar cvzf $OUT 
      return 1
  fi
    showError $ERROR
}

NOTE: I guess I never thought to check for floats or mixed types that will make the entire script bomb... in my case, I didn't want it go any further. I'm gonna play around with mrucci's solution and Duffy's regex - they seem the most robust within the bash framework...

注意：我想我从来没有想过检查会导致整个脚本炸弹的浮点数或混合类型......在我的情况下，我不希望它更进一步。我将尝试使用 mrucci 的解决方案和 Duffy 的正则表达式——它们似乎是 bash 框架中最强大的......

回答by Shirish Shukla

Below is a Script written by me and used for a script integration with Nagios and it is working properly till now

下面是我写的一个脚本，用于与 Nagios 的脚本集成，到目前为止它工作正常

#!/bin/bash
# Script to test variable is numeric or not
# Shirish Shukla
# Pass arg1 as number
a1=
a=$(echo $a1|awk '{if( > 0) print ; else print "*-1"}')
b=$(echo "scale=2;$a/$a + 1" | bc -l 2>/dev/null)
if [[ $b > 1 ]]
then
    echo " is Numeric"
else
    echo " is Non Numeric"
fi

EG:

例如：

# sh isnumsks.sh   "-22.22"
-22.22 is Numeric

# sh isnumsks.sh   "22.22"
22.22 is Numeric

# sh isnumsks.sh   "shirish22.22"
shirish22.22 is Non  Numeric