如何在Bash中查找变量是否为空

Question

如何在Bash中检查变量是否为空？

Answer 1

在bash中,至少以下命令测试$ var是否为空:

if [[ -z "$var" ]]; then
   # Do what you want
fi

这个命令man test是你的朋友.

Answer 2

推定bash:

var=""

if [ -n "$var" ]; then
    echo "not empty"
else
    echo "empty"
fi

Answer 3

我也见过

if [ "x$variable" = "x" ]; then ...

这显然非常强大且与shell无关.

此外,"空"和"未设置"之间存在差异.请参见如何判断bash shell脚本中是否未定义字符串？.

Answer 4

if [ ${foo:+1} ]
then
    echo "yes"
fi

yes如果设置了变量,则打印.${foo:+1}设置变量时返回1,否则返回空字符串.

Answer 5

if [[ "$variable" == "" ]] ...

Answer 6

[ "$variable" ] || echo empty
: ${variable="value_to_set_if_unset"}

Answer 7

true如果未设置变量或将变量设置为空字符串(""),则会返回此值.

if [ -z "$MyVar" ]
then
   echo "The variable MyVar has nothing in it."
elif ! [ -z "$MyVar" ]
then
   echo "The variable MyVar has something in it."
fi

Answer 8

该问题询问如何检查变量是否为空字符串,并且已经给出了最佳答案.
但是我经过一段时间在PHP中进行编程后才登陆这里,我实际上搜索的是像在bash shell中工作的php中的空函数一样的检查.
在阅读完答案之后,我意识到我在bash中没有正确思考,但无论如何在那个时刻像php中的空函数在我的bash代码中会非常方便.
由于我认为这可能发生在其他人身上,我决定在bash中转换php空函数

根据php手册:
如果变量不存在或者其值是以下值之一,则该变量被视为空:

• ""(空字符串)
• 0(0为整数)
• 0.0(0作为浮点数)
• "0"(0作为字符串)
• 一个空数组
• 声明的变量,但没有值

当然,无法在bash中转换null和false情况,因此省略它们.

function empty
{
    local var="$1"

    # Return true if:
    # 1.    var is a null string ("" as empty string)
    # 2.    a non set variable is passed
    # 3.    a declared variable or array but without a value is passed
    # 4.    an empty array is passed
    if test -z "$var"
    then
        [[ $( echo "1" ) ]]
        return

    # Return true if var is zero (0 as an integer or "0" as a string)
    elif [ "$var" == 0 2> /dev/null ]
    then
        [[ $( echo "1" ) ]]
        return

    # Return true if var is 0.0 (0 as a float)
    elif [ "$var" == 0.0 2> /dev/null ]
    then
        [[ $( echo "1" ) ]]
        return
    fi

    [[ $( echo "" ) ]]
}

用法示例:

if empty "${var}"
    then
        echo "empty"
    else
        echo "not empty"
fi

演示:
以下代码段:

#!/bin/bash

vars=(
    ""
    0
    0.0
    "0"
    1
    "string"
    " "
)

for (( i=0; i<${#vars[@]}; i++ ))
do
    var="${vars[$i]}"

    if empty "${var}"
        then
            what="empty"
        else
            what="not empty"
    fi
    echo "VAR \"$var\" is $what"
done

exit

输出:

VAR "" is empty
VAR "0" is empty
VAR "0.0" is empty
VAR "0" is empty
VAR "1" is not empty
VAR "string" is not empty
VAR " " is not empty

在bash逻辑中说过,这个函数中的零检查可能会导致问题,任何使用此函数的人都应该评估这个风险,并且可能决定将这些检查切掉而只留下第一个.