Bash if条件在比较特定字符串时不匹配
匹配字符串:
Red Hat Enterprise Linux Server release 7.2 (Maipo)
代码:
machineOSVersion="Red Hat Enterprise Linux Server release 7.2 (Maipo)"
modifiedOSVersion="CentOS Linux release 7 OR Red Hat Enterprise Linux Server release 7.2 (Maipo)"
if [[ ${machineOSVersion} =
我正试图在我的.zshrc中进行操作系统检查。在Ubuntu上运行时,我无法获得与Ubuntu的字符串比较的正确匹配。
片段:
function get_linux_distro()
{
echo `awk -F= '/^NAME/{print $2}' /etc/os-release`
}
function is_os_ubuntu()
{
set -x
local dist=`get_linux_distro`
if [[ ${dist} = "Ubuntu"* ]]; then # <<< string
当makefile需要在不同的操作系统上运行,并且应该根据操作系统正确设置各种设置(转义、路径分隔符等)时,就会出现问题。第一种方法是使用Windows COMSPEC:
ifneq ($(COMSPEC)$(ComSpec),)
## in windows
else
## in linux
endif
这是Cygwin的误报,因为它会看到Windows的环境变量并将Cygwin检测为Windows。然后我们尝试Linux PWD:
ifeq ($(PWD),)
## in windows
else
## in linux, cygwin
endif
然而,由于集成了非现场工具,我们在wi
我试图从这里运行代码,和我发现自己被这个错误信息困住了:
qt.qpa.xcb: could not connect to display
qt.qpa.plugin: Could not load the Qt platform plugin "xcb" in "" even though it was found.
This application failed to start because no Qt platform plugin could be initialized. Reinstalling the application may fix
我正在制作一个节目,讨论你最喜欢的主题,以及你是左脑还是右脑。
function questionGame() {
var user = prompt("Are you RIGHT brained, LEFT brained, a bit of BOTH, or NEITHER?").toUpperCase();
switch (user){
case 'RIGHT':
var user1 = prompt("You like Art?").toUpperCase();
if (user1 = 'YES'
很难理解..。如何检查字符串状态?
x := $$( if [ $(HOME) = "/root" ]; then echo "IS ROOT"; else echo "IS OTHER"; fi )
这是行不通的:
ifeq ($(HOME),"/root")
x = "IS ROOT"
else
x = "IS OTHER"
endif
make --version说:
GNU Make 4.2.1
Built for x86_64-pc-linux-gnu
我已经创建了一个脚本来创建/删除本地用户。
代码:
#!/bin/bash
# This script can create or remove a local linux user.
# --------------------------------------------------------
# Ask if the user wants to create or delete user!
read -p 'Do you want to create or delete user? (c/d) : ' TO_DO
# Account creation
if [[ &
我的问题是,我不明白为什么当我运行脚本时,如果我写"yes“或"no”,它总是说我正在退出脚本。我知道它不承认$reply和“是”是相等的,但是为什么呢?(我是bash编程的新手)。谢谢!
#!/bin/bash
clear
echo 'This script will install: Firefox 17.0.1 (language: enGB or itIT or enUS) and flash 11 in Firefox17.0.1, continue?'
read reply
if (( "$reply" = "yes"
我的问题是,为什么这些表达式是假的?
Python 2.6.4 (r264:75706, Dec 7 2009, 18:45:15)
[GCC 4.4.1] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> num = raw_input("Choose a number: ")
Choose a number: 5
>>> print num
5
>
考虑以下版本go1.18beta2Linux/amd64 64的代码片段
type Vector[T comparable] struct {
data_ []T
}
func (v *Vector[T]) Contains(e T) bool {
for _, x := range v.data_ {
if x == e {
return true
}
}
return false
}
func TestVect
我这样做是为了一个班级,但我对它有问题。我是Linux的新手,真的很艰难。我尝试输入3个值(M,R,T),找出它们是否大于、小于或等于2000,然后打印一条语句。我不确定我做得对不对。我得到了问题,可以输入,但不确定它是否完全有效。
#!/bin/sh
clear
echo -n "What is the value of M?"
read $M
sleep 3
echo -n "What is the value of R?"
read $R
echo -n "What is the value of T?"
read $T
A=$M+