我正在尝试编译Linux源代码,但总是找不到qt,下面是一个输出
[root@localhost linux-2.6.32.5]# yum install qt3
Loaded plugins: refresh-packagekit
Setting up Install Process
Package qt3-3.3.8b-25.fc11.i586 already installed and latest version
Nothing to do
[root@localhost linux-2.6.32.5]# make xconfig
CHECK qt
*
* Unable to
我有一个简单的bash脚本
#!/bin/bash -x
for line in `tail /home/user/line`
do
echo $line
done
但是不知道为什么它会回响所有的单词,而不是下面的每一行都是输出。
++ tail /home/user/line
+ for line in '`tail /home/user/line`'
+ echo Linux
Linux
+ for line in '`tail /home/user/line`'
+ echo is
is
+ for line in '`tail
我将直接举一个例子:
docker run --rm -i alpine /bin/sh -vs <<EOF
echo BEFORE
cat /etc/*elease
echo AFTER
EOF
输出:
echo BEFORE
cat /etc/*elease
echo AFTER
BEFORE
3.9.3
NAME="Alpine Linux"
ID=alpine
VERSION_ID=3.9.3
PRETTY_NAME="Alpine Linux v3.9"
HOME_URL="https://alpinelinux.org/"
有谁知道如何在Linux中通过sed删除下面字符串中的模式"@TechCrunch:“?
str="0,RT @TechCrunch: The Tyranny Of Government And Our Duty Of Confidentiality As Bloggers."
因此,期望的输出将是:
"0,RT The Tyranny Of Government And Our Duty Of Confidentiality As Bloggers."
我试过很多方法,但都没成功,例如:
echo $str | sed 's/@[a-zA-Z]
echo $var命令可以很容易地调用bash中变量的值,如下所示
user@linux:~$ a=1; b=2; c=a+b
user@linux:~$ echo $a $b $c
1 2 a+b
user@linux:~$
我想要实现的是用x中的实际值替换a,b,c
user@linux:~$ a=1; b=2; c=a+b
user@linux:~$ for i in a b c; do echo "$i = x"; done
a = x
b = x
c = x
user@linux:~$
通过使用类似的for循环,我希望能得到这样的输出
a = 1
b = 2
c
我有一个简单的脚本如下。
names=("windows" "ubuntu" "raspbian" "debian" "kali linux")
echo "names array length: ${#names[@]}"
for n in ${names[@]}; do
echo $n
done
我的意图和希望是:
names array length: 5
windows
ubuntu
raspbian
debian
kali linux
但结果却是:
names array len
这是剧本
user@linux:~$ cat script.sh
#!/bin/bash
for i in `seq $#`
do
echo $i
done
user@linux:~$
输出
user@linux:~$ ./script.sh a b c
1
2
3
user@linux:~$
Desired输出
我想得到这样的论证价值.而不仅仅是数字
user@linux:~$ ./script.sh a b c
1 - a
2 - b
3 - c
user@linux:~$
在bash脚本中
if [ 1 ]
then
echo "Yes"
else
echo "No"
fi
输出:Yes
它表示“1”被视为真值。
但在代码中:
word = Linux
letter = nuxi
if echo "$word" | grep -q "$letter"
then
echo "Yes"
else
echo "No"
fi
输出:No
但是echo "$word" | grep -q "$letter"将返回1,那
紧随其后
https://stackoverflow.com/questions/1809899/how-can-i-assign-the-output-of-a-function-to-a-variable-using-bash和如何将命令的输出分配给变量?
我写
function getos(){
# http://stackoverflow.com/a/27776822/1637673
case "$(uname -s)" in
Darwin)
OS='mac'
;;
Linu
我有一个shell脚本,可以在这样的文件夹中打印appimage文件名
#! /bin/bash
Dir="$HOME/Applications/"
Dir2="$HOME/Downloads/"
cd -P "$Dir"
for f in *.AppImage; do
z=$(echo $f | head -n1 | awk '{print $1;}')
echo $z
done
现在输出如下:
Altus-4.8.5-x86_64.AppImage
GitHubDesktop-linux-3.2.0-linux1.Ap