主页 / user-166917

Dave's questions

Martin Hope
Dave
Asked: 2023-06-15 21:38:01 +0800 CST
  • 8

我正在使用 bash 外壳。如果我有行号文件

1
4
7
9

和另一个行文件，其中第一个元素是一个数字，后跟一串文本

1,Michael Jordan
2,Karl Malone,
3,Charles Barkley
4,Greg Anthony
5,Chris Mullen
6,Reggie Miller
7,Billy Owens
8,David Robinson
9,Shaquille O'Neal
10,John Stockton

awk仅当第一个数字属于第一个文件时，如何编写命令从第二个文件中提取行？在上面的例子中，我希望这个结果

1,Michael Jordan
4,Greg Anthony
7,Billy Owens
9,Shaquille O'Neal

我试过这个

awk 'FNR==NR{a[$1];next} $1 in a' /tmp/first_file /tmp/second_file > /tmp/third_file

但它在第三个文件中什么也没有产生。

shell-script
Martin Hope
Dave
Asked: 2022-02-04 17:55:55 +0800 CST
  • 0

我正在使用 ag v 2.2。如何配置 ag 以便它搜索以“。”开头的文件 除了所有其他文件？我注意到我有一个像这样的文件

$ cat client/.env.production 
REACT_APP_PROXY=https://map.chicommons.coop

但是当我使用该文件中的术语搜索“ag”时，该术语不会出现......

$ ag 'map\.chicom' .
web/directory/settings.py
28:ALLOWED_HOSTS = ['127.0.0.1', 'localhost', 'dev.chicommons.coop', 'map.chicommons.coop']

client/config.js
3:    ? "map.chicommons.com"
bash search
Martin Hope
Dave
Asked: 2021-12-12 12:23:22 +0800 CST
  • 0

我想在远程服务器（Cent OS 7）上执行命令并使用该服务器上定义的环境变量。但是，当我运行此命令时

ssh [email protected] "PGPASSWORD=$DB_PASS psql -U $DB_USER -d $DB_NAME -c 'COPY myapp_currencyprice to STDOUT WITH (DELIMITER \",\", FORMAT CSV, HEADER)' > /tmp/prices.csv"

似乎该命令正在尝试使用来自 localhost 而不是来自“remotehost.com”的“$DB_PASS”和其他环境变量。从远程服务器引用环境变量的正确方法是什么？

bash ssh
Martin Hope
Dave
Asked: 2021-09-16 08:44:04 +0800 CST
  • 1

我在 Mac OS Big Sur 上运行虚拟 Ubuntu 16.04 Linux 实例（使用 vagrant）。我可以很好地运行以下两个命令（一个用于 ssh 进入虚拟服务器，第二个用于在其中运行命令）......

$ ssh myvirtual.local
Warning: Permanently added '10.0.4.19' (ECDSA) to the list of known hosts.
Last login: Wed Sep 15 16:37:57 2021 from 10.0.4.1
$ foreman start -f Procfile.debug
16:38:26 rails.1      | started with pid 27884
16:38:26 worker.1     | started with pid 27885
16:38:26 scheduler.1  | started with pid 27887
...

我希望能够将这两个命令结合起来，所以我尝试了

$ ssh -t myvirtual.local 'foreman start -f Procfile.debug’
Warning: Permanently added '10.0.4.19' (ECDSA) to the list of known hosts.
bash: foreman: command not found
Connection to 10.0.4.19 closed.

我很困惑我需要运行什么其他设置才能有一个命令来模拟我最初正在做的事情。有什么想法吗？

bash ssh
Martin Hope
Dave
Asked: 2020-03-06 08:48:45 +0800 CST
  • 0

我正在使用 bash shell。我已经定义了以下变量...

localhost:tmp davea$ echo $json
{"id": "abc", "name": "dave"}

我想将此 var 的值插入到 ruby​​ 命令中，所以我尝试了这个...

localhost:tmp davea$ ruby -rjson -e 'j = JSON.parse($json); puts j["Instances"][0]["ImageId"]'
Ignoring bindex-0.5.0 because its extensions are not built.  Try: gem pristine bindex --version 0.5.0
Ignoring byebug-11.0.1 because its extensions are not built.  Try: gem pristine byebug --version 11.0.1
Ignoring byebug-10.0.2 because its extensions are not built.  Try: gem pristine byebug --version 10.0.2
Ignoring byebug-10.0.1 because its extensions are not built.  Try: gem pristine byebug --version 10.0.1
Ignoring byebug-10.0.0 because its extensions are not built.  Try: gem pristine byebug --version 10.0.0
Ignoring childprocess-1.0.1 because its extensions are not built.  Try: gem pristine childprocess --version 1.0.1
Ignoring debase-0.2.4 because its extensions are not built.  Try: gem pristine debase --version 0.2.4
Ignoring eventmachine-1.2.5 because its extensions are not built.  Try: gem pristine eventmachine --version 1.2.5
Ignoring ffi-1.11.1 because its extensions are not built.  Try: gem pristine ffi --version 1.11.1
Ignoring ffi-1.9.25 because its extensions are not built.  Try: gem pristine ffi --version 1.9.25
Ignoring ffi-1.9.23 because its extensions are not built.  Try: gem pristine ffi --version 1.9.23
Ignoring ffi-1.9.18 because its extensions are not built.  Try: gem pristine ffi --version 1.9.18
Ignoring hiredis-0.6.1 because its extensions are not built.  Try: gem pristine hiredis --version 0.6.1
Ignoring json-2.1.0 because its extensions are not built.  Try: gem pristine json --version 2.1.0
Ignoring nio4r-2.4.0 because its extensions are not built.  Try: gem pristine nio4r --version 2.4.0
Ignoring nio4r-2.3.1 because its extensions are not built.  Try: gem pristine nio4r --version 2.3.1
Ignoring nio4r-2.3.0 because its extensions are not built.  Try: gem pristine nio4r --version 2.3.0
Ignoring nio4r-2.2.0 because its extensions are not built.  Try: gem pristine nio4r --version 2.2.0
Ignoring nokogiri-1.10.4 because its extensions are not built.  Try: gem pristine nokogiri --version 1.10.4
Ignoring nokogiri-1.8.2 because its extensions are not built.  Try: gem pristine nokogiri --version 1.8.2
Ignoring nokogiri-1.8.1 because its extensions are not built.  Try: gem pristine nokogiri --version 1.8.1
Ignoring pg-1.1.4 because its extensions are not built.  Try: gem pristine pg --version 1.1.4
Ignoring pg-0.21.0 because its extensions are not built.  Try: gem pristine pg --version 0.21.0
Ignoring puma-3.12.1 because its extensions are not built.  Try: gem pristine puma --version 3.12.1
Ignoring puma-3.11.4 because its extensions are not built.  Try: gem pristine puma --version 3.11.4
Ignoring puma-3.11.3 because its extensions are not built.  Try: gem pristine puma --version 3.11.3
Ignoring puma-3.11.2 because its extensions are not built.  Try: gem pristine puma --version 3.11.2
Ignoring rmagick-2.16.0 because its extensions are not built.  Try: gem pristine rmagick --version 2.16.0
Ignoring ruby-debug-ide-0.7.0 because its extensions are not built.  Try: gem pristine ruby-debug-ide --version 0.7.0
Ignoring ruby-filemagic-0.7.2 because its extensions are not built.  Try: gem pristine ruby-filemagic --version 0.7.2
Ignoring sqlite3-1.4.1 because its extensions are not built.  Try: gem pristine sqlite3 --version 1.4.1
Ignoring sqlite3-1.3.13 because its extensions are not built.  Try: gem pristine sqlite3 --version 1.3.13
Ignoring websocket-driver-0.6.5 because its extensions are not built.  Try: gem pristine websocket-driver --version 0.6.5
Ignoring websocket-native-1.0.0 because its extensions are not built.  Try: gem pristine websocket-native --version 1.0.0
/Users/davea/.rvm/gems/ruby-2.4.0/gems/json-2.1.0/lib/json/pure/parser.rb:135:in `convert_encoding': nil is not like a string (TypeError)
    from /Users/davea/.rvm/gems/ruby-2.4.0/gems/json-2.1.0/lib/json/pure/parser.rb:78:in `initialize'
    from /Users/davea/.rvm/gems/ruby-2.4.0/gems/json-2.1.0/lib/json/common.rb:156:in `new'
    from /Users/davea/.rvm/gems/ruby-2.4.0/gems/json-2.1.0/lib/json/common.rb:156:in `parse'
    from -e:1:in `<main>'

看来我的 $json 没有被翻译。将其插入上述命令的正确方法是什么？

environment-variables bash
Martin Hope
Dave
Asked: 2019-02-20 14:45:29 +0800 CST
  • 1

我正在使用 CentOS 7。我想使用 Pip 安装 virtualenv 。所以我像这样安装了Pip ...

[laredotornado@server Python-3.7.2]$ sudo yum install python-pip httpd mod_wsgi
[sudo] password for laredotornado: 
Loaded plugins: fastestmirror
Loading mirror speeds from cached hostfile
 * base: mirror.fileplanet.com
 * extras: mirror.us-midwest-1.nexcess.net
 * updates: mirror.fileplanet.com
No package python-pip available.
Package httpd-2.4.6-88.el7.centos.x86_64 already installed and latest version
Package mod_wsgi-3.4-18.el7.x86_64 already installed and latest version
Nothing to do

但是当我实际尝试使用 pip 时，我被告知找不到它...

[laredotornado@server Python-3.7.2]$ sudo pip install virtualenv
sudo: pip: command not found

如何在我的 Linux 发行版上正确安装 pip？

centos python
Martin Hope
Dave
Asked: 2018-08-07 11:15:38 +0800 CST
  • 7

我正在使用 bash。我有一个 CSV 文件，其中包含两列数据，大致如下所示

 num_logins,day
 253,2016-07-01
 127,2016-07-02

我想交换第一列和第二列（使日期列成为第一列）。所以我尝试了这个

awk ' { t = $1; $1 = $2; $2 = t; print; } ' /tmp/2016_logins.csv

但是，结果输出相同。为了让事情正确切换，我在上面的 awk 语句中遗漏了什么？

shell-script text-processing
Martin Hope
Dave
Asked: 2018-06-01 12:17:42 +0800 CST
  • 7

我正在使用带有 bash shell 的 CentOS 7。我认为对二进制文件进行 base64 编码就像

[rails@server lib]$ cat mybinary.file | base64 > /tmp/output.base64

但是，我注意到当我查看文件长度时，它不是四的倍数

[rails@server lib]$ ls -al /tmp/output.base64 
-rw-rw-r-- 1 rails rails 92935 May 31 15:50 /tmp/output.base64

我不知道我所做的是否有效，但是当我尝试使用 JS 库解码文件时，我收到一个错误，抱怨字符串长度不是四的倍数，所以我想知道我上面所做的是否正确，或者是否有其他方法可以做到。

centos character-encoding
Martin Hope
Dave
Asked: 2018-05-30 10:45:00 +0800 CST
  • 3

我在 Amazon Linux 上使用 bash shell。我在 shell 脚本中有一个命令

wget -O - "http://localhost:8088/subco/books/$e_id/segments/$segment_id?product=$product_id&audience=teacher" > /dev/null

读到这个——https://superuser.com/questions/321240/how-do-you-redirect-wget-to-standard-out/321241，我被引导相信我可以输出 wget 的结果（它检索到的) 到我的屏幕并抑制所有其他输出。然而，相反，我得到的是

--2018-05-29 18:39:49--  http://localhost:8088/subco/books/C2644BB08F394E209A26175BD2C89F5A/segments/C2F62E7002964DD396E381DB331129A4?product=D399B9C5F6204EDE80A002930CC0D02F&audience=teacher
Resolving localhost (localhost)... 127.0.0.1
Connecting to localhost (localhost)|127.0.0.1|:8088... connected.
HTTP request sent, awaiting response... 200 OK
Length: unspecified [text/html]
Saving to: ‘STDOUT’

如何将 wget 的结果输出到屏幕并抑制所有其他信息（例如“发送的 HTTP 请求，等待响应...... 200 OK”部分）？

bash shell-script
Martin Hope
Dave
Asked: 2018-05-30 09:25:12 +0800 CST
  • 12

我在 CentOS 7 上使用 bash shell。我想从 shell 脚本运行 MySQL 查询并遍历每一行结果。如果返回 4 行，我想我可以像这样捕获数组中的四行：

query="select p.id, p.ebook_id, es.id FROM ...";
echo "$query" > /tmp/query.sql
mysql -u user --password=pass db_id < /tmp/query.sql > /tmp/query.csv

linesIN=`cat /tmp/query.csv | sed 's/\t/,/g'`
arraylength=${#linesIN[@]}
echo $arraylength

但是，即使我可以看到返回的多个结果行，也$arraylength总是输出。1如何修改上述内容以正确创建结果数组，其中数组中的每个元素代表结果集中的一行？

bash shell-script
Martin Hope
Dave
Asked: 2018-05-15 17:43:59 +0800 CST
  • 0

我正在使用 CentOS 7。我正在尝试在脚本变量中捕获进程的“主 PID”。这是我的脚本

#!/bin/sh

set -e

PID="$APP_ROOT/shared/pids/puma.pid"
echo "before ..."
MASTER_PID=`pgrep -f '^([^ ]*/)?puma '`
echo "after ..."
xxx

但是，永远不会打印出“after ...”这个词。这条线有点东西

MASTER_PID=`pgrep -f '^([^ ]*/)?puma '`

这导致事情表现不佳，尤其是在没有运行 puma 进程的情况下。有谁知道重写上述内容的方法，以便我可以捕获我的主 PID，或者如果它不存在，至少允许执行传递到下一行？

shell-script shell
Martin Hope
Dave
Asked: 2018-05-15 12:04:53 +0800 CST
  • 20

我正在使用 CentOS 7。我想获取在端口 3000 上运行的进程的 PID（如果存在）。我想获取此 PID，以便将其保存到 shell 脚本中的变量中。到目前为止我有

[rails@server proddir]$ sudo ss -lptn 'sport = :3000'
State      Recv-Q Send-Q                           Local Address:Port                                          Peer Address:Port
Cannot open netlink socket: Protocol not supported
LISTEN     0      0                                            *:3000                                                     *:*                   users:(("ruby",pid=4861,fd=7),("ruby",pid=4857,fd=7),("ruby",pid=4855,fd=7),("ruby",pid=4851,fd=7),("ruby",pid=4843,fd=7))

但我无法弄清楚如何在没有所有这些额外信息的情况下单独隔离 PID。

linux shell-script
Martin Hope
Dave
Asked: 2018-05-10 13:40:04 +0800 CST
  • 3

我正在使用 CentOS 7。我正在尝试编写一个脚本来启动和停止 puma 进程，但如果 taht 是正确的术语，我无法弄清楚如何获得“主”PID。在下面的命令中

[rails@server myproject_production]$ ps aux | grep puma
rails    15767  0.0  1.2 437904 13612 ?        Sl   17:20   0:00 puma 3.11.4 (tcp://0.0.0.0:3000,unix:///home/rails/myproject_production/shared/sockets/puma.sock) [myproject_production]
rails    15779  0.6  7.6 1061248 80688 ?       Sl   17:20   0:05 puma: cluster worker 1: 15767 [myproject_production]
rails    15781  0.6  7.7 1061248 80876 ?       Sl   17:20   0:05 puma: cluster worker 2: 15767 [myproject_production]
rails    15785  0.6  7.4 1061964 78488 ?       Sl   17:20   0:05 puma: cluster worker 3: 15767 [myproject_production]
rails    15880  0.7  7.4 1059612 78592 ?       Sl   17:22   0:05 puma: cluster worker 0: 15767 [myproject_production]
rails    17106  0.0  0.1 112612  1064 pts/0    S+   17:33   0:00 grep --color=auto puma

主 PID 为“15767”。如果我杀死所有其他美洲狮进程将死亡。如何编写命令以将 taht 放入脚本变量？

shell-script centos
Martin Hope
Dave
Asked: 2018-05-09 07:24:41 +0800 CST
  • 24

我正在使用 CentOS 7。如何找出服务无法启动的原因？我创建了这个服务

[rails@server ~]$ sudo cat /usr/lib/systemd/system/nodejs.service
[Unit]
Description=nodejs server

[Service]
User=rails
Group=rails
ExecStart=/home/rails/NodeJSserver/start.sh
ExecStop=/home/rails/NodeJSserver/stop.sh

[Install]
WantedBy=multi-user.target

该文件指向此

[rails@server ~]$ cat /home/rails/NodeJSserver/start.sh
#!/bin/bash

forever start /home/rails/NodeJSserver/server.js

我可以自己运行这个文件。但是当我尝试将它作为服务的一部分运行时，我注意到我的 nodeJS 服务器没有启动。即使我检查“sudo systemctl --state=failed”，我也看不到任何错误......

[rails@server ~]$ sudo systemctl enable NodeJSserver
[rails@server ~]$ sudo systemctl start NodeJSserver
[rails@server ~]$
[rails@server ~]$
[rails@server ~]$ forever list
info:    No forever processes running
[rails@server ~]$
[rails@server ~]$
[rails@server ~]$ sudo systemctl --state=failed
  UNIT                           LOAD   ACTIVE SUB    DESCRIPTION
● nginx.service                  loaded failed failed The nginx HTTP and reverse proxy server
● systemd-sysctl.service         loaded failed failed Apply Kernel Variables
● systemd-vconsole-setup.service loaded failed failed Setup Virtual Console

LOAD   = Reflects whether the unit definition was properly loaded.
ACTIVE = The high-level unit activation state, i.e. generalization of SUB.
SUB    = The low-level unit activation state, values depend on unit type.

3 loaded units listed. Pass --all to see loaded but inactive units, too.
To show all installed unit files use 'systemctl list-unit-files'.

如何找出我的服务无法启动的原因？

centos boot
Martin Hope
Dave
Asked: 2018-05-08 11:57:04 +0800 CST
  • 0

我正在使用 CentOS 7。我想将 nodeJS 脚本作为服务运行，但以“rails”用户身份运行该服务。作为 root，我创建了文件 /usr/lib/systemd/system/myservice.service，其内容

[Unit]
Description=mydir nodejs server

[Service]
ExecStart=/home/rails/mydir/start.sh
ExecStop=/home/rails/mydir/stop.sh

[Install]
WantedBy=multi-user.target

我还需要做什么才能让服务以我想要的用户身份运行？

centos systemd
Martin Hope
Dave
Asked: 2017-12-08 07:38:41 +0800 CST
  • 5

我正在使用亚马逊 Linux。我已经对文件设置了全局读取权限，但我似乎无法以普通用户身份访问它：

[myuser@mymachine ~]$ ls -al /usr/java/jboss/standalone/deployments/myproject.war/css/reset.css
ls: cannot access /usr/java/jboss/standalone/deployments/myproject.war/css/reset.css: Permission denied
[myuser@mymachine ~]$ sudo ls -al /usr/java/jboss/standalone/deployments/myproject.war/css/reset.css
-rwxrwxr-x 1 jboss jboss 771 Oct 29 18:51 /usr/java/jboss/standalone/deployments/myproject.war/css/reset.css
[myuser@mymachine ~]$ whoami
myuser

请注意，当我运行“sudo”时，我可以访问它。我想保留 jboss 用户拥有的文件。如何在读取模式下让我的（或其他任何人的用户）可以访问该文件？

permissions users
Martin Hope
Dave
Asked: 2017-12-07 13:11:51 +0800 CST
  • 0

我正在使用 bash 外壳。我有一个 CSV 文件，其中每一行的标记都用逗号分隔。我想取出第二列和第三列并从中形成新的字符串（SQL 语句）。我想我可以为此目的使用 awk，所以我尝试了......

localhost:mydir davea$ awk -F ',' -v OFS=',' "REPLACE INTO my_table (ID, NAME, HOURS) VALUES ('$2', '$2', '$3');" types.csv
awk: syntax error at source line 1
 context is
    REPLACE INTO my_table >>>  (ID, <<<
awk: bailing out at source line 1

但如您所见，我遇到了一个错误。我遗漏了什么吗？如何从 CSV 文件中的每一行形成我的新字符串？

bash shell-script