I've just started Api-Platform framework and while executing:
php bin/schema generate-types src/ app/config/schema.yml
I get this:
C:\wamp\www\sf2-api>php bin/schema generate-types src/ app/config/schema.yml
dir=$(d=${0%[/\\]*}; cd "$d"; cd "../vendor/api-platform/schema-generator/bin" &
& pwd)
# See if we are running in Cygwin by checking for cygpath program
if command -v 'cygpath' >/dev/null 2>&1; then
# Cygwin paths start with /cygdrive/ which will break windows PHP,
# so we need to translate the dir path to windows format. However
# we could be using cygwin PHP which does not require this, so we
# test if the path to PHP starts with /cygdrive/ rather than /usr/bin
if [[ $(which php) == /cygdrive/* ]]; then
dir=$(cygpath -m $dir);
fi
fi
dir=$(echo $dir | sed 's/ /\ /g')
"${dir}/schema" "$#"
I am using Symfony 2.7.8 on window7.
I have the same issue on ubunbu 14.04.
Finally, I replace the bin directory with the one in blog-api.
Updated:
The bin-api-platform is the one generated by api-platform.
The bin-blog-api is the one I copy from blog-api. This works fine.
Use :
php vendor/api-platform/schema-generator/bin/schema generate-types src/app/config/schema.yml
instead of :
php bin/schema generate-types src/ app/config/schema.yml
The correct syntax is:
php vendor/api-platform/schema-generator/bin/schema generate-types src/ app/config/schema.yml
Related
When using WP CLI in docker, I need to execute it as root.
I need to add the flag --allow-root directly in .bashrc and I am trying to figure out why it doesn't work.
FROM webdevops/php-dev:7.3
# configure postfix to use mailhog
RUN postconf -e "relayhost = mail:1025"
# install wp cli
RUN curl -O https://raw.githubusercontent.com/wp-cli/builds/gh-pages/phar/wp-cli.phar && \
chmod +x wp-cli.phar && \
mv wp-cli.phar /usr/local/bin/wp && \
echo 'wp() {' >> ~/.bashrc && \
echo '/usr/local/bin/wp "$#" --allow-root' >> ~/.bashrc && \
echo '}' >> ~/.bashrc
WORKDIR /var/www/html/
my .bashrc
# ~/.bashrc: executed by bash(1) for non-login shells.
# Note: PS1 and umask are already set in /etc/profile. You should not
# need this unless you want different defaults for root.
# PS1='${debian_chroot:+($debian_chroot)}\h:\w\$ '
# umask 022
# You may uncomment the following lines if you want `ls' to be colorized:
# export LS_OPTIONS='--color=auto'
# eval "`dircolors`"
# alias ls='ls $LS_OPTIONS'
# alias ll='ls $LS_OPTIONS -l'
# alias l='ls $LS_OPTIONS -lA'
#
# Some more alias to avoid making mistakes:
# alias rm='rm -i'
# alias cp='cp -i'
# alias mv='mv -i'
wp() {
/usr/local/bin/wp "$#" --allow-root
}
when I try to execute any wp command I get this error:
Error: YIKES! It looks like you're running this as root. You probably meant to run this as the user that your WordPress installation exists under.
If you REALLY mean to run this as root, we won't stop you, but just bear in mind that any code on this site will then have full control of your server, making it quite DANGEROUS.
If you'd like to continue as root, please run this again, adding this flag: --allow-root
If you'd like to run it as the user that this site is under, you can run the following to become the respective user:
sudo -u USER -i -- wp <command>
It looks like that command line doesn't consider what I input into .bashrc
Guys, do you have any suggestion how to fix this problem?
You are struggling with the classic conundrum: What goes in bashrc and what in bash_profile and which one is loaded when?
The extreme short version is:
$HOME/.bash_profile: read at login shells. Should always source $HOME/.bashrc. Should only contain environmental variables that can be passed on to other functions.
$HOME/.bashrc: read only for interactive shells that are not login
(eg. opening a terminal in X). Should only contain aliases and functions
How does this help the OP?
The OP executes the following line:
$ sudo -u USER -i -- wp <command>
The flag -i of the sudo-command initiates a login-shell
-i, --login: Run the shell specified by the target user's password database entry as a login shell. This means that login-specific resource files such as .profile, .bash_profile or .login will be read by the shell. If a command is specified, it is passed to the shell for execution via the shell's -c option. If no command is specified, an interactive shell is executed.
So the OP initiates a login-shell which only reads the .bash_profile. The way to solve the problem is now to source the .bashrc file in there as is strongly recommended.
# .bash_profile
if [ -n "$BASH" ] && [ -r ~/.bashrc ]; then
. ~/.bashrc
fi
more info on dot-files:
http://mywiki.wooledge.org/DotFiles
man bash
What's the difference between .bashrc, .bash_profile, and .environment?
About .bash_profile, .bashrc, and where should alias be written in?
related posts:
Run nvm (bash function) via sudo
Can I run a command loaded from .bashrc with sudo?
I recently had the same problem. In my Dockerfile, I was running:
RUN wp core download && wp plugin install woocommerce --activate --allow-root
I looked at the error message, and thought that from the way it was worded, the --allow-root gets ignored the first time you use it. So I added it to the first wp command, and It worked.
RUN wp core download --allow-root && wp plugin install woocommerce --activate --allow-root
The problem is that ~/.bashrc is not being sourced. It will only be sourced in an interactive Bash shell.
You might get better results doing it via executables. Something like this:
# install wp cli
RUN curl -O https://raw.githubusercontent.com/wp-cli/builds/gh-pages/phar/wp-cli.phar && \
chmod +x wp-cli.phar && \
mv wp-cli.phar /usr/local/bin/wp-cli.phar && \
echo '#!/bin/sh' >> /usr/local/bin/wp && \
echo 'wp-cli.phar "$#" --allow-root' >> /usr/local/bin/wp && \
chmod +x /usr/local/bin/wp
/Users/ello/.zshrc:source:3: no such file or directory:
/Users/ello/Projects/config/env.sh
Ello-MacBook-Pro% /Users/ello/.zshrc:source
zsh: no such file or directory: /Users/ello/.zshrc:source
Ello-MacBook-Pro% /Users/ello/.zshrc
zsh: permission denied: /Users/ello/.zshrc
Ello-MacBook-Pro%
This has been happening, after I foolishly edited the .zshrc file. All that remains in the file now, after attempting to reset the shell, is this:
# Created by newuser for 5.3.1
# Add env.sh
How do I undo everything, reinstall zsh, or remake the .zshrc file?
This is on macOS Sierra.
Edit: I reinstalled oh-my-zsh, leading to this message:
ain() {
# Use colors, but only if connected to a terminal, and that terminal
# supports them.
if which tput >/dev/null 2>&1; then
ncolors=$(tput colors)
fi
if [ -t 1 ] && [ -n "$ncolors" ] && [ "$ncolors" -ge 8 ]; then
RED="$(tput setaf 1)"
GREEN="$(tput setaf 2)"
YELLOW="$(tput setaf 3)"
BLUE="$(tput setaf 4)"
BOLD="$(tput bold)"
NORMAL="$(tput sgr0)"
else
RED=""
GREEN=""
YELLOW=""
BLUE=""
BOLD=""
NORMAL=""
fi
# Only enable exit-on-error after the non-critical colorization
stuff,
# which may fail on systems lacking tput or terminfo
set -e
CHECK_ZSH_INSTALLED=$(grep /zsh$ /etc/shells | wc -l)
if [ ! $CHECK_ZSH_INSTALLED -ge 1 ]; then
printf "${YELLOW}Zsh is not installed!${NORMAL} Please install zsh
first!\n"
exit
fi
unset CHECK_ZSH_INSTALLED
if [ ! -n "$ZSH" ]; then
ZSH=~/.oh-my-zsh
fi
if [ -d "$ZSH" ]; then
printf "${YELLOW}You already have Oh My Zsh installed.${NORMAL}\n"
printf "You'll need to remove $ZSH if you want to re-install.\n"
exit
fi
# Prevent the cloned repository from having insecure permissions.
Failing to do
# so causes compinit() calls to fail with "command not found:
compdef" errors
# for users with insecure umasks (e.g., "002", allowing group
writability). Note
# that this will be ignored under Cygwin by default, as Windows ACLs
take
# precedence over umasks except for filesystems mounted with option
"noacl".
umask g-w,o-w
printf "${BLUE}Cloning Oh My Zsh...${NORMAL}\n"
hash git >/dev/null 2>&1 || {
echo "Error: git is not installed"
exit 1
}
# The Windows (MSYS) Git is not compatible with normal use on cygwin
if [ "$OSTYPE" = cygwin ]; then
if git --version | grep msysgit > /dev/null; then
echo "Error: Windows/MSYS Git is not supported on Cygwin"
echo "Error: Make sure the Cygwin git package is installed and is
first on the path"
exit 1
fi
fi
env git clone --depth=1 https://github.com/robbyrussell/oh-my-zsh.git
$ZSH || {
printf "Error: git clone of oh-my-zsh repo failed\n"
exit 1
}
printf "${BLUE}Looking for an existing zsh config...${NORMAL}\n"
if [ -f ~/.zshrc ] || [ -h ~/.zshrc ]; then
printf "${YELLOW}Found ~/.zshrc.${NORMAL} ${GREEN}Backing up to
~/.zshrc.pre-oh-my-zsh${NORMAL}\n";
mv ~/.zshrc ~/.zshrc.pre-oh-my-zsh;
fi
zsh itself does not have a default user configuration. So the default ~/.zshrc is actually no ~/.zshrc.
But as you tagged the question with oh-my-zsh I would assume that you want to restore the default oh-my-zsh configuration. For this it should be sufficient to copy templates/zshrc.zsh-template from your oh-my-zsh installation path, usually ~/.oh-my-zsh:
cp ~/.oh-my-zsh/templates/zshrc.zsh-template ~/.zshrc
You may want to backup your current ~/.zshrc beforehand. Although it may have some problems now, you still might want to look up some settings once you reverted to default.
There is no such thing as "default". The best you can do, is check if your system has /etc/skel/.zshrc. If yes copy that into your home.
When you log in first time, your home is populated with everything from /etc/skel.
My dumass decided to just put a crash command into the zsh file. Now when I open the terminal, it just kernel panics. so I just deleted the config file using rm -f ~/.zshrc* and by default, it just got replaced with another copy. So good luck.
You can copy .zshrc template from
https://github.com/ohmyzsh/ohmyzsh/blob/master/templates/zshrc.zsh-template
And copy and paste all content in to ~/.zshrc
[MS Windows Friendly Solution - If terminal(using vim editor) steps are confusing]
Actually, there is no default .zshrc file, but if you need to edit is as a simple notepad, do these:
Goto /Users/ Folder via Finder App.
Click Shift + Command + . (Dot) to view hidden system files.
Look on .zshrc file, double click to open, then it will open in a notepad(TextEdit.app) in default.
Clear whichever lines to be removed.
Retype/Edit the file as per the Paths to be added.
Hit Command + s to save and exit.
Make it your default shell using this command:
chsh -s $(which zsh)
I am using mkdir to create directories under FreeBSD 10.2. I know -p option enables me to create a/b/c very easily (mkdir -p a/b/c). Now I want a to have two son directories b and c(a/c,a/b). Is it possible to do that by using only one mkdir command ? I have searched the net and found :
mkdir -p project/{lib/ext,bin,src,doc/{html,info,pdf},demo/stat/a}
which claims to generate the following result:
project/
project/lib/ext
project/bin
project/src
project/doc/html
project/doc/info
project/doc/pdf
project/demo/stat/a
However, this doesn't work in FreeBSD. Anybody can explain ? Thanks
It works fine here (FreeBSD 10.2-STABLE amd64) using the default tcsh shell;
> mkdir -p project/{lib/ext,bin,src,doc/{html,info,pdf},demo/stat/a}
> find .
.
./project
./project/lib
./project/lib/ext
./project/bin
./project/src
./project/doc
./project/doc/html
./project/doc/info
./project/doc/pdf
./project/demo
./project/demo/stat
./project/demo/stat/a
It does not work in the Bourne shell, sh.
Is it possible to check if a directory exists and delete if it does,in Unix using a single command? I have situation where I use ANT 'sshexec' task where I can run only a single command in the remote machine. And I need to check if directory exists and delete it...
Why not just use rm -rf /some/dir? That will remove the directory if it's present, otherwise do nothing. Unlike rm -r /some/dir this flavor of the command won't crash if the folder doesn't exist.
Assuming $WORKING_DIR is set to the directory... this one-liner should do it:
if [ -d "$WORKING_DIR" ]; then rm -Rf $WORKING_DIR; fi
(otherwise just replace with your directory)
Try:
bash -c '[ -d my_mystery_dirname ] && run_this_command'
This will work if you can run bash on the remote machine....
In bash, [ -d something ] checks if there is directory called 'something', returning a success code if it exists and is a directory. Chaining commands with && runs the second command only if the first one succeeded. So [ -d somedir ] && command runs the command only if the directory exists.
Here is another one liner:
[[ -d /tmp/test ]] && rm -r /tmp/test
&& means execute the statement which follows only if the preceding
statement executed successfully (returned exit code zero)
I recommend opening documentation of rm command.
If open then you will see that there is a
-f flag does what you want. Example: rm -f -R ./my_folder
Can PowerShell on Windows by itself or using simple shell script, list files and directory this way: (or using Mac OS X or Ubuntu's shell script)
audio
mp3
song1.mp3
some other song.mp3
audio books
7 habits.mp3
video
samples
up.mov
cars.mov
Unix's ls -R or ls -lR can't seem to list it in a tree structure unfortunately.
You can use tree.com for listing like indented like shown above. Note that tree.com only works with the filesystem. If you ever have a need to display structure for other providers like WSMan or RegEdit, you can use the Show-Tree function that comes with the PowerShell Community Extensions.
In Linux, you can use:
ls -R directory | grep ":$" | sed -e 's/:$//' -e 's/[^-][^\/]*\//--/g' -e 's/^/ /' -e 's/-/|/'
or for the current directory:
ls -R | grep ":$" | sed -e 's/:$//' -e 's/[^-][^\/]*\//--/g' -e 's/^/ /' -e 's/-/|/'
You can put this "small" command in a script: look here
you can use Unix's tree command, or if you are on Windows, the GNU windows tree.
Windows has a tree command:
C:\folder>tree . /F
Folder PATH listing for volume sys
Volume serial number is F275-CBCA
C:\FOLDER.
│ file01.txt
│
├───Sub folder
│ chart-0001.png
│ chart-0002.png
└───────chart-0004.png
The /F parameter is what tells it to show files. You can execute this from Powershell
This is probably what you're looking for:
ls -R | tree
It's not installed by default on Ubuntu. So, to install it:
sudo apt-get install tree