英文:
How to lunch symfony command via shell script
问题
我有一个Symfony命令，我想要从一个shell脚本中启动它。
示例：symfony mycommand。
我在项目文件夹中有一个shell脚本，并将Symfony命令作为参数传递给它。
这个sh脚本：
time=$1
php_script_path=$2
log_file="script.log"
# 无限循环
while true; do
# 循环运行PHP脚本，直到达到超时时间
while timeout $time php $php_script_path; do
# 仅当PHP脚本在超时时间内成功退出时执行此代码
echo "$(date): PHP脚本成功执行" >> $log_file
done
# 仅当PHP脚本在超时时间内未成功退出时执行此代码
if [ $? -eq 124 ]; then
echo "$(date): PHP脚本超时" >> $log_file
else
echo "$(date): PHP脚本完成时出现错误" >> $log_file
fi
done
英文:
i have a symfony command that i want to lunch it from a shell script.
exemple: symfony mycommand.
i have a script shell figuring in the project folder and i'm passing to it the symfony command to execute as an argument.
the sh script:
time=$1
php_script_path=$2
log_file="script.log"
# Infinite loop
while true; do
# Loop over the PHP script until the timeout is reached
while timeout $time php $php_script_path; do
# This code will only execute if the PHP script exits successfully within the timeout period
echo "$(date): PHP script executed successfully" >> $log_file
done
# This code will only execute if the PHP script did not exit successfully within the timeout period
if [ $? -eq 124 ]; then
echo "$(date): PHP script timed out" >> $log_file
else
echo "$(date): PHP script completed with an error" >> $log_file
fi
done
答案1
得分: 1
你可以使用与cron命令相同的方式。请在以下PHP脚本中设置您的项目路径。
time=$1
php_script_path="/var/www/project/bin/console app:your_command_name"
log_file="script.log"
# 无限循环
while true; do
# 在超时时间内循环执行PHP脚本
while timeout $time php $php_script_path; do
# 仅当PHP脚本在超时时间内成功退出时执行此代码
echo "$(date): PHP脚本成功执行" >> $log_file
done
# 仅当PHP脚本在超时时间内未能成功退出时执行此代码
if [ $? -eq 124 ]; then
echo "$(date): PHP脚本超时" >> $log_file
else
echo "$(date): PHP脚本完成并出现错误" >> $log_file
fi
done
英文:
You can use the same as the cron command. Please set your project path in below PHP script.
time=$1
php_script_path="/var/www/project/bin/console app:your_command_name"
log_file="script.log"
# Infinite loop
while true; do
# Loop over the PHP script until the timeout is reached
while timeout $time php $php_script_path; do
# This code will only execute if the PHP script exits successfully within the timeout period
echo "$(date): PHP script executed successfully" >> $log_file
done
# This code will only execute if the PHP script did not exit successfully within the timeout period
if [ $? -eq 124 ]; then
echo "$(date): PHP script timed out" >> $log_file
else
echo "$(date): PHP script completed with an error" >> $log_file
fi
done
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