linux service小例

linux service 测试

1.创建一个app

// myapp.c
// 间隔10s写入时间到文件
#include <stdio.h>
#include <time.h>
#include <unistd.h> // 引入unix标准函数定义，如sleep()int main()
{FILE *fp;time_t now;char buffer[80];// 打开文件以追加模式fp = fopen("/tmp/time_log.txt", "w");if (fp == NULL){perror("Error opening file");return (-1);}while (1){                                                                 // 无限循环time(&now);                                                   // 获取当前时间strftime(buffer, 80, "%Y-%m-%d %H:%M:%S\n", localtime(&now)); // 格式化时间// 写入时间到文件fprintf(fp, "%s", buffer);// 刷新输出缓冲区到文件，确保数据即时写入fflush(fp);// 等待3秒sleep(10);}// 注意：这个程序实际上不会到达这里，因为我们在上面使用了无限循环// 如果你想要程序在某个条件下退出，你需要在循环中添加适当的逻辑// 关闭文件fclose(fp);return 0;
}

gcc myapp.c -o myapp
sudo mv myapp /usr/local/bin/
sudo chmod 777 /usr/local/bin/myapp
cd /usr/local/bin
./myapp
tail -n 3 /tmp/time_log.txt
rm /tmp/time_log.txt

2.创建服务

#myapp.service 不要有这一行
[Unit]  
Description=My Application Service  
After=network.target  [Service]  
Type=simple  
User=c
ExecStart=/usr/local/bin/myapp  
Restart=on-failure  [Install]  
WantedBy=multi-user.target

sudo mv myapp.service /etc/systemd/system/
cd /etc/systemd/system/sudo systemctl daemon-reload
sudo systemctl start myapp.service
sudo systemctl stop myapp.service
sudo systemctl status myapp.service
sudo systemctl enable myapp.service
sudo systemctl disable myapp.service
journalctl -u myapp.servicetail -n 3 /tmp/time_log.txt
rm /tmp/time_log.txt