我已经提到了bakery.cakephp/articles/view/calling-controller-actions-from-cron-and-the-command-line 并创建了 cron_dispatcher.php 并将其放在应用程序文件夹中.>
我已经返回了一些用于 cron 作业的测试电子邮件函数,以便在我的用户控制器的测试方法中运行.
而且我在我的网络服务器的控制面板中创建了一个 Cron 作业,如
"/usr/bin/php/home4/enventur/public_html/pennystock/cron_dispatcher.php/users/test"但是它给了我一个错误没有指定输入文件."
请帮帮我,如何解决??
提前致谢
解决方案我采用了不同的方式,
请查看步骤,它可能对其他人有帮助..
Cron/Shell 使用 Cakephp 框架结构:
创建
F:websitesprojectnameappvendorsshellsfilename.php
class ClassName extends Shell {//var $uses = array('Post');//型号名称//主函数总是在shell执行时运行函数主(){邮件(nidhin@2basetechnologies",测试",测试");}}2.设置754权限到F:websitesprojectnamecakeconsolecake
将 cron 作业设置为 /home4/enventur/public_html/pennystock/cake/console/cake -app "/home4/enventur/public_html/pennystock/app" ClassName >/dev/null 2>&1
/dev/null 2>&1: 用于抑制来自服务器的警告/错误/消息
谢谢尼丁
I have referred bakery.cakephp/articles/view/calling-controller-actions-from-cron-and-the-command-line and created cron_dispatcher.php and placed it in the app folder.
I have return some test email function for the cron job to run in my users controller's test method.
And i have created a Cron job in my web server's control panel like
"/usr/bin/php/home4/enventur/public_html/pennystock/cron_dispatcher.php /users/test"But its giving me an error as "No input file specified."
Please help me, how to solve it ??
Thanks in Advance
解决方案I have done it in different way,
Please see the steps, it may helpful for others..
Cron/Shell Using Cakephp Framework Structure:
create
F:websitesprojectnameappvendorsshellsfilename.php
class ClassName extends Shell { //var $uses = array('Post'); //name of Model //Main function runs always when shell executes function main() { mail("nidhin@2basetechnologies","Test","Test"); } }2.set 754 permission to F:websitesprojectnamecakeconsolecake
Set cron job as /home4/enventur/public_html/pennystock/cake/console/cake -app "/home4/enventur/public_html/pennystock/app" ClassName >/dev/null 2>&1
/dev/null 2>&1: for Suppressing warning/error/msg from server
Thank you Nidhin
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