问题描述
我已经提到了http://bakery.cakephp.org/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
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([email protected]",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 http://bakery.cakephp.org/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
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("[email protected]","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
Thank youNidhin
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