dtest84004 2012-07-31 16:41
浏览 43
已采纳

Yii - 根据存储在表格中的数据创建链接

Ive been trying to think of / find a solution to an issue i have. Im creating a menu system for my Yii app but I need to store part of the items need for creating the URL in a table. I have a table called system_menu_item:

---------------------------------------------------------------------| item_id | dis_text | action | variables | ---------------------------------------------------------------------| 1 | edit article | document/view | array('id'=>$model->arl_id) |

What i was hoping for was:

$model = SystemMenuItem::model()->findByPk('1');
$url = yii::app()->createUrl($model->action, $model->variables);

This doesn't work.. I tried different ways of doing it all not working. Any ideas?

Thanks

  • 写回答

1条回答 默认 最新

  • dongpan8928 2012-07-31 16:47
    关注

    I'm going to guess the issue is that $model->variables evaluates to a string with the value array('id'=>$mode->arl_id).

    An ugly (and dangerous!) hack would be to read this in using eval()

    A better way would be to probably figure out a different way to store your parameters, and read them into an array after performing your DB query.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

报告相同问题?

悬赏问题

  • ¥15 如何用stata画出文献中常见的安慰剂检验图
  • ¥15 c语言链表结构体数据插入
  • ¥40 使用MATLAB解答线性代数问题
  • ¥15 COCOS的问题COCOS的问题
  • ¥15 FPGA-SRIO初始化失败
  • ¥15 MapReduce实现倒排索引失败
  • ¥15 ZABBIX6.0L连接数据库报错,如何解决?(操作系统-centos)
  • ¥15 找一位技术过硬的游戏pj程序员
  • ¥15 matlab生成电测深三层曲线模型代码
  • ¥50 随机森林与房贷信用风险模型