dongluoqiu0255 2011-10-06 05:08
浏览 41
已采纳

仅在整数时获取查询字符串

Trying to get a query string into a variable but only if it's an integer.

Code is probably a bit more complicated than it should be but this is where I'm up to-

  //get page number. default is 1. check is not empty and is a number      
     if (empty($_GET['pag'])) {$page = 1;}
     else if (is_int($_GET['pag'])){$page = $_GET['pag'];}
     else {$page = 1;}

Where am I going wrong?

  • 写回答

3条回答 默认 最新

  • dsj1961061 2011-10-06 05:11
    关注

    You probably want is_numeric() instead - is_int() doesn't test to see if a string is a numeric string.

    if (empty($_GET['pag'])) {$page = 1;}
    else if (is_numeric($_GET['pag'])){$page = (int) $_GET['pag'];}
    else {$page = 1;}
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
查看更多回答(2条)

报告相同问题?

悬赏问题

  • ¥15 GDI处理通道视频时总是带有白色锯齿
  • ¥20 用雷电模拟器安装百达屋apk一直闪退
  • ¥15 算能科技20240506咨询(拒绝大模型回答)
  • ¥15 自适应 AR 模型 参数估计Matlab程序
  • ¥100 角动量包络面如何用MATLAB绘制
  • ¥15 merge函数占用内存过大
  • ¥15 Revit2020下载问题
  • ¥15 使用EMD去噪处理RML2016数据集时候的原理
  • ¥15 神经网络预测均方误差很小 但是图像上看着差别太大
  • ¥15 单片机无法进入HAL_TIM_PWM_PulseFinishedCallback回调函数