donglu9134 2013-04-08 16:55
浏览 59
已采纳

动态命名变量

I am accepting a variable into a function.

For example, the variable is called $field and I then want to name a variable by what is inside the $field variable.

Say $field = 'randomName', I want my variable to be $randomName.

Is this possible?

Thanks.

  • 写回答

6条回答 默认 最新

  • douwo1517 2013-04-08 16:58
    关注

    Yes it is possible:

    $field = "randomName";
    $$field = "test";
    $$$field = "test 2";
    echo $randomName . "
    "; //outputs: "test"
    echo $test. "
    "; //outputs: "test 2"
    

    Check this out

    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(5条)

报告相同问题?

悬赏问题

  • ¥40 图书信息管理系统程序编写
  • ¥15 7-1 jmu-java-m02-使用二维数组存储多元线性方程组
  • ¥20 Qcustomplot缩小曲线形状问题
  • ¥15 企业资源规划ERP沙盘模拟
  • ¥15 树莓派控制机械臂传输命令报错,显示摄像头不存在
  • ¥15 前端echarts坐标轴问题
  • ¥15 ad5933的I2C
  • ¥15 请问RTX4060的笔记本电脑可以训练yolov5模型吗?
  • ¥15 数学建模求思路及代码
  • ¥50 silvaco GaN HEMT有栅极场板的击穿电压仿真问题