php - 如何构建这个 mysqli bind_params 语句然后执行它

Question

我正在编写一个函数，通过读取表结构并提交两个变量来向任何表添加记录$params并$data构建。

$params或许ssssssssssssss

$data将会="$Email, $First, $Surname, $Group_ID, $city, $bdFlag, $Sector, $Telephone, $createJobs, $artworker, $Password, $currentPass, $enteredBY, $Created"

如果我使用下面的语句，它工作正常。但我不能用我的“$data”变量替换字段变量。有谁知道我怎么能做到这一点？

// This works
$stmt->bind_param($params , $Email, $First, $Surname, $Group_ID, $city, $bdFlag, $Sector, $Telephone, $createJobs, $artworker, $Password, $currentPass, $enteredBY, $Created);

// This doesn't
$stmt->bind_param($params , $data);