Question

我有点问题。这是我的要求（工作）：

 $reponse = $bdd->prepare('select * from student where UserName = ? and Password = ? ');
 $reponse->execute(array($name,$pass));

这是工作，但是当我想用php变量替换'student'时，它不起作用：/ 看：

 $table="stackoverflow"
 $reponse = $bdd->prepare('select * from '$table' where UserName = ? and Password = ? ');
 $reponse->execute(array($name,$pass));

你有个主意吗？如果我可以用变量替换表名，那将更简单。对不起，我的英语不好。谢谢你花在我身上的时间。

Answer 1

改变这个：

 $reponse = $bdd->prepare('select * from '$table' where UserName = ? and Password = ? ');

要：

 $reponse = $bdd->prepare("select * from `$table` where UserName = ? and Password = ? ");

Answer 2

在PHP中，您可以用句点字符（。）连接两个字符串或变量。

因此，要在表名之前和之后连接表名和语句，您应该更改此行

$reponse = $bdd->prepare('select * from '$table' where UserName = ? and Password = ? ');

到

$reponse = $bdd->prepare('select * from '.$table.' where UserName = ? and Password = ? ');

Answer 3

由于表名是字符串，请尝试 -

"SELECT * FROM from \"".$table."\" WHERE ....... ";

无法在SQL查询中将表名替换为变量

3 个答案: