Question

我有很多复选框项目，每个员工都可以访问不同的部门。我想将检查项添加到数据库中。

如果选中，则显示门的访问权限的代码

 $door1 = $_POST['door1'];
 echo ' dooraccess is available for ' .$door1. '<br>'; 

 $door2 = $_POST['door2'];
 echo 'dooraccess is available for  ' .$door2. '<br>'; 

 $door3 = $_POST['door3'];
 echo 'dooraccess is available for  ' .$door3. '<br>'; 

 $door4 = $_POST['door4'];
 echo 'dooraccess is available for  ' .$door4. '<br>'; 

 $door5 = $_POST['door5'];
 echo 'dooraccess is available for  ' .$door5. '<br>'; 

 $door6 = $_POST['door6'];
 echo 'dooraccess is available for  ' .$door6. '<br>'; 

 $door7 = $_POST['door7'];
 echo 'dooraccess is available for  ' .$door7. '<br>'; 

 $door8 = $_POST['door8'];
 echo 'dooraccess is available for  ' .$door8. '<br>'; 

 $door9 = $_POST['door9'];
 echo 'dooraccess is available for  ' .$door9. '<br>'; 

 $door10 = $_POST['door10'];
 echo 'dooraccess is available for  ' .$door10. '<br>'; 

 $door11 = $_POST['door11'];
 echo 'dooraccess is available for  ' .$door11. '<br>'; 

 $door12 = $_POST['door12'];
 echo 'dooraccess is available for  ' .$door12. '<br>';

如果未检查的值，则它将显示如下错误：

未定义的索引：[因为未检查门7]中的门7

将这些值与其他值一起插入数据库php mysql database

的代码

 if(isset($_POST['submit']))
  { 
   $query = "INSERT INTO form_details1(firstname,secondname,location,designation,fileno,doa,doj,cardtype1,cardcolor,cardtype2,door1,door2,door3,door4,door5,door6,door7,door8,door9,door10,door11,door12,door13) VALUES('$firstname','$secondname','$location','$designation','$fileno','$doa','$doj','$cardtype1',$cardcolor,'$cardtype2','$door1','$door2','$door3','$door4','$door5','$door6','$door7','$door8','$door9','$door10','$door11','$door12')"; //query
  if(mysqli_query($connect,$query)){ //check query executed or not
     echo 'inserted' . '<br>' ;
   }

查询未插入数据库，因为查询具有未选中的值

我想使用if循环或任何其他方式检查所有选中的项目，并仅输入已检查的项目。我怎么能这样做？

我的表格似乎是

+---------+---------+-------+
|   Id    |  door1  | door2 |
+---------+---------+-------+
| Value 1 | Value 2 |   123 |
| Value 2 | NULL    |   123 |
+---------+---------+-------+

The uncheck box will be empty in the table .

mysql field datatype for array having values that has been selected
应该是什么

Answer 1

您需要使用复选框数组。请查看此stackoverflow答案。

Get $_POST from multiple checkboxes

Answer 2

将PHP代码更改为：

isWindows()

Answer 3

首先检查是否设置了复选框： var str = "Hello, playground" let newTypeString = NSString(string: str)

然后您的查询有效：员工可以访问的门有1，其他人有0。

优化：

始终使用此处所述的参数化查询：Parameterized Query
更容易遍历您的复选框，您可以设置如下名称： $door1 = isset($POST['door1'])? 1 : 0 ; 然后循环：

foreach（$ _ POST ['门']为$ num =＆gt; $ value）{

}

列出项目

规范化数据库（在你放置所有门的地方制作一个桌子门。然后用一张外键制作一张桌子用于员工桌和门牌。如果有一个带有员工ID和第一门ID的条目，然后这名员工可以进入第一扇门。如果需要，可以更容易地添加新门。）

已经检查过的复选框项需要添加到php mysql数据库中

3 个答案: