基于HTML表单输入在sql中移动记录

时间:2018-05-22 10:24:29

标签: php html sql sql-server forms

我正在尝试根据表单的输入将记录移动到数据库,但似乎遇到了问题。我已经尝试过以下但它没有在数据库上执行命令。它需要读取一个值作为决策,并根据决定进行查询。任何人都可以在这里发现问题吗?我在SQL Server中运行了查询,但它确实有效。

HTML表单代码:

<!--Other Form-->
 <center>
  <div class="container">
    <form action="dLG.php" onsubmit="return confirm('Are you sure you wish to delete ?');">
      <div class="row">
        <div class="col-25">
          <label for="License Group ID">Enter the License Group ID that you wish to delete</label>
        </div>
        <div class="col-75">
          <input type="number" name="LicenseGroupID" placeholder="LicenseGroupID">
        </div>
      </div>
      <br>
      <div class="row">
        <input type="submit" value="Delete License Group">
      </div>
    </form>
  </div>
</center>

php:

<html>
<?php 
$serverName = "x";
$options = array(  "UID" => "x",  "PWD" => "x",    "Database" => "x");
$conn = sqlsrv_connect($serverName, $options);
if( $conn === false )
 {
 echo "Could not connect.\n";
 die( print_r( sqlsrv_errors(), true));
 }  
$License_Group_ID = $_POST['LicenseGroupID']; 

$query = "DELETE FROM dbo.tbl_license_group
        OUTPUT 
        deleted.License_Group_ID,
        deleted.Vendor_Name,
        deleted.License_Version,
        deleted.Edition,
        deleted.Operating_System,
        deleted.Quantity            
        INTO dbo.tbl_license_group_delete(License_Group_ID, Vendor_Name, License_Version,Edition,Quantity)
        WHERE  License_Group_ID = '$License_Group_ID'; " ;        
       $params1 = array($License_Group_ID);                       
       $result = sqlsrv_query($conn,$query,$params1);
       sqlsrv_close($conn);
       ?>
       </html>

1 个答案:

答案 0 :(得分:0)

默认表单方法是GET,而不是POST。如果您想使用POST(在这种情况下应该使用),请使用:

<form action=".." method="post"  onsubmit="...">
...
</form>

第二个问题是$ query字符串。它应该使用这样的参数:

$query = "DELETE ... WHERE  License_Group_ID = ?; " ;

PS:使用PDO比使用sqlsrv_ *函数更好。