Question

我正在尝试在我的数据库问题文本中插入/更新两个值并输入。在添加一行时插入了questiontext，并在需要时成功更新。但是，我在插入类型或更新它时都没有成功。有什么帮助吗？

 case 'Addquiz':

         $sql = "SELECT id,questiontext,type FROM questioninfo ORDER BY type DESC ";

         $result = mysqli_query($con,$sql);
         $selectedtable  = "<form method='post' action=''>\n";
         $selectedtable .= "<table class='sortable'>\n<tr><th>Question</th><th>Type</th></tr>\n";

         while($row = mysqli_fetch_assoc($result)) { 

             $rowID = $row['id']; 
             $text = $row['questiontext'];
             $type = $row['type']; 

            $selectedtable .= "<tr>
<td><input type='text' name='QuestionText[$rowID]' value='$text'></td><td><select name='type[$rowID]'><option selected='selected'></option><option value='$type'>Performace</option><option value='$type'>Loyalty</option></select></td></tr>\n"; 

         }
         $selectedtable .= "</table>\n"; 
         $selectedtable .= "<input type='submit' name='submit' value='Update' style='width:80px; height:30px; text-align:center; padding:0px;'>\n";
         $selectedtable .= "<input type='submit' name='addquestion' value='Add Question' style='width:140px; height:30px; text-align:center; padding:0px;'>\n";
         $selectedtable .= "</form>\n";

         if(isset($_POST['submit'])) {   
             foreach($_POST['QuestionText'] as $rowID => $text) { 

                 $sql = "UPDATE questioninfo 
                         SET questiontext = '$text', 
                             type = '$type' 
                         WHERE id = '$rowID'"; 
                  mysqli_query($con,$sql);
             } 

          }
          if(isset($_POST['addquestion'])) {   
              $sql="INSERT INTO `questioninfo` (`ID`) VALUES (NULL)";
               mysqli_query($con,$sql);
        }


    break;

Answer 1

实际上我觉得你对网页的基本生命周期感到有点困惑。

第一次加载页面时可能是因为点击了菜单或链接，因此没有数据需要处理。如果按下某个表单<input type='submit' ....>按钮，您只会尝试处理用户输入。

因此，处理表单的代码的基本布局应该是这样的： -

<?php

    if ( $_SERVER["REQUEST_METHOD"] == 'POST' ) { // or 'GET'

        // The user has pressed the submit button

        // Check that all required fields are present in $_POST/$_GET

        // check for which button was pressed if more than one button exists
        // Do any database access/updates etc based on validated inputs      
        // Store any error message in an array for example to be used 
        // in the main HTML generating phase

        // set a flag or 2 so in the HTML generating phase you know 
        // what flavor of page you want the user to see 
        // based on what the just did.
    } // end of user input processing


    // So now we generate the HTML for the initial page ( no user input )
    // or possibly tailor what we output depending upon
    // what the user entered and we processed above
    // and any flags we set above to control what this 
    // screen should look like

如果你仔细观察，你的脚本正在尝试处理那些case 'Addquiz':中实际无法获取的数据，因为当你生成的按钮实际被按下而字段中有数据时，它实际上不会运行这个案例它将运行另一个，因为在这种情况下您创建的按钮将导致另一个案例完全运行。

将值从下拉列表插入数据库

1 个答案: