0

我想使用 PHP 从 MYsql 中检索三个值，并使用 $_Get 函数将这三个值发送到其他表单，我正在获取前两个输入的值（至少在我所需的文本框之外）。我无法获得第三个值（P_age）

    // in form where I retrieve values from Database and send it to other using "?" and "&"

          $connection=mysql_connect("localhost", "root","");
          mysql_select_db("Aravind_Database",$connection);
          $v= mysql_query("Select * from patient_details order by P_id");
          while ($row = mysql_fetch_array($v))
          {           
          echo $row['P_id'];      
          echo "<tr><td>".$row['P_id']."</td><td>".$row['P_name']."</td><td>".$row['P_age']."</td><td>"."<a href=Editing_Patient_Detail.php>Edit</a></td><td><a href=Delete_Patient_Detail.php?P_id=".$row['P_id']."&P_name=".$row['P_name']."&P_age=".$row['P_age'].">Delete</a>";
          }
          mysql_close( $connection);


// form where I try to retrieve values through $_Get


    $connection=mysql_connect("localhost", "root","");
               mysql_select_db("Aravind_Database",$connection);
               if ($_GET['P_id']!= '')
              {     
                            $v= mysql_query(" Select * from patient_details order by P_id");
                            while ($row = mysql_fetch_array($v))
                            {   
                                  if ($row['P_id'] == $_GET['P_id'])
                                  {
                                   ?> <input type="text" value='<?php echo $_GET['P_id'] ?>' disabled="true" /><br>
 <input type="text" width="145" value='<?php  echo $_GET['P_name'] ?>' disabled="true" /><br>
 <input type="text" value   ='<?php  echo $_GET['P_age']?>' disabled="true" /><br>
 <?php 
                                   break;
                                  }
                            }
                            mysql_close($connection);
              } 


    enter code here
4

1 回答 1

0

我会包括

name="myinput_name"

在输入标签中以及禁用时为真... GET 和 POST 忽略它...我认为您应该将其设为 readonly="readonly" 而不是 disabled

于 2012-05-09T10:43:48.547 回答