dtv11049 2017-11-25 16:59
浏览 58

更新/编辑值时，从输入字段的占位符中获取数据库的值

In this code there is one input field which is used for update data. data is updating perfectly. but i want to show the previous data in my input field when i will click 'Add' button. So that when i click add button i can see my previous data in the input field and then i can update.

I want to show biotext in the input field from database when I update data. i used $biotext in the input field. but the previous data is not showed.The input field shows ->
Notice: Undefined variable: biotext in /Applications/XAMPP/xamppfiles/htdocs/Social/edit.php on line 76
I don't know what is the problem. please help me to find out the problem. thanks in advance.

<?php


// edit bio
// edit bio

if(isset($_GET['edit_id']) && !empty($_GET['edit_id']))
    {
        $id = $_GET['edit_id'];
        $stmt= $db->select('SELECT biotext FROM bio WHERE userid =:userid',array('userid'=>$id));

    }
    

    if(isset($_POST['update'])) 

  {
      $userid = $_SESSION['userid'];
      $biotext = $_POST['biotext'];

      if(empty($biotext))
    {
      $errMSG = "Please update your bio.";
    }

    else 
    {
       
      $user_bio = $db->update("UPDATE bio SET biotext=:biotext WHERE userid=:userid",array('biotext'=>$biotext,'userid'=>$userid));
          
    }
   
  }

?>
<!DOCTYPE html>
<html>
<head>
    <title></title>
</head>
<body>

    <?php 
if(isset($_GET['edit_id']) && !empty($_GET['edit_id']))
{


?>
          <div class="row main-row">
            <div class="col-md-11 ">
              <?php
      if(isset($errMSG))
      {
        echo $errMSG;
      }

      ?>

        <form method="post" name="form1" action="edit.php">
          <table width="25%" border="0">
            <tr> 
              <td>Add your bio</td>
              <td><input type="text" name="biotext" value="<?php echo $biotext; ?>"></td> 
            </tr>

            <tr> 
            <td></td>
            <td><input type="submit" name="update" value="Add"></td>
            </tr>
          </table>
        </form>
            
          </div>
          </div>
   <?php 

}


// edit bio
// edit bio
?>

</div>
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1条回答 默认 最新

  • duanquanzhi5560 2017-11-25 18:38
    关注

    Hope I was able to understand your concern.
    Here, you can use the below code I've written. Change the connection details and edit the query with the session variable you were fetching the user id from.

    <?php
define('DB_SERVER', 'localhost');
define('DB_USERNAME', 'root');
define('DB_PASSWORD', 'root');
define('DB_DATABASE', 'stackDB');
$con = mysqli_connect(DB_SERVER,DB_USERNAME,DB_PASSWORD,DB_DATABASE);

if (!$con)
{
    die('Could not connect: ' . mysqli_error());
} 

$result= mysqli_query($con, "SELECT biotext FROM bio WHERE userid = 1234");
$fieldValDemo = mysqli_fetch_row($result);
?>

<!DOCTYPE html>
<html>
<head>
    <title></title>
</head>
<body>
<div class="row main-row">
    <div class="col-md-11 ">
        <form method="post" name="form1" action="edit.php">
            <table width="25%" border="0">
                <tr> 
                    <td>Add your bio</td>
                    <td><input type="text" name="biotext" value="<?php echo $fieldValDemo[0]; ?>"></td> 
                </tr>

                <tr> 
                    <td></td>
                    <td><input type="submit" name="update" value="Add"></td>
                </tr>
            </table>
        </form>
    </div>
</div>
</body>
</html>


    Stored in DB - enter image description here

    Fetched from DB and shown to user before editing enter image description here

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