Question

我已经构建了一个更新文本和图像的后端系统。值将从admin.php页面上的URL传递到索引页面。正在使用表格来显示文本和图像。更新的文本和图像应显示在表格中。出于某些奇怪的原因，当我更新内容然后单击编辑内容按钮时，文本和图像总是消失。内容在数据库中更新没有问题。我没有发现任何错误。如何在没有内容消失的情况下获取更新的文本和图像以显示在表中？我感谢你的帮助，我还在学习PHP。

这是代码：

admin.php的

<?php
  require_once('authorize.php');
?>

<!DOCTYPE html>
<html>
<head>
<style>
a:link {
    text-decoration: none;
}
</style>

  <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
  <title>Remove Thumbnail Administration</title>
  <link rel="stylesheet" type="text/css" href="style.css" />
</head>
<body>
  <h2>Administration</h2>
  <p>Below is a list of all thumbnails. Use this page to remove thumbnails as needed.</p>
  <hr />

<?php
  require_once('appvars.php');
  require_once('connectvars.php');

  // Connect to the database 
  $conn = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);

  // Retrieve the data from MySQL
  $query = "SELECT * FROM table1 ORDER BY name ASC, caption ASC";
  //$query = "SELECT * FROM table1 ORDER BY name DESC, caption ASC";     
  $data = mysqli_query($conn, $query);

 /// Loop through the array of data, formatting it as HTML 
  echo '<table>';
  echo '<tr><th>Name</th><th>Caption</th><th>Action</th></tr>';
  while ($row = mysqli_fetch_array($data)) { 
    // Display the thumbnails data
    echo '<tr class="scorerow"><td><strong>' . $row['name'] . '</strong></td>';
    echo '<td>' . $row['caption'] . '</td>';

    //edit link

      echo '<td><a href="index.php?id=' . $row['id'] . '&amp;image=' . $row['image1'] . '&amp;name=' . $row['name'] . 
    '&amp;caption=' . $row['caption'] .
      '&amp;video=' . $row['video'] . '">Edit </a>'; 




    echo '</td></tr>';
  }
  echo '</table>';   
  echo "<br><br>";

  mysqli_close($conn);
?>

</body> 
</html>

的index.php

<!DOCTYPE html>
<html>
<head>
  <title>Edit Conent</title>
  <link rel="stylesheet" type="text/css" href="style.css" />
  <style>
      .bigger_textbox {
   width: 400px;
   height: 40px;
 }

     </style> 

</head>
<body>
  <h3>Edit Conent</h3>

<?php
  require_once('appvars.php');
  require_once('connectvars.php');

  $vid="";
  $vname="";
  $vcaption="";
  $vvideo="";
  $id ="";

  // Connect to the database
  $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);

  if(isset($_POST["button_edit"])){
     $id = $_POST["id"];
    $name = $_POST['name']; 
    $caption = $_POST['caption']; 
    $video = $_POST['video'];

     $old_picture = $_POST['old_picture'];
     if(!empty($_FILES["new_picture"]["name"])) {
    $new_picture = $_FILES["new_picture"]["name"];
    $qry = mysqli_query($dbc,"Update table1 Set image1='$new_picture', name='$name', caption='$caption', video='$video' Where id='$id'");

        $target_dir = "images/";
        $target_file = $target_dir . basename($_FILES["new_picture"]["name"]);
       $imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);
    if (move_uploaded_file($_FILES["new_picture"]["tmp_name"],$target_file)){
    if (isset($old_picture) && ($old_picture != $new_picture)) {
              @unlink("images/" . $old_picture);
              echo "<h1>image was uploaded</h1>";

     }
     }
    }
    else{
 $qry = "Update table1 Set name='$name', caption='$caption', video='$video' Where id='$id'";
}

$qryUpdate = mysqli_query($dbc,$qry);
    }

else if(isset($_GET["edit"])){
    $qry = mysqli_query($dbc,"Select * From table1 Where id='".$_GET["edit"]."'");
    while($row=mysqli_fetch_array($qry,MYSQLI_ASSOC)){
        $vid=$row["id"];
        $old_picture=$row["image1"];
        $vname=$row["name"];
        $vcaption=$row["caption"];
        $vvideo=$row["video"];


    }
}


?>



<!DOCTYPE html>
<html>
<head>
<title>Edit</title>
</head>
<body>
<form action='<?php echo $_SERVER["PHP_SELF"]; ?>' method="post" enctype="multipart/form-data" >
    <table>
    <tr>
            <td>Product ID</td>
            <td><input type="text" name="id" value="<?php echo $vid;?>"></td></tr>
        <tr>
            <td>Name</td>
            <td><input type="text" class="bigger_textbox" name="name" value="<?php echo $vname;?>"></td></tr>
        <tr><td>Caption</td>
        <td><input type="text" class="bigger_textbox" name="caption"  value="<?php echo $vcaption;?>"></td></tr>
        <tr><td>Video</td>
        <td><input type="text" class="bigger_textbox" name="video"  value="<?php echo $vvideo;?>"></td></tr>
        <input type="hidden" name="old_picture" value="<?php if (!empty($old_picture)) echo $old_picture; ?>" />
        <tr><td>Picture</td>
        <td><input type="file" name="new_picture" ></td></tr>
        <?php if (!empty($old_picture)) {
        echo '<img class="profile" src="images/' . $old_picture . '" alt="image" style=width:150px;height:xpx;">';
      } 
      ?>
        <tr><td colspan="2">
        <input type="submit" name="button_edit" value="Edit Content"></td></tr> </table>
</form>
<table border=1>
    <tr><th>Name</th><th>Caption</th>
    <th>Video</th><th>image</th>  <th>Action</th></tr>
    <?php
     if (isset($_GET["id"])) {
    $qry =mysqli_query($dbc, "Select * From table1 Where id='".$_GET["id"]."'");
    while($row=mysqli_fetch_array($qry,MYSQLI_ASSOC)) {
        echo '<tr><td>'.$row["name"].'</td>';
        echo '<td>'.$row["caption"].'</td>';
        echo '<td>'.$row["video"].'</td>'; 
        echo '<td><img src="images/'.$row["image1"].'" style=width:100px;height:xpx;"/></td>'; 

        echo '<td><a href="?id='.$row["id"].'&edit='.$row["id"]. '&video='.$row["video"].'&image='.$row["image1"].'">Edit</a> </td></tr>';
        }


    }



    ?>
</table>

<p><a href="admin.php">&lt;&lt; Back to admin page</a></p>

<br><br><br>
</body>
</html>

Answer 1

在index.php之上，您将清除变量的值。

试试这个：

<?php 
    isset($vid) ? $vid : "";
    isset($vname) ? $vname : "";
    isset($vcaption) ? $vcaption : "";
    isset($vvideo) ? $vvideo : "";
    isset($id) ? $id : "";

这是一个快速反应;但是，您需要重新组织代码。将HTML部分分成部分，其中静态部分（页眉和页脚可以包含在一行中）。另外，不要在每一行上使用echo，而是使用$ output。=“结果数据”;连接数据，然后回显$ output。

为什么内容在更新后会从表中消失

1 个答案: