Question

我正在尝试创建一个用户可以编辑预先存在的帖子的功能。当用户进入edit.php时，会向他们显示一个表单，向他们显示与该帖子关联的现有数据。然后，他们可以对任何字段（描述，类别，添加其他图像等）进行更改，并且在点击提交按钮时，更新的信息将显示在帖子页面上。

我的问题实际上是让它更新信息。表单将显示预先存在的信息，我可以对任何字段进行更改。但是，当我按提交时，我被带到帖子列表，但我所做的更改尚未在SQL表中更新。

点击提交时没有任何错误。一切都在顺利进行，除了事实上数据库中没有实际更新的事实。

我一直在寻找几个不同的网站以获取有关此事的帮助，我已经尝试了我的public partial class Form2 : Form { BackgroundWorker bgWorker = new BackgroundWorker(); public Form2() { InitializeComponent(); bgWorker.DoWork += BgWorker_DoWork; bgWorker.WorkerReportsProgress = true; bgWorker.WorkerSupportsCancellation = true; bgWorker.ProgressChanged += BgWorker_ProgressChanged; bgWorker.RunWorkerCompleted += BgWorker_RunWorkerCompleted; } private void BgWorker_RunWorkerCompleted(object sender, RunWorkerCompletedEventArgs e) { this.DialogResult = DialogResult.OK; } private void BgWorker_ProgressChanged(object sender, ProgressChangedEventArgs e) { if(progressBar1.Maximum >= e.ProgressPercentage) { progressBar1.Value = e.ProgressPercentage; } } private void BgWorker_DoWork(object sender, DoWorkEventArgs e) { int i = 0; while (i <= 100) { if (bgWorker.CancellationPending) { bgWorker.CancelAsync(); return; } i++; System.Threading.Thread.Sleep(100); bgWorker.ReportProgress(i); } } protected override void OnLoad(EventArgs e) { base.OnLoad(e); bgWorker.RunWorkerAsync(); } }查询的几种变体，可能是我错误地调用了它。这是我在尝试我发现的其他几个例子后正在进行的迭代：

UPDATE

我对PHP很新，所以我很可能会制作一些我没注意到的简单语法错误。或者它可能是我的代码的其他部分，我没有正确执行。如果有人能看看我的代码并帮助我指出正确的方向，我将非常感激。

另外，我想补充一点是，我知道我的代码很容易被注入。我现在唯一关心的是让这个功能起作用。在我开始工作之后，我会处理任何安全措施。

PHP

if($title && $price && $description && $category){
    $editquery = "UPDATE post SET title='$title', price='$price', description='$description', category='$category' WHERE post_id='$id'";
    $edquery = $db->prepare($editquery);
    $edquery->bind_result("ssss", $title, $price, $description, $category);
    $edquery->execute();
    if($edquery){
      echo "Updated!";
    }else{
      echo "error";
    }
  }else{
    echo "missing data";
  }

HTML表单这是HTML中与此功能相关的唯一部分。

<?php

if(!isset($_GET['id'])){
  header('Location: modify.php');
  exit();
}else{
  $id = $_GET['id'];
}

include('../includes/db_connect.php');

if(!is_numeric($id)){
  header('Location: inventory.php');
}

if(isset($_POST['submit'])){

  $title = $_POST['title'];
  $price = $_POST['price']; 
  $description = $_POST['description'];
  $category = $_POST['category'];
  $title = $db->real_escape_string($title);
  $price = $db->real_escape_string($price);
  $description = $db->real_escape_string($description);

     if($title && $price && $description && $category){
    $editquery = "UPDATE post SET title='$title', price='$price', description='$description', category='$category' WHERE post_id='$id'";
    $edquery = $db->prepare($editquery);
    $edquery->bind_result("ssss", $title, $price, $description, $category);
    $edquery->execute();
    if($edquery){
      echo "Updated!";
    }else{
      echo "error";
    }
  }else{
    echo "missing data";
  }

    $postid = $db->insert_id;

  for($i=0; $i<count($_FILES["images"]["name"]); $i++)
  {
   $filetmp = $_FILES["images"]["tmp_name"][$i];
   $filename = $_FILES["images"]["name"][$i];
   $filetype = $_FILES["images"]["type"][$i];
   $filepath = "images/".$filename;

   move_uploaded_file($filetmp, $filepath);

   $sql = "INSERT INTO images (img_name, img_path, img_type, post_id) VALUES 
   ('$filename', '$filepath', '$filetype', '$postid')";
   $result = mysqli_query($db, $sql);
  }
 }        


?>

修改

无论我的代码发生了什么，我按“提交”后都无法看到任何回显的语句：

        <form action="<?php echo $_SERVER['PHP_SELF']?>" method="POST" enctype="multipart/form-data">
              <?php

              $editsql = "SELECT * FROM post INNER JOIN categories ON categories.category_id=post.category_id WHERE post_id=' ".$id." '";
              $editquery = $db->query($editsql);
              if($editquery->num_rows !=1){
                header('Location: inventory.php');
                exit();
              }

              $editrow = $editquery->fetch_object();

              echo "<div class='form-group'>";
                  echo "<label>Title*</label>";
                  echo "<input type='text' name='title' class='form-control' value='".$editrow->title."' required>";
              echo "</div>";
              echo "<div class='form-group'>";
                  echo "<label>Price*</label>";
                  echo "<input type='text' name='price' class='form-control' value='".$editrow->price."'required>";
              echo "</div>";
              echo "<div class='form-group'>";
                  echo "<label>Category</label>";
                  echo "<select name='category' class='form-control'>";
                                echo "<option value='".$editrow->category_id."'>".$editrow->category."</option>";
                                $catquery = $db->query("SELECT * FROM categories");
                                while($row = $catquery->fetch_object()){
                                  echo "<option value='".$row->category_id."'>".$row->category."</option>";
                                }             
                   echo "</select>";
              echo "</div>";

              echo "<div class='form-group'>";
                  echo "<label>Description*</label>";
                  echo "<textarea type='textarea' name='description' class='form-control' required>".$editrow->description."</textarea>";
              echo "</div>";
              echo "<div class='form-group'>";
                  echo "<label>Image(s)</label>";
                  echo "<input type='hidden' name='size' value='1000000'>";
                  echo "<input multiple='multiple' name='images[]' type='file'/>";
              echo "</div>";
              echo "<div class='required'>";
                  echo "* indicates a required field";
              echo "</div>";
              echo "<button type='submit' name='submit' value='submit' class='btn btn-default'>EDIT POST</button>"
              ?>
        </form>

这是否可能导致问题？

    if($query){
      echo "product updated";
    }else{
      echo "error";
    }
  }else{
    echo "missing data";
  }

或者我需要使用隐藏的输入吗？

if(!isset($_GET['id'])){
  header('Location: modify.php');
  exit();
}else{
  $id = $_GET['id'];
}

编辑2

我已将其分为两个文件（edit.php和submitedit.php），以使echo "<button type='submit' name='submit' value='submit' class='btn btn-default'>EDIT POST</button>"和$_GET彼此分开。但是，我仍然遇到数据库无法更新的相同问题。

edit.php 我只展示了PHP和相关的HTML表单

$_POST

submitedit.php

<?php
session_start();
$msg = "";

if(!isset($_GET['id'])){
  header('Location: delete.php');
  exit();
}else{
  $id = $_GET['id'];
}

include('../includes/db_connect.php');

if (!isset($_SESSION['user_id'])) {
  header('Location: login.php');
  exit();
}

if(!is_numeric($id)){
  header('Location: inventory.php');
}

?>

<!-- WHERE THE HTML STARTS -->

 <form action="submitedit.php" method="POST" enctype="multipart/form-data">
                  <?php

                  $editsql = "SELECT * FROM post INNER JOIN categories ON categories.category_id=post.category_id WHERE post_id='$id'";
                  $editquery = $db->query($editsql);
                  if($editquery->num_rows !=1){
                    header('Location: inventory.php');
                    exit();
                  }

                  $editrow = $editquery->fetch_object();

                  echo "<div class='form-group'>";
                      echo "<label>Title*</label>";
                      echo "<input type='text' name='title' class='form-control' value='".$editrow->title."' required>";
                  echo "</div>";
                  echo "<div class='form-group'>";
                      echo "<label>Price*</label>";
                      echo "<input type='text' name='price' class='form-control' value='".$editrow->price."'required>";
                  echo "</div>";
                  echo "<div class='form-group'>";
                      echo "<label>Category</label>";
                      echo "<select name='category' class='form-control'>";
                                    echo "<option value='".$editrow->category_id."'>".$editrow->category."</option>";
                                    $catquery = $db->query("SELECT * FROM categories");
                                    while($row = $catquery->fetch_object()){
                                      echo "<option value='".$row->category_id."'>".$row->category."</option>";
                                    }             
                       echo "</select>";
                  echo "</div>";

                  echo "<div class='form-group'>";
                      echo "<label>Description*</label>";
                      echo "<textarea type='textarea' name='description' class='form-control' required>".$editrow->description."</textarea>";
                  echo "</div>";
                  echo "<div class='form-group'>";
                      echo "<label>Image(s)</label>";
                      echo "<input type='hidden' name='size' value='1000000'>";
                      echo "<input multiple='multiple' name='images[]' type='file'/>";
                  echo "</div>";
                  echo "<div class='required'>";
                      echo "* indicates a required field";
                  echo "</div>";
                  echo "<button type='submit' name='submit' value='submit' class='btn btn-default'>EDIT POST</button>"
                  ?>


                </form>

Answer 1

当您尝试从GET数组中读取id时，使用POST方法发送表单。将其更改为$_POST['id']，您已全部设置

在使用HTML表单提交时更新SQL表时出现问题

1 个答案: