Question

我在获取将数据插入数据库的正确代码时遇到问题。而且我也不知道如何将图像上传到数据库中。以下是我的FashionAddResult.php代码，

$userid=$_SESSION['userid'];
//check for blanks
if(!empty($_POST['fashionname'])) {
    $fashionname = $_POST['fashionname'];
} else {
    $fashionname = null;
    echo '<p><font color="red">Please enter the Fashion Name!</font></p>';
}
if(!empty($_POST['description'])) {
    $description = $_POST['description'];
} else {
    $description = null;
    echo '<p><font color="red">Please enter the Fashion Description!</font></p>';
}
if(!empty($_POST['imagefile'])) {
    $imagefile = $_POST['imagefile'];
} else {
    $imagefile = null;
    echo '<p><font color="red">Please enter the Fashion Image!</font></p>';
}
if($fashionname != null && $description != null && $imagefile != null){
    //TODO 1: Connect to forumdb database
    $stmt = new mysqli("localhost", "root", null, "fashiondb");


    //TODO 2: Prepare the statement to update subject and message in forummessage  
    $stmt = $mysqli->prepare("insert into fashion(fashionname,description,imagefile) values (?,?,?)");

    if (!$stmt = $mysqli->prepare($sql))
    {
    die('Query failed: (' . $mysqli->errno . ') ' . $mysqli->error);
    }

    //TODO 3: Bind the values   
    $stmt->bind_param('sss', $fashionname, $description, $imagefile);
    if (!$stmt->bind_param('sss', $fashionname, $description, $imagefile))
    {
    die('Binding parameters failed: (' . $stmt->errno . ') ' . $stmt->error);
    }
    //TODO 4: Execute the statement
    $result = $stmt->execute();
    if (!$stmt->execute())
    {
    die('Execute failed: (' . $stmt->errno . ') ' . $stmt->error);
    }
    //TODO 5: If execute is successful, display update successful message
    //else display error message
    if($result == true && $stmt->affected_rows>0){
        echo '<p>Your fashion name has been added!</p>';
    }
    else{
        echo '<p>Fashion information is not adeed!</p>';
        //echo "result=$result<br/>row=$stmt->row_affected<br/>";
    }
    //TODO 6: close the statement
    $stmt->close();

    //TODO 7: close $mysqli
    $mysqli->close();
}
?>

FashionAdd.php代码如下，

<form enctype="multipart/form-data" action="FashionAddResult.php" method="post">
<p>Fashion Name: <input type="text" name="fashionname"></p>
<p>Fashion Description:<br/></p>
<textarea name="description" rows="10" cols="75">
</textarea><br>
Please choose a file: <input name="imagefile" type="file" /><br/>
<br/>
<input type="submit" value="Upload Fashion!" />
</form>

当我在图像中已有图像时，显示图像未上传。 请帮我解决错误以及如何将图片上传到数据库:) !!谢谢！

Answer 1

您有SQL错误。

此：

insert into fashion(fashionname,description,imagefile values (?,?,?)where userid=?

应该是：

insert into fashion (fashionname, description, imagefile, userid) values (?,?,?,?)

您还应该绑定4个参数，而不仅仅是3个。

此：

$stmt->bind_param('sss', $fashionname, $description, $imagefile,$userid);

应该是：

$stmt->bind_param('sssi', $fashionname, $description, $imagefile, $userid);

要看到这一点，你应该做的事情如下：

try
{

  //TODO 2: Prepare the statement to update subject and message in forummessage  
  if (!$stmt = $mysqli->prepare("insert into fashion (fashionname,description,imagefile,userid) values (?,?,?,?)"))
  {
    throw new Exception($mysqli->error);
  }

  //TODO 3: Bind the values   
  if (!$stmt->bind_param('sssi', $fashionname, $description, $imagefile,$userid))
  {
    throw new Exception($stmt->error);
  }

  //TODO 4: Execute the statement
  if (!$result = $stmt->execute())
  {
    throw new Exception($stmt->error);
  }

}

catch (Exception $e)
{
  echo $e->getMessage();
}

Answer 2

您的TODO 2缺少“）”

$stmt = $mysqli->prepare("insert into fashion(fashionname,description,imagefile values (?,?,?)where userid=?");

应该是 - 注意关闭后的imagefile：

$stmt = $mysqli->prepare("insert into fashion(fashionname,description,imagefile) values (?,?,?)where userid=?");

表单不发送所有字段和查询不正常工作

2 个答案: