使用 PDO 将图像位置上传到数据库并将图像文件上传到目录？

Question

我能否就这个示例代码获得一些帮助，尝试在将链接位置存储在数据库中的同时将图像添加到文件目录，但使用 PDO 而不是旧方法。

它基于我在网上找到的一个示例，该示例带有 dbconnect.php、save.php、addstudent.php 和其他一些不需要此查询的示例。

    <form method="post" name="frmStudent" action="save.php">
    <input type="hidden" name="pid" value="<?php echo $ppid; ?>"/>
        <table>
            <tr><td>First Name</td><td>:</td><td><input type="text" name="fname" required="required" value="<?php echo $pfname; ?>"/></td></tr>
            <tr><td>Last Name</td><td>:</td><td><input type="text" name="lname" required="required" value="<?php echo $plname; ?>"/></td></tr>
            <tr><td>Contact No.</td><td>:</td><td><input type="tel" name="contact" required="required" value="<?php echo $pcontact; ?>"/></td></tr>
            <tr><td>Email</td><td>:</td><td><input type="email" name="email" required="required" value="<?php echo $pemail; ?>"/></td></tr>
             <tr><td>Image</td><td>:</td><td><input type="file" name="email" required="required" value="<?php echo $pimg_url; ?>"/></td></tr>
            <tr><td></td><td></td><td><input type="submit" class="myButton" value="Save"/></td></tr>
        </table>
    </form>

这是保存到数据库的代码

<?php   
error_reporting(0);
    include ("dbconnection.php");
    $fname=$_POST['fname'];
    $lname=$_POST['lname'];
    $contact=$_POST['contact'];
    $email=$_POST['email'];
    $img_url=$_POST['img_url'];
    $id=$_POST['pid'];
    if($id==null){
            $sql="INSERT INTO student(fname,lname,contact,email,img_url)values(:fname,:lname,:contact,:email,:img_url)";
            $qry=$db->prepare($sql);
            $qry->execute(array(':fname'=>$fname,':lname'=>$lname,':contact'=>$contact,':email'=>$email,':img_url'=>$img_url));
    }else{
            $sql="UPDATE student SET fname=?, lname=?, contact=?, email=?, img_url=? where id=?";
            $qry=$db->prepare($sql);
            $qry->execute(array($fname,$lname,$contact,$email,$img_url,$id));   
    }
    echo "<script language='javascript' type='text/javascript'>alert('Successfully Saved!')</script>";
    echo "<script language='javascript' type='text/javascript'>window.open('index.php','_self')</script>";
?>

任何关于如何做到这一点的见解将不胜感激，谢谢。

Answer 1

对于无法获得 Meda 答案的人，只需将 enctype="multipart/form-data" 添加到 <form> 元素即可。

Answer 2

首先修复图像的HTML 输入

<tr>
    <td>Image</td><td>:</td>
    <td><input type="file" name="image" required="required" value=""/></td>
</tr>

然后扩展您的PHP 代码：

<?php   
error_reporting(0);
include ("dbconnection.php");

if(is_uploaded_file($_FILES['image']['tmp_name'])){ 
    $folder = "upload/"; 
    $file = basename( $_FILES['image']['name']); 
    $full_path = $folder.$file; 
    if(move_uploaded_file($_FILES['image']['tmp_name'], $full_path)) { 
        echo "succesful upload, we have an image!";
        $fname=$_POST['fname'];
        $lname=$_POST['lname'];
        $contact=$_POST['contact'];
        $email=$_POST['email'];
        $img_url= $full_path;
        $id=$_POST['pid'];
        if($id==null){
            $sql="INSERT INTO student(fname,lname,contact,email,img_url)values(:fname,:lname,:contact,:email,:img_url)";
            $qry=$db->prepare($sql);
            $success = $qry->execute(array(':fname'=>$fname,':lname'=>$lname,':contact'=>$contact,':email'=>$email,':img_url'=>$full_path));
        }else{
            $sql="UPDATE student SET fname=?, lname=?, contact=?, email=?, img_url=? where id=?";
            $qry=$db->prepare($sql);
            $success = $qry->execute(array($fname,$lname,$contact,$email,$full_path,$id));   
        }

        if($success){
            echo "<script language='javascript' type='text/javascript'>alert('Successfully Saved!')</script>";
            echo "<script language='javascript' type='text/javascript'>window.open('index.php','_self')</script>";
        }else{
            echo 'db transaction failed';
        }
    } else { 
       echo "upload received! but process failed";
    } 
}else{ 
    echo "upload failure ! Nothing was uploaded";
} 
?>