如何使用php将图像存储在mysql数据库中

Question

如何在 MySQL 数据库中存储和显示图像。到目前为止，我只编写了从用户那里获取图像并将它们存储在文件夹中的代码，我到目前为止编写的代码是： HTML 文件

<input type="file" name="imageUpload" id="imageUpload">

PHP 文件

    $target_dir = "uploads/";
$target_file = $target_dir . basename($_FILES["imageUpload"]["name"]);
$uploadOk = 1;
$imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);


if (move_uploaded_file($_FILES["imageUpload"]["tmp_name"], $target_file)) {
    echo "The file ". basename( $_FILES["imageUpload"]["name"]). " has been uploaded.";
} else {
    echo "Sorry, there was an error uploading your file.";}

Answer 1

我找到了答案，对于那些正在寻找相同事物的人来说，我就是这样做的。 您不应该考虑将图像上传到数据库，而是可以将上传文件的名称存储在数据库中，然后检索文件名并使用它来显示图像。

HTML 代码

<input type="file" name="imageUpload" id="imageUpload">

PHP 代码

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

    //Process the image that is uploaded by the user

    $target_dir = "uploads/";
    $target_file = $target_dir . basename($_FILES["imageUpload"]["name"]);
    $uploadOk = 1;
    $imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);

    if (move_uploaded_file($_FILES["imageUpload"]["tmp_name"], $target_file)) {
        echo "The file ". basename( $_FILES["imageUpload"]["name"]). " has been uploaded.";
    } else {
        echo "Sorry, there was an error uploading your file.";
    }

    $image=basename( $_FILES["imageUpload"]["name"],".jpg"); // used to store the filename in a variable

    //storind the data in your database
    $query= "INSERT INTO items VALUES ('$id','$title','$description','$price','$value','$contact','$image')";
    mysql_query($query);

    require('heading.php');
    echo "Your add has been submited, you will be redirected to your account page in 3 seconds....";
    header( "Refresh:3; url=account.php", true, 303);
}

显示图像的代码

while($row = mysql_fetch_row($result)) {
    echo "<tr>";
    echo "<td><img src='uploads/$row[6].jpg' height='150px' width='300px'></td>";
    echo "</tr>\n";
}

Answer 2

if(isset($_POST['form1']))
{
    try
    {


        $user=$_POST['username'];

        $pass=$_POST['password'];
        $email=$_POST['email'];
        $roll=$_POST['roll'];
        $class=$_POST['class'];



        if(empty($user)) throw new Exception("Name can not empty");
        if(empty($pass)) throw new Exception("Password can not empty");
        if(empty($email)) throw new Exception("Email can not empty");
        if(empty($roll)) throw new Exception("Roll can not empty");
        if(empty($class)) throw new Exception("Class can not empty");


        $statement=$db->prepare("show table status like 'tbl_std_info'");
        $statement->execute();
        $result=$statement->fetchAll();
        foreach($result as $row)
        $new_id=$row[10];


        $up_file=$_FILES["image"]["name"];

        $file_basename=substr($up_file, 0 , strripos($up_file, "."));
        $file_ext=substr($up_file, strripos($up_file, ".")); 
        $f1="$new_id".$file_ext;

        if(($file_ext!=".png")&&($file_ext!=".jpg")&&($file_ext!=".jpeg")&&($file_ext!=".gif"))
        {
            throw new Exception("Only jpg, png, jpeg or gif Logo are allow to upload / Empty Logo Field");
        }
        move_uploaded_file($_FILES["image"]["tmp_name"],"../std_photo/".$f1);


        $statement=$db->prepare("insert into tbl_std_info (username,image,password,email,roll,class) value (?,?,?,?,?,?)");

        $statement->execute(array($user,$f1,$pass,$email,$roll,$class));


        $success="Registration Successfully Completed";

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

Answer 3

插入图片zh

-当我们使用插入查询在数据库中插入图像时

$Image = $_FILES['Image']['name'];
    if(!$Image)
    {
      $Image="";
    }
    else
    {
      $file_path = 'upload/';
      $file_path = $file_path . basename( $_FILES['Image']['name']);    
      if(move_uploaded_file($_FILES['Image']['tmp_name'], $file_path)) 
     { 
}
}

Answer 4

<!-- 
//THIS PROGRAM WILL UPLOAD IMAGE AND WILL RETRIVE FROM DATABASE. UNSING BLOB
(IF YOU HAVE ANY QUERY CONTACT:rahulpatel541@gmail.com)


CREATE TABLE  `images` (
  `id` int(100) NOT NULL AUTO_INCREMENT,
  `name` varchar(100) NOT NULL,
  `image` longblob NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB ;

-->
<!-- this form is user to store images-->
<form action="index.php" method="post"  enctype="multipart/form-data">
Enter the Image Name:<input type="text" name="image_name" id="" /><br />

<input name="image" id="image" accept="image/JPEG" type="file"><br /><br />
<input type="submit" value="submit" name="submit" />
</form>
<br /><br />
<!-- this form is user to display all the images-->
<form action="index.php" method="post"  enctype="multipart/form-data">
Retrive all the images:
<input type="submit" value="submit" name="retrive" />
</form>



<?php
//THIS IS INDEX.PHP PAGE
//connect to database.db name is images
        mysql_connect("", "", "") OR DIE (mysql_error());
        mysql_select_db ("") OR DIE ("Unable to select db".mysql_error());
//to retrive send the page to another page
if(isset($_POST['retrive']))
{
    header("location:search.php");

}

//to upload
if(isset($_POST['submit']))
{
if(isset($_FILES['image'])) {
        $name=$_POST['image_name'];
        $email=$_POST['mail'];
        $fp=addslashes(file_get_contents($_FILES['image']['tmp_name'])); //will store the image to fp
        }
                // our sql query
                $sql = "INSERT INTO images VALUES('null', '{$name}','{$fp}');";
                            mysql_query($sql) or die("Error in Query insert: " . mysql_error());
} 
?>



<?php
//SEARCH.PHP PAGE
    //connect to database.db name = images
         mysql_connect("localhost", "root", "") OR DIE (mysql_error());
        mysql_select_db ("image") OR DIE ("Unable to select db".mysql_error());
//display all the image present in the database

        $msg="";
        $sql="select * from images";
        if(mysql_query($sql))
        {
            $res=mysql_query($sql);
            while($row=mysql_fetch_array($res))
            {
                    $id=$row['id'];
                    $name=$row['name'];
                    $image=$row['image'];

                  $msg.= '<a href="search.php?id='.$id.'"><img src="data:image/jpeg;base64,'.base64_encode($row['image']). ' " />   </a>';

            }
        }
        else
            $msg.="Query failed";
?>
<div>
<?php
echo $msg;
?>