Question

如何在MySQL数据库中存储和显示图像。直到现在我只编写了代码来从用户那里获取图像并将它们存储在一个文件夹中，我写的代码到现在为止： 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，对于那些在这里寻找同样事物的人，我是如何做到的。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，然后检索文件名并将其用于想要显示图像的位置。

HTML CODE

<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);
}

显示图片的代码

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

插入图像zh

-当我们使用插入查询将图像插入数据库时

$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;
?>

如何使用php在mysql数据库中存储图像

4 个答案: