php 如何使用php将图像存储在mysql数据库中
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时间：2020-08-25 23:02:37 来源:igfitidea点击:
How to store images in mysql database using php
提问by Vipul Behl
How can i store and display the images in a MySQL database. Till now i have only written the code to get the images from the user and store them in a folder, the code that i wrote till now is: HTML FILE
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一个文件夹中的代码，我编写到现在的代码是： HTML FILE
<input type="file" name="imageUpload" id="imageUpload">
PHP FILE
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( $_FILES["imageUpload"]["name"]). " has been uploaded.";
} else {
echo "Sorry, there was an error uploading your file.";}
回答by Vipul Behl
I found the answer, For those who are looking for the same thing here is how I did it. You should not consider uploading images to the database instead you can store the name of the uploaded file in your database and then retrieve the file name and use it where ever you want to display the image.
我找到了答案，对于那些在这里寻找同样事情的人，我就是这样做的。您不应该考虑将图像上传到数据库，而是可以将上传文件的名称存储在数据库中，然后检索文件名并在您想要显示图像的任何地方使用它。
HTML CODE
代码
<input type="file" name="imageUpload" id="imageUpload">
PHP CODE
PHP代码
if(isset($_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($query);
require('heading.php');
echo "Your add has been submited, you will be redirected to your account page in 3 seconds....";
header( "Refresh:3; url=account.php", true, 303);
}
CODE TO DISPLAY THE IMAGE
显示图像的代码
while($row = mysql_fetch_row($result)) {
echo "<tr>";
echo "<td><img src='uploads/$row[6].jpg' height='150px' width='300px'></td>";
echo "</tr>\n";
}
回答by Ali Azam
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();
}
}
回答by zalak patel
insert image zh
插入图片zh
-while we insert image in database using insert query
- 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'], $file_path))
{
}
}
回答by rahul patel
<!--
//THIS PROGRAM WILL UPLOAD IMAGE AND WILL RETRIVE FROM DATABASE. UNSING BLOB
(IF YOU HAVE ANY QUERY CONTACT:[email protected])
CREATE TABLE `images` (
`id` int(100) NOT NULL AUTO_INCREMENT,
`name` varchar(100) NOT NULL,
`image` longblob NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB ;
-->
<!-- this form is user to store images-->
<form action="index.php" method="post" enctype="multipart/form-data">
Enter the Image Name:<input type="text" name="image_name" id="" /><br />
<input name="image" id="image" accept="image/JPEG" type="file"><br /><br />
<input type="submit" value="submit" name="submit" />
</form>
<br /><br />
<!-- this form is user to display all the images-->
<form action="index.php" method="post" enctype="multipart/form-data">
Retrive all the images:
<input type="submit" value="submit" name="retrive" />
</form>
<?php
//THIS IS INDEX.PHP PAGE
//connect to database.db name is images
mysql_connect("", "", "") OR DIE (mysql_error());
mysql_select_db ("") OR DIE ("Unable to select db".mysql_error());
//to retrive send the page to another page
if(isset($_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;
?>
