php 通过php上传MySql数据库中的图片
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/10837556/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
To upload a image in MySql database via php
提问by Neeraj
I am currently working on a website that needs to upload the images of different products by its users. I am implementing it by using MySql database via php.
我目前正在开发一个需要用户上传不同产品图片的网站。我正在通过 php 使用 MySql 数据库来实现它。
My code for a basic form for taking input from users is:
我用于获取用户输入的基本表单的代码是：
<form enctype="multipart/form-data" action="testimage1.php" method="post" name="changer">
<input name="MAX_FILE_SIZE" value="102400" type="hidden">
<input name="image" accept="image/jpeg" type="file">
<input value="Submit" type="submit">
</form>
My database table is:
我的数据库表是：
mysql> CREATE TABLE tbl_images (
> id tinyint(3) unsigned NOT NULL auto_increment,
> image blob NOT NULL,
> PRIMARY KEY (id)
> );
testimage1.php has the following code:-
testimage1.php 有以下代码：-
$username = "root";
$password = "";
$host = "localhost";
$database = "thinstrokes";
$link = mysql_connect($host, $username, $password);
if (!$link) {
die('Could not connect: ' . mysql_error());
}
// Select your database
mysql_select_db ($database);
if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) {
// Temporary file name stored on the server
$tmpName = $_FILES['image']['tmp_name'];
// Read the file
$fp = fopen($tmpName, 'r');
$data = fread($fp, filesize($tmpName));
$data = addslashes($data);
fclose($fp);
// Create the query and insert
// into our database.
$query = "INSERT INTO tbl_images ";
$query .= "(image) VALUES ('$data')";
$results = mysql_query($query, $link) or die(mysql_error());
// Print results
print "Thank you, your file has been uploaded.";
}
else {
print "No image selected/uploaded";
}
On submitting the form I am getting an error: No image selected/uploaded
在提交表单时，我收到一个错误： No image selected/uploaded
I am not getting the error... and I've already asked for this before as:
我没有收到错误......我之前已经要求过这个：
But until now I am not successful in storing the image in the database.
但是直到现在我还没有成功地将图像存储在数据库中。
采纳答案by Ahoura Ghotbi
Your script is working just fine, This is what I test:
你的脚本工作得很好，这是我测试的：
<?
if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) {
// Temporary file name stored on the server
$tmpName = $_FILES['image']['tmp_name'];
// Read the file
$fp = fopen($tmpName, 'r');
$data = fread($fp, filesize($tmpName));
$data = addslashes($data);
fclose($fp);
// Create the query and insert
// Print results
print "Thank you, your file has been uploaded.";
}
else {
print "No image selected/uploaded";
}
?>
<form enctype="multipart/form-data" action="" method="post" name="changer">
<input name="MAX_FILE_SIZE" value="102400" type="hidden">
<input name="image" accept="image/jpeg" type="file">
<input value="Submit" type="submit">
</form>
And it works just fine, if you want to see it in action I can send you the link.
它工作得很好，如果你想看到它的实际效果，我可以给你发送链接。
It must be something else that is ruining your code (*note that I removed the DB queries to avoid getting mysql errors but the script was working even with them there.
一定是其他东西破坏了您的代码（*请注意，我删除了数据库查询以避免出现 mysql 错误，但即使在那里脚本也能与它们一起工作。
回答by Sanjiv Kumar
//Here is the solution for your problem
//这里是你的问题的解决方案
<form enctype="multipart/form-data" action="" method="post" name="changer">
<input name="MAX_FILE_SIZE" value="102400" type="hidden">
<input name="image" accept="image/jpeg" type="file">
<input value="Submit" type="submit">
</form>
<?php
// connection to database
include 'includes/connection.php';
?>
<?php
if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) {
// Temporary file name stored on the server
$tmpName = $_FILES['image']['tmp_name'];
// Read the file
$fp = fopen($tmpName, 'r');
$data = fread($fp, filesize($tmpName));
$data = addslashes($data);
fclose($fp);
$result = mysql_query("INSERT INTO image (image)VALUES ( '$data')", $connection);
if(!$result)
{
die("Database query failed: ". mysql_error());
}
// Print results
print "Thank you, your file has been uploaded.";
}
else
{
print "No image selected/uploaded";
}
?>
<?php
//close connection
include 'includes/close.php';
?>
