如何使用PHP代码将图像上传到MySQL数据库
/*-------------------
IMAGE QUERY
---------------*/
$file =$_FILES['image']['tmp_name'];
if(!isset($file))
{
echo 'Please select an Image';
}
else
{
$image_check = getimagesize($_FILES['image']['tmp_name']);
if($image_check==false)
{
echo 'Not a Valid Image';
}
else
{
$image = file_get_contents ($_FILES['image']['tmp_name']);
$image_name = $_FILES['image']['name'];
if ($image_query = mysql_query ("insert into product_images values (1,'$image_name',$image )"))
{
echo $current_id;
//echo 'Successfull';
}
else
{
echo mysql_error();
}
}
}
/*-----------------
IMAGE QUERY END
---------------------*/
<form action='insert_product.php' method='POST' enctype='multipart/form-data' ></br>
File : <input type='file' name= 'image' >
</form>
错误消息：您的SQL语法有错误；请检查与MySQL服务器版本相对应的手册，以获得在第1行使用Next‘的正确语法。
慕标琳琳浏览 737回答 3
3回答
-
动漫人物
这是通过MySQL数据库上传和显示图像的完美代码。<html><body><form method="post" enctype="multipart/form-data"><input type="file" name="image"/> <input type="submit" name="submit" value="Upload"/></form><?php if(isset($_POST['submit'])) { if(getimagesize($_FILES['image']['tmp_name'])==FALSE) { echo " error "; } else { $image = $_FILES['image']['tmp_name']; $image = addslashes(file_get_contents($image)); saveimage($image); } } function saveimage($image) { $dbcon=mysqli_connect('localhost','root','','dbname'); $qry="insert into tablename (name) values ('$image')"; $result=mysqli_query($dbcon,$qry); if($result) { echo " <br/>Image uploaded."; header('location:urlofpage.php'); } else { echo " error "; } }?></body></html>
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