Php 如何使用平铺在数据库中显示图像
我正试图为我的网页上传文件与瓷砖我不能得到它显示在网站上Php 如何使用平铺在数据库中显示图像,php,upload,Php,Upload,我正试图为我的网页上传文件与瓷砖我不能得到它显示在网站上 if (isset($_POST["submit"])) { $title = $_POST["title"]; #file name with a random number so that similar dont get replaced $pname = rand(1000,10000)."-".$_FILES["file"]["name"]; #temporary file name to store file
if (isset($_POST["submit"]))
{
$title = $_POST["title"];
#file name with a random number so that similar dont get replaced
$pname = rand(1000,10000)."-".$_FILES["file"]["name"];
#temporary file name to store file
$tname = $_FILES["file"]["tmp_name"];
#upload directory path
$uploads_dir = 'images';
#TO move the uploaded file to specific location
move_uploaded_file($tname, $uploads_dir.'/'.$pname);
#sql query to insert into database
$sql = "INSERT into fileup(title,image) VALUES('$title','$pname')";
if(mysqli_query($conn,$sql)){
echo "File Sucessfully uploaded";
}
else{
echo "Error";
}
}
在这里,我将图像的路径保存到数据库中,然后检索图像源vai DB 用于上传照片
$folder=“../uploads/”;
$destFile=$folder。basename($_文件[“照片”][“名称]);
$sourdeFile=$\u文件[“照片”][“tmp\u名称”];
如果(移动上传的文件($sourdeFile,$destFile)){
echo“文件已上传”;
$photo=basename($_文件[“photo”][“name”]);
}否则{
echo$_文件['photo']['error'];
$photo=“images/default.png”;
}
$sql=“插入汽车(照片)值(“$photo”)”;
if(mysqli_查询($con,$sql)){
回响
"
警报(“您的图像已成功上传”);
location='newindex.php';
";
}否则{
echo“Error:.mysqli_Error($con);
}
查看上载的图像
$sql=“选择*自车”;
$result=mysqli\u查询($con,$sql);
$row=mysqli\u fetch\u数组($result);
“class=“img缩略图”style=“最大宽度:400px;最大高度:350px;"/>
在这里,照片将被上传到名为uploads的文件夹中。在上传开始时,图像的名称将保存在DB表中。然后要查看上传的图像
img src
将使用默认路径名和数据库中的照片名称。希望您能从中获取一些信息。另外,请注意,这只是一个示例。使用SQL
请使用参数化查询以避免SQL注入
。谢谢您面临的问题是什么?
$folder ="../uploads/";
$destFile = $folder . basename($_FILES["photo"]["name"]);
$sourdeFile = $_FILES["photo"]["tmp_name"];
if(move_uploaded_file($sourdeFile,$destFile)){
echo "File has been uploaded";
$photo = basename($_FILES["photo"]["name"]);
}else{
echo $_FILES['photo']['error'];
$photo = "images/default.png";
}
$sql= "INSERT INTO car(photo) VALUES('$photo') ";
if(mysqli_query($con,$sql)){
echo
"
<script>
alert ('Your Image successfully upladed');
window.location='newindex.php';
</script>
";
}else{
echo "Error:".mysqli_error($con);
}
$sql=" SELECT * FROM car";
$result=mysqli_query($con,$sql);
$row=mysqli_fetch_array($result);
<img src="Your default path name/<?=$row["photo"];?>" class="img-thumbnail" style="max-width:400px;max-height:350px;"/>