如何在PHP（MYSQLi）中显示数据库映像

Question

您能帮我吗，我知道这个问题已经被问过无数次了。 我尝试自己做，但没有成功。 我没有任何线索。

这是将图像上传到数据库的代码：

<?php
$con = mysqli_connect('localhost','root','root', 'toevoegen');
if (mysqli_connect_errno())
{echo "Failed to connect to MySQL: " . mysqli_connect_error();}

function GetImageExtension($imagetype)
 {
   if(empty($imagetype)) return false;
   switch($imagetype)
   {
       case 'image/bmp': return '.bmp';
       case 'image/gif': return '.gif';
       case 'image/jpeg': return '.jpg';
       case 'image/png': return '.png';
       default: return false;
   }
 }
if (!empty($_FILES["uploadedimage"]["name"])) {

$file_name=$_FILES["uploadedimage"]["name"];
$temp_name=$_FILES["uploadedimage"]["tmp_name"];
$imgtype=$_FILES["uploadedimage"]["type"];
$ext= GetImageExtension($imgtype);
$imagename=$_FILES["uploadedimage"]["name"];
$target_path = "images/".$imagename;


if(move_uploaded_file($temp_name, $target_path)) {
$query_upload="INSERT into images_tbl ( images_path ) VALUES ('".$target_path."')";
mysqli_query($con, $query_upload) or die("error in $query_upload == ----> ".mysqli_error());  
}else{

   exit("Error While uploading image on the server");
} 

}

?>

<form action="saveimage.php" enctype="multipart/form-data" method="post">

<table style="border-collapse: collapse; font: 12px Tahoma;" border="1"      cellspacing="5" cellpadding="5">
<tbody><tr>
<td>
<input name="uploadedimage" type="file">
</td>
</tr>
<tr>
<td>
<input name="Upload Now" type="submit" value="Upload Image">
</td>
</tr>
</tbody></table>
</form>

这是显示代码的开头：

$con = mysqli_connect('localhost', 'root', 'root', 'toevoegen');
$query = mysqli_query($con,"SELECT 'images_path' FROM 'images_tbl'");
while($row = mysqli_fetch_assoc($query)){

echo "<div class='tile' style='background-image: url(\"$row[images_path]\")'>";
}

你能帮我完成这个吗？

Answer 1

我通过执行此操作来修复它：

<!DOCTYPE html>
<html>
<head>
<link rel="stylesheet" type="text/css" href="series.css">
</head>
<body>

<?php 
include_once("mysqliconnect.php");

$query = "SELECT images_path FROM images_tbl ORDER BY images_ID DESC";
$query = mysqli_real_escape_string($con,$query);

if($result = mysqli_query($con,$query)){
while($row = mysqli_fetch_object($result)){

    echo   

    '<div><img src="'. $row->images_path . '" border=0 class="tile" id="hover"></div>';

}

mysqli_free_result($result);
}
mysqli_close($con);
?>


</body>
</html>

如何在PHP（MYSQLi）中显示数据库映像

问题描述

1 个解决方案

解决方案1
0 2016-04-15 12:35:09

如何在PHP（MYSQLi）中显示数据库映像

问题描述

1 个解决方案

解决方案1 0 2016-04-15 12:35:09

解决方案1
0 2016-04-15 12:35:09