Question

这是我尝试过的另一堆代码：

form.html

<form method='post' action='upload2.php' enctype='multipart/form-data'>
Select file :<input type='file' name='filename' size='10' />
<input type='submit' value='upload'>

upload.php的

  <?include "connect.php";
   $name = $_FILES['filename']['name'];
   $image = $_FILES['filename']['tmp_name'];

   $sql = "INSERT INTO files ". "VALUES ('', '$name', '$image')";

   mysql_query($sql) or die(mysql_error());?>

Display.php的

 <?php
  include "connect.php";
  $fid = $_FILES['fid'];

  // do some validation here to ensure id is safe
  $sql = "SELECT * FROM files WHERE fid='$fid'";
  $result = mysql_query($sql) or die(mysql_error());
  $row = mysql_fetch_assoc($result);
  $image = $row['content'];

  echo $image;?>

这里我调用display.php来查看文件：

  <?
  echo "<img src=display.php?fid=1  />";
  ?>

它只显示空白页

在我忘记之前我已经忘了..这是我的表：

 CREATE TABLE `files` ( `fid` INT( 11 ) NOT NULL AUTO_INCREMENT PRIMARY KEY COMMENT
 'unique id', `name` VARCHAR( 30 ) NOT NULL COMMENT 'file name', `content` BLOB NOT
  NULL COMMENT 'actual file' ) ENGINE = MYISAM COMMENT = 'Uploaded files'

这次只是一个空白页面图像成功保存在数据库中但我无法在页面中显示它。我的错误...我已经通过这里查看了解决方案，但我仍然无法使其工作..它要么不显示任何内容，要么显示损坏的图像

Answer 1

您可以将图像插入mysql数据库

$image = addslashes(file_get_contents($_FILES['filename']['tmp_name']));
$sql = "INSERT INTO files ". "VALUES ('', '$name', '$image')";

从表中获取图片并使用<img src>

嵌入页面

$sql = "SELECT * FROM files WHERE fid='$fid'";
$result = mysql_query($sql) or die(mysql_error());
$row = mysql_fetch_assoc($result);
$image = $row['content'];

echo '<img src="data:image/jpeg;base64,'.base64_encode( $image ).'"/>';

PS。未推荐使用mysql_ *函数。相反，您可以使用mysqli_ *或pdo_ *函数。

Answer 2

感谢大家的帮助......现在我明白了我的错误..问题是我的文件没有转移到.. @madforstrength感谢链接 the tutorial

首先是form.html

  <!DOCTYPE html>
  <html>
  <body>

  <form action="upload.php" method="post" enctype="multipart/form-data">
   Select image to upload:
   <input type="file" name="fileToUpload" id="fileToUpload">
   <input type="submit" value="Upload Image" name="submit">
   </form>

   </body>
   </html>

和上传脚本

 <?php
 $target_dir = "uploads/";
 $target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
  $uploadOk = 1;
  $imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);
  // Check if image file is a actual image or fake image
  if(isset($_POST["submit"])) {
$check = getimagesize($_FILES["fileToUpload"]["tmp_name"]);
if($check !== false) {
    echo "File is an image - " . $check["mime"] . ".";
    $uploadOk = 1;
} else {
    echo "File is not an image.";
    $uploadOk = 0;
}
}
// Check if file already exists
if (file_exists($target_file)) {
echo "Sorry, file already exists.";
$uploadOk = 0;
}
// Check file size
if ($_FILES["fileToUpload"]["size"] > 500000) {
echo "Sorry, your file is too large.";
$uploadOk = 0;
}
// Allow certain file formats
if($imageFileType != "jpg" && $imageFileType != "png" && $imageFileType != "jpeg"
&& $imageFileType != "gif" ) {
echo "Sorry, only JPG, JPEG, PNG & GIF files are allowed.";
$uploadOk = 0;
}
// Check if $uploadOk is set to 0 by an error
if ($uploadOk == 0) {
echo "Sorry, your file was not uploaded.";
// if everything is ok, try to upload file
} else {
if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)) {
    echo "The file ". basename( $_FILES["fileToUpload"]["name"]). " has been
uploaded.";
} else {
    echo "Sorry, there was an error uploading your file.";
}
}?>

这段代码我提供了设置的文件夹将文件保存到

$target_dir = "uploads/";

现在只是尝试在几个部分中应用数据库信息显示需要写的文件，接近这个的东西

谢谢大家

Answer 3

您需要将图像文件从临时文件夹移动到主图像目录，然后从那里提供服务。

可以找到一个完整的示例here

如何从我的数据库中显示图像

3 个答案: