Question

我在DB中插入图像时遇到问题。该表具有以下结构：

ID-＆GT; INT（3） - ＆GT;自动增量
名称 - ＆GT; VARCHAR（30）
extension-＆gt; VARCHAR（10）[可能太短]
img-＆GT; MEDIUMBLOB

插入图像的PHP代码是：

if($_FILES['file']['error']==0){
        $result = is_uploaded_file($_FILES['file']['tmp_name']);
        if(!$result){
            echo "Upload failed";
        }else{
            $type = explode("/", $_FILES['file']['type']);
            $extension = $type[1];

            $name = $_FILES['file']['name'];
            $img = $_FILES['file']['tmp_name'];
            $img = file_get_contents($_FILES['file']['tmp_name']);
            $img = addslashes ($img);
        }

        $sql = "INSERT INTO images (name, extension, img) VALUES ('$name', '$extension', '$img')";
        $result = $mysqli->query($sql);
        if($result){
            echo "insertion was successful";
        }else{
            echo "insertion failed: ".$mysqli->error;
        }

这就是我试图看到img的方式：

$sql = "SELECT name, extension, img FROM images WHERE id='1'";
$result = $mysqli->query($sql);
if($result){
    $a = $result->fetch_assoc();
    header ("Content-type: image/".$a['estensione']);
    echo $a['img'];

}else{
    echo "AAAAAAAAA<hr>";
    echo $mysqli->error;
}

插入正常，但我无法查看图像。另外，还有另一种在Db上传图像的方法吗？

Answer 1

图片标记示例中的第一个（您要显示结果的页面）

标签中的

从fetch_image_frm_db.php页面获取图像，并显示

<img src="fetch_image_frm_db.php?id=<?php echo $id_of_row;?>"/>

fetch_image_frm_db.php页面

$id=$_GET['id'];

$query = "SELECT * FROM images WHERE id=$id";

$result=mysql_query($query) or die('Error, query failed'.mysql_error()); 
$row=mysql_fetch_array($result);

header("Content-type:image/jpeg");
stripslashes ($row['img']);

echo $row['img'];

* * 在db中存储图像不是一种好习惯

Answer 2

我认为你在第5行的代码中拼错了扩展名：

$sql = "SELECT name, extension, img FROM images WHERE id='1'";
$result = $mysqli->query($sql);
if($result){
    $a = $result->fetch_assoc();
    header ("Content-type: image/".$a['extension']);
    echo $a['img'];

}else{
    echo "AAAAAAAAA<hr>";
    echo $mysqli->error;
}

Answer 3

好的，我这样解决了...... 索引页面的代码是：

<?php
  $mysqli = new mysqli("127.0.0.1","root","root","prova");
  if($mysqli->connect_errno){
     echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
  }

if(isset($_POST['submit'])){
 if(isset($_FILES['file'])){
    if($_FILES['file']['error']==0){
        $result = is_uploaded_file($_FILES['file']['tmp_name']);
        if(!$result){
            echo "Impossibile eseguire l'upload";
        }else{
            $type = $_FILES['file']['type'];
            $nome = $_FILES['file']['name'];
            $img = file_get_contents($_FILES['file']['tmp_name']);
            $img = addslashes ($img);

        }

        $sql = "INSERT INTO immagini (nome, estensione, img) VALUES ('$nome', '$type', '$img')";
        $result = $mysqli->query($sql);
        if($result){
            echo "Insertion was succesfully executed";
        }else{
            echo "Insertion failed: ".$mysqli->error;
        }
    }

 }
}
    echo "<br><br><hr>";
    $query = "SELECT * FROM immagini";

    $result=$mysqli->query($query) or die('Error, query failed'.mysql_error());

    while($row = mysqli_fetch_array($result)){
        echo "<p><img height='250' width='250' src=\"fetch_img.php?id=$row[id]\">";
    }

?>

这是脚本fetch_img.php，taht有效打印图像：

<?php
    $mysqli = new mysqli("127.0.0.1","root","root","prova");
    if($mysqli->connect_errno){
        echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
    }

    if($_GET['id']==""){
        header("Location: index.php");
    }

    $id=$_GET['id'];

    $query = "SELECT * FROM immagini WHERE id=$id";

    $result=$mysqli->query($query) or die('Error, query failed'.mysql_error()); 
    //var_dump($result);
    $row= $result->fetch_assoc();

    header("Content-type: image/jpeg");
    $rs_immagine=imagecreatefromstring($row['img']);
    imagejpeg( $rs_immagine, null, 100);
    imagedestroy($rs_immagine);

?>

感谢所有人，同时也建议将数据库中存储文件（尤其是图像）存储在内部......我是第一个说这个的人，但我必须这样做。

在MySQL DB中插入/查看图像

3 个答案: