PHP / MySQL：圖像成功保存到數據庫但未能在網頁上顯示

Question

我的系統中有一個非常奇怪的問題。 我已經創建了一個系統來將圖像上傳到數據庫並顯示它。 問題是，圖片上傳成功，但是會返回“上傳失敗！”的信息。 然后，已上傳的圖片不顯示。 下面是我的代碼：

    <body>

        <div class="wrapperDiv">
            <form action="" method="post" id="form" enctype="multipart/form-data">
                Upload image : 
                <input type="file" name="uploadFile" value="" />
                <input type="submit" name="submitBtn" value="Upload" />
            </form>


            <?php
            $last_insert_id = null;
            include('db2.php');
            if(isset($_POST['submitBtn']) && !empty($_POST['submitBtn'])) {
                if(isset($_FILES['uploadFile']['name']) && !empty($_FILES['uploadFile']['name'])) {
                    //Allowed file type
                    $allowed_extensions = array("jpg","jpeg","png","gif");

                    //File extension
                    $ext = strtolower(pathinfo($_FILES['uploadFile']['name'], PATHINFO_EXTENSION));

                    //Check extension
                    if(in_array($ext, $allowed_extensions)) {
                        //Convert image to base64
                        $encoded_image = base64_encode(file_get_contents($_FILES['uploadFile']['tmp_name']));
                        $encoded_image = $encoded_image;
                        $query = "INSERT INTO tbl_images SET encoded_image = '".$encoded_image."'";
                        $sql = $conn->prepare($query);
                        $sql -> execute();
                        //$results = $sql -> fetchAll(PDO::FETCH_OBJ);
                        echo "File name : " . $_FILES['uploadFile']['name'];
                        echo "<br>";
                        if($sql->rowCount() > 1 ) {
                            echo "Status : Uploaded";
                            $last_insert_id = $conn-> lastInsertId(); 
                        } else {
                            echo "Status : Failed to upload!";
                        }
                    } else {
                        echo "File not allowed";
                    }
                }

                if($last_insert_id) {
                    $query = "SELECT encoded_image FROM tbl_images WHERE id= ". $last_insert_id;
                    $sql = $conn->prepare($query);
                    $sql -> execute();
                    if($sql->rowCount($sql) == 1 ) {
                        //$row = mysqli_fetch_object($result);
                        while($row = $sql->fetch(PDO::FETCH_ASSOC)) {
                        echo "<br><br>";
                        echo '<img src="'.$row->encoded_image.'" width="250">';

                        }
                    }
                }
            }
            ?>
        </div>
    </body>

有人能幫我嗎？ 謝謝！

Answer 1

您首先做錯了一些事情，在存儲在數據庫中時對圖像進行了編碼，因此您必須再次對其進行解碼，並且標簽中的 src 獲取一個 url 而不是圖像內容，只是回顯如下內容：

header('Content-type: image/jpeg');
echo base64_decode($row->encoded_image);

或者

<img src="data:image/png;base64,'.$row->encoded_image.'" width="250">

但總而言之，將圖像存儲在數據庫中不是一個好的選擇，您的數據庫變得太重並且無法快速響應並且內存過多您可以將圖像名稱存儲在數據庫中並將文件形式移動到您的服務器中的特殊位置可以這樣顯示。

echo '<img src="specialRoot/'.$row->image_name.'" width="250">';

Answer 2

將圖像存儲在文件夾中.. 我在根目錄中創建了上傳文件夾，您可以在任何地方創建文件夾並在獲取圖像的同時寫下您的路徑..

<body>

    <div class="wrapperDiv">
        <form action="" method="post" id="form" enctype="multipart/form-data">
            Upload image : 
            <input type="file" name="uploadFile" value="" />
            <input type="submit" name="submitBtn" value="Upload" />
        </form>


        <?php
        $last_insert_id = null;
        include('db2.php');
        if(isset($_POST['submitBtn']) && !empty($_POST['submitBtn'])) {
            if(isset($_FILES['uploadFile']['name']) && !empty($_FILES['uploadFile']['name'])) {
                //Allowed file type
                $allowed_extensions = array("jpg","jpeg","png","gif");
                $name = $_FILES['uploadFile']['name'];
                $target_dir = "uploads/";   //give path of your folder where images are stored.
                $target_file = $target_dir . basename($_FILES["uploadFile"]["name"]);
                $imageFileType = strtolower(pathinfo($target_file,PATHINFO_EXTENSION));

                //Check extension
                if( in_array($imageFileType,$allowed_extensions) ){
                    //Convert image to base64

                    $image_base64 = base64_encode(file_get_contents($_FILES['uploadFile']['tmp_name']) );
                    $encoded_image = 'data:image/'.$imageFileType.';base64,'.$image_base64;

                    //$encoded_image = base64_encode($_FILES['uploadFile']['tmp_name']);
                    //$encoded_image = $encoded_image;
                    $query = "INSERT INTO tbl_images SET encoded_image = '".$encoded_image."'";
                    $sql = $conn->prepare($query);
                    $result = $sql -> execute();
                    move_uploaded_file($_FILES['uploadFile']['tmp_name'],$target_dir.$name);
                    echo "File name : " . $_FILES['uploadFile']['name'];
                    echo "<br>";

                    if($result == 1) {
                        echo "Status : Uploaded";
                        $last_insert_id = $conn->insert_id;
                    } else {
                        echo "Status : Failed to upload!";
                    }
                } else {
                    echo "File not allowed";
                }
            }

            if($last_insert_id) {

               $query = "SELECT encoded_image FROM tbl_images WHERE id= ". $last_insert_id;
                $result = $conn->query($query);
                while($row = $result->fetch_assoc()){
                    echo '<img src="'.$row['encoded_image'].'" width="250">';
                }
            }
        }
        ?>
    </div>
</body>

PHP / MySQL：圖像成功保存到數據庫但未能在網頁上顯示

問題描述

2 個解決方案

解決方案1
0 2019-12-30 07:11:53

解決方案2
-1 2019-12-30 08:01:41

PHP / MySQL：圖像成功保存到數據庫但未能在網頁上顯示

問題描述

2 個解決方案

解決方案1 0 2019-12-30 07:11:53

解決方案2 -1 2019-12-30 08:01:41

解決方案1
0 2019-12-30 07:11:53

解決方案2
-1 2019-12-30 08:01:41