php - how to retrieve image from database using mysqli prepare? -
i'm trying convert codes mysqli prepare , i'm having trouble retrieving images database particular data. let example, id 1 post image of in tha same page.
can please check codes guys? i'm trying view image particular data. can me guys? thank you, code:
this code in form:
<form action="upload_photo.php" method="post" enctype="multipart/form-data"> <div class="col-md-6"> <div class="alert alert-info" role="alert"><span class="glyphicon glyphicon-user" aria-hidden="true"></span> add/update photo</div> <input type="hidden" name="id" value="<?= $id; ?>" /> <input type="file" name="image"><br> <input type="submit" value="upload image" name="submit"> </div> <div class="row"> <div class="col-md-6 col-md-3"> <a href="#" class="thumbnail"> <img src="image_view.php?id=$id" alt="..."> </a> </div> updated photo </div> </form> this code php upload image:
<?php include '../session.php'; require_once 'config.php'; if (isset($_post['submit'])) { $imagename = mysqli_real_escape_string($conn, $_files["image"]["name"]); $imagedata = mysqli_real_escape_string($conn, file_get_contents($_files["image"]["tmp_name"])); $imagetype = mysqli_real_escape_string($conn, $_files["image"]["type"]); if (substr($imagetype, 0,5) == "image") { $query = "update `crew_info` set `image_name` = ?, `updated_photo` = ? `id` = ?"; $stmt = mysqli_prepare($conn, $query); mysqli_stmt_bind_param($stmt, 'ssi', $imagename, $imagedata, $_post['id']); mysqli_stmt_execute($stmt); $id = $_post['id']; header("location: ../admin/view_all_info.php?id=$id"); } else { echo "image not uploaded!"; } } ?> this code retrieving image not working me:
<?php include '../session.php'; require_once 'config.php'; if (isset($_post['id'])) { $id = mysqli_real_escape_string($_post['id']); $query = "select * `crew_info` `id` = ?"; $stmt = mysqli_prepare($conn, $query); mysqli_stmt_bind_param($stmt, 's', $_post['id']); mysqli_stmt_execute($stmt); mysqli_stmt_bind_result($stmt, $id, $first_name, $middle_name, $last_name, $age, $month, $day, $year, $birth_place, $gender, $martial_status, $religion, $nationality, $email_address, $address_1, $address_2, $course, $school_graduated, $remarks, $date_added, $crew_status, $image_name, $updated_photo); while (mysqli_stmt_fetch($stmt)) { sprintf("%s", $updated_photo); } header("content-type: image/jpeg"); sprintf("%s", $updated_photo); } else { echo "bad"; } ?> the image uploading cannot retrieve it. please me guys thank you
there few problems here:
- in html use
<img src="image_view.php?id=$id" alt="...">. makes request server, need$_get['id']instead of$_post['id']in script output image. - you should not use
mysqli_real_escape_string()when use prepared statement, mysqli take care of when execute prepared statement.additionaly, in script use image, not sending database connection first parameter, make fail well.edit: don't seem using escaped value can remove line.
apart should probably:
- store image file in file-system. if content not sensitive / protected, can put somewhere in root of web-server can include link it. make database maintenance , backups lot easier.
- only need when display image if use method using now. make code easier read , new columns not introduce bugs when do:
select updated_photo crew_info id = ?.
