CODEWITHSUNDEEP
phpmysqlmysqliimage-uploading
mar
mar

Reputation: 121

uploading pictures +picture information from php form to mysql database

I have this code for a form that uploads pictures to my website and saves the information to a mysql database:

<form method='post'>
        Album Name: <input type="text" name="title" /> 
        <input type="submit" name="submit" value="create" />
    </form>
<h4>Add Photo</h4>
<form enctype="multipart/form-data" method="post">
        <?php
        require_once 'config.php'; 
    $mysqli = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
    if(isset($_POST['upload'])){
        $caption = $_POST['caption'];
        $albumID = $_POST['album'];
        $file = $_FILES ['file']['name'];
        $file_type = $_FILES ['file']['type'];
        $file_size = $_FILES ['file']['size'];
        $file_tmp = $_FILES ['file']['tmp_name'];
        $random_name = rand();

        if(empty($file)){
            echo "Please enter a file <br>";
        } else{
             move_uploaded_file($file_tmp, 'uploads/'.$random_name.'.jpg');
    $ret = mysqli_prepare($mysqli, "INSERT INTO photos (caption, image_url, date_taken)
    VALUES(?, ?, NOW())");
    $filename = "uploads/" + $random_name + ".jpeg";
    mysqli_stmt_bind_param($ret, "ss", $caption, $filename);
    mysqli_stmt_execute($ret);
    echo "Photo successfully uploaded!<br>";
    }
    }
    ?>
    Caption: <br>
    <input type="text" name="caption">
    <br><br>
    Select Album: <br>
    <select name="album">
    <?php
    $mysqli = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
    $result = $mysqli->query("SELECT * FROM albums");
    while ($row = $result->fetch_assoc()) {
        $albumID = $row['albumID'];
        $title = $row['title'];
        echo "<option value='$albumID'>$title</option>";
    }
    ?>
    </select>
    <br><br>
    Select Photo: <br>
    <input type="file" name="file">
    <br><br>
    <input type="submit" name="upload" value="Upload">
</form>

This successfully uploads the picture to my 'uploads' folder as well as my mysql database, however, I would like to put in image URL "uploads/(random name generated).jpg" I have failed to do this with my current code, the information recorded in the 'image_url' column of my photos table is just the random number generated. without the "uploads/" in the beginning and ".jpg" in the end.

I should mention that the schema for my photos table is: caption, image_url, date_taken, imageID

Any help will be very much appreciated!! thank you in advance

Upvotes: 2

Views: 864

Answers (1)

Funk Forty Niner
Funk Forty Niner

Reputation: 74217

You are using + (plus) signs to concatenate with, in this line:

$filename = "uploads/" + $random_name + ".jpeg";

PHP uses dots/periods to concatenate with, rather than plus signs which is JS/C language syntax:

$filename = "uploads/" . $random_name . ".jpeg";

Error checking would have signaled the syntax error.

Upvotes: 1

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