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PHP

Victor Bielski
Victor Bielski
2,325 Points
Posted by Victor Bielski
Victor Bielski
2,325 Points

Search field - PHP & MySQL

Hello,

I'm currently working on a search field, but I'm having some troubles and I can't quite figure out, what the problem is. At the moment, users are able to upload documents like .pdf and .docx. And then the documents get stored in a table called "files" in my database.

The "files" table contains following:

ID name title forfatter (author) size downloads (for later user)

Search.php contains following:

<div class="container">
    <div class="row">
        <div class="col-12">
            <form action="" method="post">
                <input type="text" class="form-control" name="search" placeholder="Search">
                <button type="submit" name="submit-search"></button>
            </form>
        </div>
    </div>
</div>

<div class="container"> <div class="row"> <div class="col-12"> <?php if (isset($_POST['submit-search'])) { $search = $_POST['search']; $sql = "SELECT * FROM files WHERE name LIKE '%".$search."%'"; $query = mysqli_query($conn, $sql);

                if (mysqli_num_rows($query)) {
                    echo "Sorry, no results!";
                } else {
                    while ($fetch = mysqli_fetch_assoc($query)) {
                        $id = $fetch['id'];
                        $name = $fetch['name'];

                        echo $id. " - " .$name."< /br>";
                    }
                }
            }
        ?>
    </div>
</div>

</div>

If anyone knows what I'm doing wrong, help me out :)

1 Answer

Victor Bielski
Victor Bielski
2,325 Points
Victor Bielski
Victor Bielski
2,325 Points

if (isset($_POST['submit-search'])) { $search = $_POST['search']; $sql = "SELECT * FROM files WHERE name LIKE '%".$search."%'"; $query = mysqli_query($conn, $sql);

                if (mysqli_num_rows($query)) {
                    echo "Sorry, no results!";
                } else {
                    while ($fetch = mysqli_fetch_assoc($query)) {
                        $id = $fetch['id'];
                        $name = $fetch['name'];

                        echo $id. " - " .$name."< /br>";
                    }
                }
            }