提交表单（过滤内容）后，结果表将显示在同一页面上。 我怎么做？

Question

I would like to ask for some advice on how to go about coding (with regards to my problem).

Anyways, I have a search form which has a select form and a text box. After submitting, a table appears based on the filtered results from the form.

Do I have to use a session or do I just need to have $_POST variables in my php code? Or is it javascript?

If it's $_POST or $_SESSION, do I need to create another php file? I've done that but it displays a lot of errors on my form page.

I want the table to appear on the same page.

My form

                    <div class="form-group">
                        Find
                        <select name="selected" class="form-control" required data-error="Please select from the list">
                            <option value="">--Select--</option>
                            <option value="story">Stories</option>
                            <option value="poem">Poems</option>
                            <option value="user">Users</option>
                        </select>
                        <div class="help-block with-errors"></div>

                    </div>


                    <div class="form-group">
                        With
                        <input type="text" class="form-control" id="contains" name="contains" required

                               data-error="Please enter a value"/>
                        <div class="help-block with-errors"></div>
                    </div>
                    <div class="col-md-6 col-xs-6">
                        <input type="submit" class="btn btn-primary btn-block" id="submitBtn"
                               value="Search"/></div>
                    <div class="col-md-6 col-xs-6">
                        <input type="reset" class="btn btn-default btn-block" value="Clear Fields"/></div>
                </form>

I've created a separate php file to store all my $_SESSIONS and $_POST but not really sure to go about it.

<?php
session_start();
include("dbconn.php");
$selected = $_POST['selected'];
$searchContent = $_POST['content'];
$_SESSION['selected'] = $selected;
$_SESSION['contains'] = $searchContent;

if ($selected == "story" || $selected == "poem") {
    if (isset($_SESSION['username'])) {
        $querySearchSP = "select * from storypoem where content like '%$searchContent%' and type='poem'";
    } else if (!isset($_SESSION['username'])) {
        $querySearchSP = "select * from storypoem where content like '%$searchContent%' and type='poem' and progress='Completed'";
    }
    $resultSearchSP = mysqli_query($link, $querySearchSP);
    $searchResultSP = mysqli_num_rows($resultSearchSP);

    while ($rowStoryPoem = mysqli_fetch_array($searchResultSP)) {
        $arrStoryPoem[] = $rowStoryPoem;
        $_SESSION['title'] = $rowStoryPoem['title'];
        $_SESSION['progress'] = $rowStoryPoem['progress'];
        $_SESSION['currentlyCompleted'] = $rowStoryPoem['currentlycompleted'];
    }
} else if ($selected == "user") {
    $querySearchUser = "select * from user u, storypoem sp where content like '%$searchContent%' and u.storypoem_id=sp.id";
    $resultSearchUser = mysqli_query($link, $querySearchUser);
    $searchResultUser = mysqli_num_rows($resultSearchUser);

    while ($rowUser = mysqli_fetch_array($searchResultUser)) {
        $arrUser[] = $rowUser;
        $_SESSION['username'] = $rowUser['username'];
        $_SESSION['works'] = $rowStoryPoem['title'];
    }
}
?> 

Your help is much appreciated. Thanks.

Answer 1

For same page follow this approach:

<form action="" name="frm">
// your html elements
<input type="submit" name="btn" value="Search" />
</form>

<?php
if(isset($_REQUEST['btn']))
{
    // put your mysql code here and for that response create a result table by using looping on the result set
}
?>