Live search results showing on new page - php

Question

I hope someone can help me, I have an live Ajax Jquery search bar. Now my search bar shows no more then 5 results. What I want is an button below the live results with 'More results', and if you click this button you go to a new page and see all te results of your search.
What I have tried is to put an submit button below the row results, but then he put all the names of the results into the search insteady of the text from the input.
PHP:
<form action="./s/" method="GET">
<div class="searchbox">
<!-- SEARCHBOX INPUT -->
<input type="text" name="s" class="searchbox-input" placeholder="Search trough 1,000 games in our database.." name="name" />
<div class="searchbox-line"></div>
<div class="searchbox-icon">
<input class="img-searchbox-icon" name="" value="" type="submit">
</div>
<div class="results"></div>
</div>
</form>
MySQL:
if(isset($_REQUEST["term"])){
// Prepare a select statement
$sql = "SELECT * FROM games WHERE name LIKE ? ORDER BY name LIMIT 5";
if($stmt = mysqli_prepare($link, $sql)){
// Bind variables to the prepared statement as parameters
mysqli_stmt_bind_param($stmt, "s", $param_term);
// Set parameters
$param_term = '%' . $_REQUEST["term"] . '%';
// Attempt to execute the prepared statement
if(mysqli_stmt_execute($stmt)){
$result = mysqli_stmt_get_result($stmt);
// Check number of rows in the result set
if(mysqli_num_rows($result) > 0){
// Fetch result rows as an associative array
while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
echo '<a href="' . $row["url"] . '">';
echo '<p class="results-name">' . $row["name"] . '</a></p>';
echo '<p class="results-platform">';
echo '<img class="platform" src="img/platform-icons/' . $row["platform1"] . '.png" />';
echo '<img class="platform" src="img/platform-icons/' . $row["platform2"] . '.png" />';
echo '<img class="platform" src="img/platform-icons/' . $row["platform3"] . '.png" />';
echo '<img class="platform" src="img/platform-icons/' . $row["platform4"] . '.png" />';
echo '<img class="platform" src="img/platform-icons/' . $row["platform5"] . '.png" />';
echo '<img class="platform" src="img/platform-icons/' . $row["platform6"] . '.png" />';
echo '</p>';
}
} else{
echo '<p class="results-gamename">No matches found</p>';
}
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}

Answer 1

Alright, I'll take a shot at this. In your HTML file (the one you originally named 'PHP') you've declared the form like this:
<form action="./s/" method="GET">
And you've given your search input the following name:
<input type="text" name="s" class="..." placeholder="..." name="name" />
Hence, when you send through the form, your input field will be sent through with the name 'name', since you've double declared its name. It'll fallback to the last 'name' attribute you've given to the element.