PHP Searchbar only showing 1 result - php

Question

I am trying to create a search bar in php.
I want to retrieve data from my database.
The problem is only that if I search I get only 1 result.
How do I solve this?
Or is there another way in which I get this working?
My code:
<?php
include("config.php");
session_start();
$zoekopdracht = $_GET['search'];
$categorie = $_GET['categorie'];
$zoekopdracht = mysqli_real_escape_string($db,$zoekopdracht);
$sql = ("SELECT * FROM activiteiten WHERE `inhoud` LIKE '%$zoekopdracht%'");
$result = mysqli_query($db,$sql);
$row = mysqli_fetch_array($result,MYSQLI_ASSOC);
$activiteit_id = $row['activiteit_id'];
$naamsql = ("SELECT * FROM `activiteiten` WHERE activiteit_id = {$activiteit_id} AND optie = 'naam'");
$naamresult = mysqli_query($db,$naamsql);
$naamrow = mysqli_fetch_array($naamresult,MYSQLI_ASSOC);
$beschrijvingsql = ("SELECT * FROM `activiteiten` WHERE activiteit_id = {$activiteit_id} AND optie = 'beschrijving'");
$beschrijvingresult = mysqli_query($db,$beschrijvingsql);
$beschrijvingrow = mysqli_fetch_array($beschrijvingresult,MYSQLI_ASSOC);
result bar:
<?php
if(mysqli_num_rows($result) > 0){
?>
<div class="Row" id="main" style="border: 2px solid #51BAA4;border-radius: 4px; height: 0px; width: 100%;padding: 0em 0 0em 0; margin: 0 auto; display: table;">
<div class="Column" id="plaatje" style="padding: 0em 0 0em 0;">
<left><img src="images/noimage.png" height="100px" width="100px" alt=""></left>
</div>
<div class="Column" id="titel">
<div style="text-align: left;position:relative;margin:0px;width:10%;height:40px;padding-top: 3px; font-size:125%; display: -webkit-box;-webkit-line-clamp: 1;">
<left><h3><?php echo $naamrow['inhoud'];?></h3>
</div>
<div style="text-align: left; height: 0px;width: 450px; word-wrap: break-word;white-space: initial;position:absolute;">
<p style=" overflow: hidden;display: -webkit-box;-webkit-line-clamp: 2;-webkit-box-orient: vertical"><?php echo "<b>"; echo $naamrow['inhoud']; echo "</b>, "; echo $beschrijvingrow['inhoud']; ?></p>
</div>
</div>
</div>
<?php
}
?>

Answer 1

You are process only first record there. You need to repeat for all the return value.
//$row = mysqli_fetch_array($result,MYSQLI_ASSOC);
$naamrows = [];
$beschrijvingrow = [];
while($row = mysqli_fetch_array($result,MYSQLI_ASSOC)) {
$activiteit_id = $row['activiteit_id'];
$naamsql = ("SELECT * FROM `activiteiten` WHERE activiteit_id = {$activiteit_id} AND optie = 'naam'");
$naamresult = mysqli_query($db,$naamsql);
$naamrow = mysqli_fetch_array($naamresult,MYSQLI_ASSOC);
$beschrijvingsql = ("SELECT * FROM `activiteiten` WHERE activiteit_id = {$activiteit_id} AND optie = 'beschrijving'");
$beschrijvingresult = mysqli_query($db,$beschrijvingsql);
$beschrijvingrow = mysqli_fetch_array($beschrijvingresult,MYSQLI_ASSOC);
// Store you data in array to be use later
array_push($naamrows,$naamrow);
array_push($beschrijvingrows,$beschrijvingrow);
}
Now change you template to iterate the array to render all.
<?php foreach ($naamrows as $key => $naamrow) { ?>
<div class="Row" id="main"
style="border: 2px solid #51BAA4;border-radius: 4px; height: 0px; width: 100%;padding: 0em 0 0em 0; margin: 0 auto; display: table;">
<div class="Column" id="plaatje" style="padding: 0em 0 0em 0;">
<left><img src="images/noimage.png" height="100px" width="100px" alt=""></left>
</div>
<div class="Column" id="titel">
<div
style="text-align: left;position:relative;margin:0px;width:10%;height:40px;padding-top: 3px; font-size:125%; display: -webkit-box;-webkit-line-clamp: 1;">
<left>
<h3><?php echo $naamrow['inhoud'];?></h3>
</div>
<div
style="text-align: left; height: 0px;width: 450px; word-wrap: break-word;white-space: initial;position:absolute;">
<p style=" overflow: hidden;display: -webkit-box;-webkit-line-clamp: 2;-webkit-box-orient: vertical">
<?php echo "<b>"; echo $naamrow['inhoud']; echo "</b>, "; echo $beschrijvingrows[$key]['inhoud']; ?></p>
</div>
</div>
</div>
<?php } ?>