Display an image from the databse to results page - php

Question

Is it possible for a user to input let's say 'Red Car' and it will find the image in the database and in php/html and in the search results it will show the image?
My code for "Search"
<?php
$query = $_GET['query'];
$min_length = 3;
if(strlen($query) >= $min_length){ // if query length is more or equal minimum length then
$query = htmlspecialchars($query);
// changes characters used in html to their equivalents, for example: < to >
$query = mysql_real_escape_string($query);
// makes sure nobody uses SQL injection
$raw_results = mysql_query("SELECT * FROM articles
WHERE (`title` LIKE '%".$query."%') OR (`text` LIKE '%".$query."%')") or die(mysql_error());
// * means that it selects all fields, you can also write: `id`, `title`, `text`
// articles is the name of our table
// '%$query%' is what I'm looking for, % means anything, for example if $query is Hello
// it will match "hello", "Hello man", "gogohello", if you want exact match use `title`='$query'
// or if you want to match just full word so "gogohello" is out use '% $query %' ...OR ... '$query %' ... OR ... '% $query'
if(mysql_num_rows($raw_results) > 0){ // if one or more rows are returned do following
while($results = mysql_fetch_array($raw_results)){
// $results = mysql_fetch_array($raw_results) puts data from database into array, while it's valid it does the loop
echo "<div class='successmate'><h2></h2></a>
</div><br><br><br>";
echo "<div class='search69'><a href='../pages/{$results['page_name']}'><h2>{$results['title']}</h2></a><p>{$results['text']}</‌​p>";
}
}
else{ // if there is no matching rows do following
echo ("<br><br><div class='search1'><h2>No results</h2></br></br>");
}
}
else{ // if query length is less than minimum
echo ("<br><br><div class='search1'><h2>Minnimum Length Is</h2><h2>".$min_length);
}
?>
The images that I want to be found for keywords:
http://puu.sh/cCHv1/9f58d770f3.jpg
These are the names of the images in the DB:
http://puu.sh/cCHwa/a82d2cc7fe.png
Thanks!

Answer 1

Without testing I'd say you have your search function down, to display the results would be a simple matter simply put the below code where you would want to display the image results (This will only be for displaying the images, but to get it to work with other results is a simple rework)
<?php
while($data = $result->fetch_assoc()){
echo '<img src="'.$data['image_path'].'" alt="" title="" />';
}
?>
And then you can simply wrap the images in what ever container if you wish so, another way to do it could be to store the results into an' array and display it with a foreach loop
<?php
while($data = $result->fetch_assoc()){
$imageArray[] = $data['image_path'];
}
/*Code below you can place where ever you want, no need to place it directly after the
above query execute*/
foreach($imageArray as $imgPATH){
echo '<img src="'.$imgPATH.'" alt="" title="" />';
}
?>

Answer 2

The way i understand the problem, i would do it like this
Edit your database table so you also got "image_name", after image where you contain "ser_pic1.jpg"
SELECT * FROM {your_database} WHERE image_name = "Red car"
And post it like
link here