My PHP Search Function isn't Outputting Data - php

Question

So I am trying to create a way of searching my website. I've created a search bar on my index page. You can find this page here: http://seersvillage.com/v1.2/
My search form looks like this:
<form method="post" action="search.php">
<input type="text" name="search" class="search" value="Skeleton" onFocus="this.value=''">
</form>
and I have a functions.php file attatched and this page is also connected to my mysql database. I have content available to be read / searched for all ready.
Here is my search function on functions.php:
function doSearch() {
$output = '';
if(isset($_POST['search'])) {
$searchq = $_POST['search'];
$searchq = preg_replace ("#[^0-9a-z]#i","",$searchq);
$query = mysql_query("SELECT * FROM entries WHERE name LIKE '%$searchq%' or description LIKE '%$searchq%' or content LIKE '%$searchq%'") or die("Could not search");
$count = mysql_num_rows($query);
if($count == 0) {
$output = 'there was no search results!';
} else {
while($row = mysql_fetch_array($query)) {
$eName = $row['name'];
$eDesc = $row['description'];
$eCont = $row['content'];
$id = $row['id'];
$output .= '<div>'.$eName.' '.$eDesc.'</div>';
}
}
}
}
And the only thing on my search.php (excluding your usual html layout) is as follows:
<?php
include('includes/functions.php');
if(!isset($_POST['search'])) {
header("Location:index.php");
}
?>
and further down in the tags.
<?php print("$output");?>
Now I am pretty new to PHP and MySQL. However I am getting no error on my error.log file, making troubleshooting a little hard for a first timer. Any suggestions? I'm sure it's a very simple mistake, probably just misspelt something, but I just can't see it.

Answer 1

it seems that your php.ini file is set to not display errors. Add these lines of code at the beginning of your code and retry:
<?php
error_reporting(E_ALL);
ini_set("display_errors", 1);
?>

Answer 2

Your doSearch function does not return anything.
return $output;
But $output is only declared within the function. So you'll need to use
print(doSearch());
Either that or declare $output as a global variable, but we don't want to do that :)

Answer 3

function doSearch() {
$output = '';
if(isset($_POST['search'])) {
$searchq = $_POST['search'];
$searchq = preg_replace ("#[^0-9a-z]#i","",$searchq);
$query = mysql_query("SELECT * FROM entries WHERE name LIKE '%$searchq%' or description LIKE '%$searchq%' or content LIKE '%$searchq%'") or die("Could not search");
$count = mysql_num_rows($query);
if($count == 0) {
$output = 'there was no search results!';
} else {
while($row = mysql_fetch_array($query)) {
$eName = $row['name'];
$eDesc = $row['description'];
$eCont = $row['content'];
$id = $row['id'];
$output .= '<div>'.$eName.' '.$eDesc.'</div>';
}
}
//make sure your function returns $output
return $output;
}
Make sure your function returns the output and then echo out the function:
<?php echo doSearch(); ?>
This is because of how PHP variables are scoped.
...then of course we need to add in all the standard provisos ... don't use the mysql_ library it's almost as dead as a Norwegian Blue. If you use mysqli or PDO you can bind the parameters/values to a prepared statement and not only will that improve efficiency but it'll ensure your input is properly sanitised (far better than that odd ad-hoc preg_replace you're using).
You don't want to kill your script (die) when the query fails - that's just a bit weird, handle the error properly.
There are far better ways to do searches in SQL such as FULLTEXT searches.. or if you can, perhaps implement Apache Solr rather than trying to roll your own.