Php/mysql search whith custom Dispaly - php

Question

I'm trying to make a simple search engine using php & mysql . I know how get mysql result using php. Its Work.
ex : - Mysql database stored this (database table filed "meta")
I am having some difficulty with my PHP/MySQL search. It worked great
a year ago, but for some reason it has just stopped working. I have
tried searching and cannot figure it out. I have it echo my query but
it never adds in the WHERE clause. Any help is much appreciated. Form
Code: PHP Code:
Search word is searching
I Need to dispay search result like this
I have tried searching and cannot figure it out. I have it echo my
query but it never adds in the WHERE clause. Any help is much
appreciated. Form Code: PHP Code:

Answer 1

$query= mysql_query("SELECT meta FROM data WHERE LIKE '%searching%' ");
$count = mysqli_num_rows($query);
if($count >0){
while($row = mysqli_fetch_array($query)) {
echo "$row[id]'>$row[name]</option>";
$count =1;
}
else{
echo "<option value='' selected='selected'></option>";
}
mysqli_close($con);
From the code above here, there's a couple of issues that stand out:
The echo for the option doesn't have an open for , so it's not going to show.
When you're echoing information from the $row variable, you're referencing id and name, but the query is only outputting meta.
The query isn't searching on a field for the where clause, and is therefore failing
The code, from what I can see, should be changed to the following to work (assuming no other surrounding issues):
$query = mysql_query("SELECT meta, id, name FROM data WHERE meta LIKE '%searching%'");
$count = mysqli_num_rows($query);
if ($count > 0)
{
while ($row = $mysqli_fetch_array($query))
{
echo "<option value='" . $row['id'] . "'>" . $row['name'] . "</option>";
$count = 1; // not sure why this needs to be here, but it's in your code so I'm keeping it there
}
}
else
{
echo "<option value='' selected='selected'></option>";
}
mysqli_close($con);