LIKE Clause with php variable

Hi people's,

Can somebody help me with this. I cant figure this out and i bet it is realy simple.
How do you include a variable in a LIKE clause?

Here is what i want to do:

SELECT *
FROM `markers`
WHERE radiusType LIKE '%$radius%'


Where $radius is my php variable. I have tried various brackets and " '

Cheers,
Justin

$result=mysql_query("SELECT *
FROM `markers`
WHERE radiusType LIKE '%$radius%'") or die(mysql_error());

Thanks for your reply,

Although i just need to know how to include the varable as:

SELECT *
FROM `markers`
WHERE radiusType LIKE '%$radius%'

does not work. it does work when i replace $radius with the value.
and the varable is being passed on correctley as if i use:


SELECT *
FROM `markers`
WHERE radiusType = $radius


works fine.
I just need to know how to include the variable.
Cheers,
Justin

Here is my actual query



$query = sprintf("SELECT address, name, lat, lng, ( 6371 * acos( cos( radians('%s') ) * cos( radians( lat ) ) * cos( radians( lng ) - radians('%s') ) + sin( radians('%s') ) * sin( radians( lat ) ) ) ) AS distance FROM markers WHERE radiusType LIKE '%{$radius}%' AND $manufacture = '1' AND `$category` = '1' HAVING distance < '%s' ORDER BY distance LIMIT 0 , 20",
mysql_real_escape_string($center_lat),
mysql_real_escape_string($center_lng),
mysql_real_escape_string($center_lat),
mysql_real_escape_string($radius));

$result = mysql_query($query);

if (!$result) {
die("Invalid query: " . mysql_error());
}


The $radius part does not work

It works when i do this though:


$query = sprintf("SELECT address, name, lat, lng, ( 6371 * acos( cos( radians('%s') ) * cos( radians( lat ) ) * cos( radians( lng ) - radians('%s') ) + sin( radians('%s') ) * sin( radians( lat ) ) ) ) AS distance FROM markers WHERE radiusType = $radius AND $manufacture = '1' AND `$category` = '1' HAVING distance < '%s' ORDER BY distance LIMIT 0 , 20",
mysql_real_escape_string($center_lat),
mysql_real_escape_string($center_lng),
mysql_real_escape_string($center_lat),
mysql_real_escape_string($radius));

$result = mysql_query($query);

if (!$result) {
die("Invalid query: " . mysql_error());
}


It also works if i take out the % signs so it says radiusType LIKE '$radius'
i say it dont like them, but that dont help me.

Cheers,
Justin

You're having trouble because the sprintf() function is interpreting the percent signs as replacement tokens. You just need to escape the percent signs by doubling them up like this:


$query = sprintf("SELECT address, name, lat, lng, ( 6371 * acos( cos( radians('%s') ) * cos( radians( lat ) ) * cos( radians( lng ) - radians('%s') ) + sin( radians('%s') ) * sin( radians( lat ) ) ) ) AS distance
FROM markers WHERE radiusType LIKE '%%$radius%%' AND $manufacture = '1' AND `$category` = '1' HAVING distance < '%s' ORDER BY distance LIMIT 0 , 20",
mysql_real_escape_string($center_lat),
mysql_real_escape_string($center_lng),
mysql_real_escape_string($center_lat),
mysql_real_escape_string($radius));