PDO MySQL and php like statement - php

Question

When I execute this query to the DB:
SELECT * FROM `task` WHERE `date_time_from` like '%0000%'
I get a few results, now I am trying to do the same with PDO and I can not manage to get any results or errors. This is what I have done:
$dbChain = 'mysql:host='.$GLOBALS['dbhost'].';dbname='.$GLOBALS['dbname'];
try{
$dbh = new PDO($dbChain, $GLOBALS['dbuser'], $GLOBALS['dbpassword']);
$sql = "SELECT * FROM task"
. "WHERE date_time_from like CONCAT('%', :dateFrom, '%')";
$a = '0000';
$stmt = $dbh->prepare($sql);
$stmt->bindParam(':dateFrom', $a);
$stmt->execute();
$total = $stmt->rowCount();
echo $total;
while ($row = $stmt->fetch()){
var_dump($row);
}
} catch (Exception $e){
echo 'Error'.$e->getMessage();
}
The result of this is $total = 0. Can anyone tell me what am I doing wrong?
I have also tried this:
$sql = "SELECT * FROM task"
. "WHERE date_time_from like :dateFrom";
$a = "%0000%";
$stmt = $dbh->prepare($sql);
$stmt->bindParam(':dateFrom', $a);
$stmt->execute();
Same result for $total.

Answer 1

bindParam escapes the "%" in the query. It will not work as you expect...
You can, however, use bindValue like so...
$sql = "SELECT * FROM task WHERE date_time_from LIKE ?";
$stmt = $dbh->prepare($sql);
$stmt->bindValue( 1, "%0000%" );
$stmt->execute();
Alternatively, if you want 0 values from a datetime column, you can just do this:
$sql = "SELECT * FROM task WHERE date_time_from = '0000-00-00'";