SQL LIKE % inside array

Question

I know how to perform an SQL LIKE % query for a single value like so:

SELECT * FROM users WHERE name LIKE %tom%;

but how do I do this if th

Answer 1

In this code we can't get double quotes in sql query

$sql = array('0'); // Stop errors when $words is empty

foreach($words as $word){
    $sql[] = 'name LIKE %'.$word.'%'
}

$sql = 'SELECT * FROM users WHERE '.implode(" OR ", $sql);

SELECT * FROM person where name like %"Tom"% OR name like %"Smith"% OR name like %"Larry"%;
// error in double quotes and % is out side of Double quotes .

Or you can also use comma separated value:-

$name = array(Tom, Smith, Larry);

$sql="SELECT * FROM person";

    extract($_POST);
    if ($name!=NULL){
    $exp_arr= array_map('trim', explode(",",$name));
    echo var_export($exp_arr);
    //die;
        foreach($exp_arr as $val){
        $arr = "'%{$val}%'";
        $new_arr[] = 'name like '.$arr;
        }

      $new_arr = implode(" OR ", $new_arr);
      echo $sql.=" where ".$new_arr;
    }
        else {$sql.="";}

Echo sql query like this:-

SELECT * FROM person where name like '%Tom%' OR name like '%Smith%' OR name like '%Larry%';