如何编写查询以选择类似的标题？

Question

I would like to select those movies which have similar titles. I found this, but this way it dosn't work, it gives nothing. I would like to give toy story 2, toy story 3 and others with similar title like toy soldielrs, etc.

$title = "Toy Story";
$query = mysql_query("SELECT title, year, poster, LEVENSHTEIN_RATIO( ".$title.", title ) as textDiff FROM movies HAVING textDiff > 60");

I can compare strings in PHP with this function:

static public function string_compare($str_a, $str_b) 
{
    $length = strlen($str_a);
    $length_b = strlen($str_b);

    $i = 0;
    $segmentcount = 0;
    $segmentsinfo = array();
    $segment = '';
    while ($i < $length) 
    {
        $char = substr($str_a, $i, 1);
        if (strpos($str_b, $char) !== FALSE) 
        {               
            $segment = $segment.$char;
            if (strpos($str_b, $segment) !== FALSE) 
            {
                $segmentpos_a = $i - strlen($segment) + 1;
                $segmentpos_b = strpos($str_b, $segment);
                $positiondiff = abs($segmentpos_a - $segmentpos_b);
                $posfactor = ($length - $positiondiff) / $length_b;
                $lengthfactor = strlen($segment)/$length;
                $segmentsinfo[$segmentcount] = array( 'segment' => $segment, 'score' => ($posfactor * $lengthfactor));
            } 
            else 
            {
                $segment = '';
                $i--;
                $segmentcount++;
            } 
        } 
        else 
        {
            $segment = '';
            $segmentcount++;
        }
        $i++;
    }   

    // PHP 5.3 lambda in array_map      
    $totalscore = array_sum(array_map(function($v) { return $v['score'];  }, $segmentsinfo));
    return $totalscore;     
}

But how can I compare in a SELECT query or any other way?

Answer 1

You can use like queries for that: Following example will return all the records from table customer for which customer name ends with kh

select * from customer where name like '%kh'

Following example will return all the records from table customer for which customer name start with kh

select * from customer where name like 'kh%'

Following example will return all the records from table customer for which the middle world of customer name is kh

select * from customer where name like 'kh%'

if you want more specific record then add some and/or condition in your query

I recommend you to read this http://dev.mysql.com/doc/refman/5.7/en/string-comparison-functions.html#operator_like