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将布尔表达式解析为PHP中的MySql查询

将布尔表达式解析为PHP中的MySql查询

简单的查询生成器,假设PDO

        ## for simple tokenisation, the terms are separated by space here.
        ## ###############################################################
$string = "( C AND kernel ) OR ( C++ AND UML )";

function emit_term( $tag ) {
$res = " EXISTS (
                SELECT *
                FROM skill_usage su
                JOIN skill_names sn ON sn.skill_id = su.skill_id
                WHERE su.Job_id = j.job_id
                AND sn.skillname = :" . $tag . ")\n";
return $res;
}


$fixed_part ="
SELECT job_id, job_name
 FROM jobs j
 WHERE 1=1
 AND \n" ;


# $tokens = explode( ' ' , $string ); #splits on any single space
$tokens = preg_split( '/[\s]+/' , $string ); # accepts multiple whitespace
# print_r ( $tokens );

$query = $fixed_part;

$args = array();
$num = 1;
foreach ( $tokens as $tok ) {
        switch ($tok) {
        case '':  # skip empty tokens
        case ';':  # No, you should not!
        case '"':
        case "'":
        case ';':  break;
        case '(': $query .= '('; break;
        case ')': $query .= ')'; break;
        case '&':
        case 'AND': $query .= ' AND '; break;
        case '|':
        case 'OR': $query .= ' OR '; break;
        case '!':
        case 'NOT': $query .= ' NOT '; break;
        default:
                $tag = '_q' . $num ;
                $query .= emit_term ( $tag );
                $args[$tag] = $tok;
                $num += 1;
                 break;
                }
        }
$query .= ";\n\n";

echo "Query + parameters (for PDO):\n" ;
echo $query;
print_r ( $args) ;

输出

SELECT job_id, job_name
 FROM jobs j
 WHERE 1=1
 AND 
( EXISTS (
        SELECT *
        FROM skill_usage su
        JOIN skill_names sn ON sn.skill_id = su.skill_id
        WHERE su.Job_id = j.job_id
        AND sn.skillname = :_q1)
 AND  EXISTS (
        SELECT *
        FROM skill_usage su
        JOIN skill_names sn ON sn.skill_id = su.skill_id
        WHERE su.Job_id = j.job_id
        AND sn.skillname = :_q2)
) OR ( EXISTS (
        SELECT *
        FROM skill_usage su
        JOIN skill_names sn ON sn.skill_id = su.skill_id
        WHERE su.Job_id = j.job_id
        AND sn.skillname = :_q3)
 AND  EXISTS (
        SELECT *
        FROM skill_usage su
        JOIN skill_names sn ON sn.skill_id = su.skill_id
        WHERE su.Job_id = j.job_id
        AND sn.skillname = :_q4)
);

Array
(
    [_q1] => C
    [_q2] => kernel
    [_q3] => C++
    [_q4] => UML
)
php 2022/1/1 18:52:03 有322人围观

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