Question

几年来，我一直很高兴使用PDO debugger并高度推荐。

我没有找到更好的方法来帮助我调试带有绑定参数的PDO SQL语句。它将向我展示在替换绑定参数后生成的SQL语句，这是一件很好的事情。

我发布以下代码。我的问题是，当我使用SELECT引入LIKE %gail.com查询时，$ agencyEmailDomain ==＆＃34; gmail.com＆＃34;和$expandedSqlCommand = PdoDebugger::show($sqlText, $parameters);

扩展到

SELECT count(*) AS count FROM recruiters WHERE recruiter_email LIKE "%'Gmail.com'"

请注意域名周围的单引号？

我能做些什么，比如逃避％号？

我的代码是

$sqlText = 'SELECT count(*) AS count FROM recruiters 
            WHERE recruiter_email LIKE "%:agencyEmailDomain"';

$sqlParameters =  array('agencyEmailDomain' => $agencyEmailDomain);

和PDO-debugger的代码是

/**
 * Class PdoDebugger
 *
 * Emulates the PDO SQL statement in an extremely simple kind of way
 */
class PdoDebugger
{
    /**
     * Returns the emulated SQL string
     *
     * @param $raw_sql
     * @param $parameters
     * @return mixed
     */
    static public function show($raw_sql, $parameters)
    {
        $keys = array();
        $values = array();

        /*
         * Get longest keys first, so that the regex replacement doesn't
         * cut markers (ex : replace ":username" with "'joe'name"
         * if we have a param name :user )
         */
        $isNamedMarkers = false;

        if (count($parameters) && is_string(key($parameters))) 
        {
            uksort($parameters, function($k1, $k2) 
            {
                return strlen($k2) - strlen($k1);
            });

            $isNamedMarkers = true;
        }

        foreach ($parameters as $key => $value) 
        {

            // check if named parameters (':param') or anonymous parameters ('?') are used
            if (is_string($key)) 
            {
                $keys[] = '/:'.ltrim($key, ':').'/';
            } 
            else 
            {
                $keys[] = '/[?]/';
            }

            // bring parameter into human-readable format
            if (is_string($value)) 
            {
                $values[] = "'" . addslashes($value) . "'";
            } 
            elseif(is_int($value)) 
            {
                $values[] = strval($value);
            } 
            elseif (is_float($value)) 
            {
                $values[] = strval($value);
            } 
            elseif (is_array($value)) 
            {
                $values[] = implode(',', $value);
            } 
            elseif (is_null($value)) 
            {
                $values[] = 'NULL';
            }
        }

        if ($isNamedMarkers) 
        {
            return preg_replace($keys, $values, $raw_sql);
        } 
        else 
        {
            return preg_replace($keys, $values, $raw_sql, 1, $count);
        }
    }
}

如何更改以正确处理LIKE "%:agencyEmailDomain"？

使用SELECT LIKE％<string>的PHP正则表达式

0 个答案: