Question

我有城市，生日和用户表。现在，我想要做的是通过数据库查询和过滤年龄和生活城市的范围（在所选地区的城市列表中 - 这就是我检查是否计数（$ city []））＆gt ; = 1），并根据uid / entety_id精细地提取用户名。首先，我检查用户是否匹配年龄范围，然后检查他是否住在选定列表中的一个城市。如果没有选择城市，我只想过滤年龄范围。

我不知道问题是什么，但据我所知，有这个循环，我注意到根本没有执行：

for ($d = 1; $d < (count($the_resaults_array)); $d++)

这是代码：

function toodate_query_db($loest_age, $highest_age, $city){
    $sql_user_name = '';

    if((count($city[0]))>=1){
        for ($i = 0; $i < count($city); $i++) {
            $city_item_for_check = $city[$i];

            $sql = "SELECT field_data_field_birth_date.entity_id
                        FROM field_data_field_birth_date, field_data_field_city
                        WHERE(
                        (field_birth_date_value >= DATE_ADD(NOW(), INTERVAL -{$highest_age} YEAR))
                        AND 
                        (field_birth_date_value <= DATE_ADD(NOW(), INTERVAL -{$loest_age} YEAR))
                        AND 
                        (field_city_tid = {$city_item_for_check})
                        AND
                        (field_data_field_birth_date.entity_id = field_data_field_city.entity_id))";
            if(db_query($sql)->fetchColumn()){
            $the_resaults_array[$i] = db_query($sql)->fetchAll();
            }
        }

        for ($d = 1; $d < (count($the_resaults_array)); $d++) {

            $user_id = $the_resaults_array[$d];

            $sql_user_name .= '<p><strong>'.
            db_query("SELECT name FROM {users} WHERE uid = {$user_id}")->fetchColumn()
            .'</p></strong>';           
        }
    } else{
        $sql = "SELECT field_data_field_birth_date.entity_id
            FROM field_data_field_birth_date 
            WHERE
                field_birth_date_value >= DATE_ADD(NOW(), INTERVAL -{$highest_age} YEAR) 
                AND field_birth_date_value <= DATE_ADD(NOW(), INTERVAL -{$loest_age} YEAR)";


        foreach (db_query($sql) as $result) {
            $entity_id = $result->entity_id;

            $sql_user_id = db_query("SELECT uid FROM {profile} WHERE pid = {$entity_id}")->fetchColumn();

            $sql_user_name .= '<p><strong>'.
            db_query("SELECT name FROM {users} WHERE uid = {$sql_user_id}")->fetchColumn()
            .'</p></strong>';

        }}

        return $sql_user_name;
}

通过数据库查询并按用户ID拉出用户

0 个答案: