mysql_query返回value和null，但应该只有value

Question

通过从php以json表示法返回mysql_query ，我得到了一些空值。 有些字段（如bookid，名称，作者，期限）是返回值，而其他字段则不是！ 但是我在DBVisualizer中输入了相同的查询，但我只得到了价值……这是什么问题？

    echo $myuserid . "idcheck";
    $q1 = "select u.address as address, u.postcode as postcode, u.town as town, u.telnumber, u.mail, b.bookid as bookid, b.name as name, b.author as author, s.duration as duration from status s, book b, user u where s.bookid = b.bookid and s.borroweruserid = u.userid and s.borroweruserid = $myuserid";
    echo $q1;
    $sth = mysql_query($q1) or die(mysql_error());

        while($rowla = mysql_fetch_assoc($sth)) {

     $rows[] = array(
      'address' => $rowla['address'],
      'postcode' => $rowla['postcode'],
      'town' => $rowla['town'],
      'telnumber' => $rowla['telnumber'],
      'mail' => $rowla['mail'],
      'bookid' => $rowla['bookid'],
      'name' => $rowla['name'],
      'author' => $rowla['author'],
      'duration' => $rowla['duration']);
     // 'pic' => base64_encode($row['pic']  
    //$rows[] = $r;
       }
       $j1 = json_encode($rows);

更新：当将error_reporing放到php脚本时，什么都行不通（不执行查询），并且在while之后添加print_r（$ rowla）仍然得到此反馈（第二次查询的操作相同-只是有点不同）： http：// s24。 postimg.org/4wpu29zwl/Screen_Shot_2014_05_22_at_08_45_02.png

Answer 1

你想念你的SQL。 我认为。

$q1 = "SELECT u.address AS address, u.postcode AS postcode, u.town AS town, u.telnumber, u.mail, b.bookid AS bookid, b.name AS name, b.author AS author, s.duration AS duration from status s, book b, user u WHERE s.bookid = b.bookid AND s.borroweruserid = u.userid AND s.borroweruserid = '".$myuserid."'";