php – 在longtext字段上准备好的mysqli select语句将返回空

我有一个运行良好的数据库查询功能 – 除了我遇到了 mysqli预处理语句和longtext字段显然已知的问题.即使通过phpMyAdmin运行查询工作正常,longtext字段总是空的,会发生什么.根据 http://www.workinginboxershorts.com/php-mysqli-returns-empty-variables-from-longtext-column,将数据类型切换为文本可以解决问题.然而,在我的情况下,我真的更愿意离开这个领域,因为我可以预见这个额外空间有价值的时间.

我正在使用参数化查询,这显然是问题所在.这是我的功能：

// Bind results to an array
// $stmt = sql query,$out = array to be returned
function stmt_bind_assoc (&$stmt,&$out) {
  $data = mysqli_stmt_result_metadata($stmt);
  $fields = array();
  $out = array();

  $fields[0] = $stmt;
  $count = 1;

  while($field = mysqli_fetch_field($data)) {
    $fields[$count] = &$out[$field->name];
    $count++;
  }    
call_user_func_array('mysqli_stmt_bind_result',$fields);
}

// DB Query
// $query = SQL query,$params = array of parameters,$rs = whether or not a resultset is expected,$newid = whether or not to retrieve the new ID value;
// $onedimensionkey = key required to convert array into simple one dimensional array
function db_query($query,$params,$rs = true,$newid = false,$onedimensionkey = false) {
  $link = mysqli_connect(DB_SERVER,DB_USER,DB_PASS,DB_NAME);
  if (!$link) { 
    print 'Error connecting to MySQL Server. Errorcode: ' . mysqli_connect_error(); 
    exit; 
  }

  // Prepare the query and split the parameters array into bound values
  if ($sql_stmt = mysqli_prepare($link,$query)) {
    if ($params) {
      $types = '';
      $new_params = array();
      $params_ref = array();
      // Split the params array into types string and parameters sub-array
      foreach ($params as $param) {
        $types .= $param['type'];
        $new_params[] = $param['value'];
      }
      // Cycle the new parameters array to make it an array by reference
      foreach ($new_params as $key => $parameter) {
        $params_ref[] = &$new_params[$key];
      }
      call_user_func_array('mysqli_stmt_bind_param',array_merge(array($sql_stmt,$types),$params_ref));
    }
  }
  else {
    print 'Error: ' . mysqli_error($link);
    exit();
  }

  // Execute the query
  mysqli_stmt_execute($sql_stmt);

  // If there are results to retrive,do so
  if ($rs) {
    $results = array();
    $rows = array();
    $row = array();
    stmt_bind_assoc($sql_stmt,$results);
    while (mysqli_stmt_fetch($sql_stmt)) {
      foreach ($results as $key => $value) {
        $row[$key] = $value;
      }
      $rows[] = $row;
    }
    if ($onedimensionkey) {
      $i = 0;
      foreach ($rows as $row) {
        $simplearray[$i] = $row[$onedimensionkey];
        $i++;
      }
      return $simplearray;
    }
    else {
      return $rows;
    }
  }
  // If there are no results but we need the new ID,return it
  elseif ($newid) {
    return mysqli_insert_id($link);
  }

  // Close objects
  mysqli_stmt_close($sql_stmt);
  mysqli_close($link);
}

根据我发布的链接,有一个解决方法涉及完成任务的顺序,但要么我以完全不同的方式处理我的查询,要么我只是不理解重要的事情.

感谢任何能提供帮助的人！

编辑：感谢Corina的回答,我已经解决了这个问题 – 对于遇到此问题的任何其他人,您只需要在mysql_stmt_execute命令之后添加以下内容：

// Execute the query
mysqli_stmt_execute($sql_stmt);

// Store results
mysqli_stmt_store_result($sql_stmt);