php - 从SQL PHP获取所有数据

我想在mysql中的表travel_location中获取所有位置。这是我的代码

$select = $this->_db_table->select()->from(travel_location, array('*'));

    $result = $this->_db_table->fetchAll($select);


    if(count($result) == 0) {
        throw new Exception('not found',404);
    }

    while ($row1 = mysql_fetch_array($result)){
        $user_object = new Api_Model_User($row1);

        $count = 1;

        $json = array($json[$count] = array(
        'travel_location_id' => $user_object->travel_location_id,
        'city_id' => $user_object->city_id,
        'user_id' => $user_object->user_id,
        'location_name' => $user_object->location_name,
        'description' => $user_object->description,
        'longitude' => $user_object->longitude,
        'latitude' => $user_object->latitude,
        'created_time' => $user_object->created_time,
        'updated_time' => $user_object->updated_time));

        $count++;
    }

没用我打印$ row1 = mysql_fetch_array（$ result）并返回false，所以我认为这是错误的，因为这行。我该如何解决？

最佳答案

如果使用Zend_Db_Table中的fetchAll，则会得到Zend_Db_Table_Rowset。

尝试这个：

foreach ($result as $row) {
    // $row should be a Zend_Db_Table_Row object
    // you can cast to array
    $rowArray = $row->toArray();
    $user_object = new Api_Model_User($rowArray);
}

阅读有关here和here的更多信息

关于php - 从SQL PHP获取所有数据，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/16700867/