Question

我正在尝试获取我写的这个函数来从MySQL表中返回一些数据。这是我的功能。

function getCompInfoIDS($id) {
    global $mysqli;
    $query = "
Select
  computers.asset,
  computers.serial,
  rooms.building_id,
  rooms.id,
  computers.assigned_person,
  computers.computer_name,
  computers.sticker,
  operating_systems.manufacturer_id,
  operating_systems.id As id1,
  computers.memory,
  computers.hard_drive,
  computers.department,
  computers.year_purchased,
  computers.po_cs_ca,
  computers.mac_address,
  computers.group_id,
  models.manufacturer_id As manufacturer_id1,
  models.id As id2,
  computers.type,
  computers.monitor_size
From
  computers Inner Join
  rooms On computers.room_id = rooms.id Inner Join
  operating_systems On computers.os_id = operating_systems.id Inner Join
  models On computers.model = models.id
 Where
  computers.id=?";
    $stmt = $mysqli -> prepare($query);
    $stmt -> bind_param('i',$id);
    $stmt -> execute();
    $stmt -> bind_result($asset, $serial, $building_id, $room_id, $assigned_person, $computer_name, $sticker, $os_type_id, $os_id, $memory, $hard_drive, $department, $year_purchased, $po_cs_ca, $mac_address, $group_id, $model_type_id, $model_id, $comp_type, $monitor_size, $date_modified);
    while($stmt -> fetch()){
        $computer_info = array($asset, $serial, $building_id, $room_id, $assigned_person, $computer_name, $sticker, $os_type_id, $os_id, $memory, $hard_drive, $department, $year_purchased, $po_cs_ca, $mac_address, $group_id, $model_type_id, $model_id, $comp_type, $monitor_size, $date_modified);         
    }

    return $computer_info;

查询确实有效，我在phpmyadmin中使用多个ID对其进行了测试。

我一直在关注PHP manual和this site，一步一步地看看我做错了什么。我已经在我的代码中的各个不同位置完成echo "test";以查看函数失败的位置，并且我在函数中找不到单点故障，它只是没有用结果填充数组。

我在填充数组之前尝试做echo $asset;并且没有显示任何内容，所以我不相信任何数据实际上都被放入数组变量中。

Answer 1

您必须添加$stmt->store_result();，因此您的代码如下：

$stmt = $mysqli -> prepare($query);
$stmt -> bind_param('i',$id);
$stmt -> execute();
$stmt->store_result();
$stmt -> bind_result($asset, $serial, $building_id, $room_id, $assigned_person, $computer_name, $sticker, $os_type_id, $os_id, $memory, $hard_drive, $department, $year_purchased, $po_cs_ca, $mac_address, $group_id, $model_type_id, $model_id, $comp_type, $monitor_size, $date_modified);
while($stmt -> fetch()){
    $computer_info = array($asset, $serial, $building_id, $room_id, $assigned_person, $computer_name, $sticker, $os_type_id, $os_id, $memory, $hard_drive, $department, $year_purchased, $po_cs_ca, $mac_address, $group_id, $model_type_id, $model_id, $comp_type, $monitor_size, $date_modified);         
}

Answer 2

您的代码应该像这样给出错误：

Warning: mysqli_stmt::bind_result(): Number of bind variables doesn't match number of fields in prepared statement in C:\xampp\htdocs\... on line x

查询中所选列的数量与绑定结果的数量不同。您在SELECT查询中只选择了20列，但您的bind_result()中有21个变量。

您的查询中必须缺少一列。显然是你的$date_modified变量。正确填写SELECT查询，以便绑定$date_modified变量。

MySQLI Bind_param查询不返回结果

2 个答案: