MySQLi 准备好的语句查询结果在一个数组中

Question

试图将我的旧 mysql 查询转换为 mysqli 准备好的语句。除了一件事，我已经弄清楚了一切。如何获取存储为数组的查询结果？我以前是这样的：

$sql = "SELECT * FROM Users";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result) {
 // do stuff
}

现在我有以下代码。在这种情况下，我的数组是一条记录，所以我不需要遍历它，但我想将它作为一个数组保存，以便我可以引用它的字段名称。另外，我还有其他查询会返回多条记录，所以我需要迭代。

$sql = "SELECT * FROM Users 
    WHERE (LOWER(first_name)=LOWER(?) && LOWER(last_name)=LOWER(?))";
$stmt =  mysqli_stmt_init($link);
$this_user;

if (mysqli_stmt_prepare($stmt, $sql)) {

    /* Bind the input parameters to the query */
    mysqli_stmt_bind_param($stmt, 'ss', $first_name, $last_name);

    /* execute query, store results in an array */
    mysqli_stmt_execute($stmt);
    $result = mysqli_fetch_array($stmt);

    if (mysqli_num_rows($result) == 0) {
        mysqli_stmt_close($stmt);
        mysqli_close($link);
        $tag_result = "failure";
        $tag_message = "No matching user found";
        echo encodeJSONObj($tag_result, $tag_message);
        die();
    }

    if (mysqli_num_rows($result) > 1) {
        mysqli_close($link);
        $tag_result = "failure";
        $tag_message = "Multiple records found for this user.";
        echo encodeJSONObj($tag_result, $tag_message);
        die();
    }

    $this_user = mysqli_fetch_array($result);

    /* close statement */
    mysqli_stmt_close($stmt);
}

$id                     = $this_user['id'];
$first_name             = $this_user['first_name'];
$last_name              = $this_user['last_name'];
// and so on...

谁能告诉我我做错了什么？谢谢！

编辑：非常感谢 Phil，我修改了我的代码。但是，即使我的输入参数应该正好返回 1 行，我似乎仍然返回 0 行。这是我所拥有的：

$sql = "SELECT id, first_name, last_name, group_id, email, cell
            FROM Users 
            WHERE (first_name=? && last_name=?)";
$stmt =  mysqli_stmt_init($link);

if (mysqli_stmt_prepare($stmt, $sql)) {

    /* Bind the input parameters to the query */
    mysqli_stmt_bind_param($stmt, 'ss', $first_name, $last_name);

    /* execute query, bind result, and fetch value */
    mysqli_stmt_execute($stmt);
    mysqli_stmt_bind_result($stmt, $id, $first_name, $last_name,  $group_id, $email, $cell);
    mysqli_stmt_fetch($stmt);

    if (mysqli_stmt_num_rows($stmt) == 0) {
        mysqli_stmt_close($stmt);
        mysqli_close($link);
        echo "No results returned";
        die();
    }

    ...
}

当它应该找到 1 行并跳过该块时，它总是输出 No results returned。我已经盯着这个看了很长时间，但我就是看不出我做错了什么。

Answer 1

您的脚本包含许多错误（如上面的 cmets 中所述）。这是一个简单的分步...

1. 准备语句并绑定参数

   $stmt = $link->prepare($sql);
   if (!$stmt) {
       throw new Exception($link->error, $link->errno);
   }
   
   // you can error check this too but it rarely goes wrong
   $stmt->bind_param('ss', $first_name, $last_name);
   

2. 执行语句并存储结果

   if (!$stmt->execute()) {
       throw new Exception($stmt->error, $stmt->errno);
   }
   $stmt->store_result();
   

3. 您的行数检查$stmt->num_rows...

   if ($stmt->num_rows == 0) {
       // ... 
   }
   if ($stmt->num_rows > 1) {
       // ...
   }
   

4. 绑定并获取结果

   // This relies on the SELECT column ordering.
   // You should probably change your SELECT statement to
   // SELECT id, first_name, last_name FROM Users...
   $stmt->bind_result($id, $first_name, $last_name);
   $stmt->fetch();
   $stmt->close();
   $link->close();
   

   如果您想获取单个结果行作为关联数组，请尝试使用此方法

   $result = $stmt->get_result(); // note - this requires the mysqlnd driver
   $this_user = $result->fetch_array(MYSQLI_ASSOC);
   $result->free();
   $stmt->close();
   $link->close();