如何獲取mysqli OOP中affected_rows的數量？ Mysqli OOP affected_rows返回-1

Question

什麼是執行方式來獲取mysqli OOP中受影響的行數。我正在使用msqli OOP創建crud類。我得到int -1。後續代碼var_dump（$語句）的

$query = "SELECT * FROM `sk_courses`"; 
    $stmt = $this->_mysqli->prepare($query); 
    $stmt->execute(); 
    $stmt->affected_rows ; 

    var_dump($stmt->affected_rows); // output is int -1 

輸出是：

object(mysqli_stmt)[7] 
    public 'affected_rows' => null 
    public 'insert_id' => null 
    public 'num_rows' => null 
    public 'param_count' => null 
    public 'field_count' => null 
    public 'errno' => null 
    public 'error' => null 
    public 'error_list' => null 
    public 'sqlstate' => null 
    public 'id' => null 

Answer 1

這是理解mysqli_stmt文檔，其中包括下面的每個功能頁面的問題。

A SELECT查詢不會影響任何行。

查看手冊[php docs for mysqli affected-rows]。在返回值下：

-1 indicates that the query returned an error. 

但是，這實際上是一個混合的問題。

要獲取的行數，請嘗試：

$query = "SELECT * FROM `sk_courses`";  // or 
$query = "SELECT lastName, firstName, ... FROM `sk_courses`"; 

$stmt = $mysqli->prepare($query); 
$stmt->execute(); 
$stmt->store_result();      // without this line, num_rows = 0 
print $stmt->num_rows; 

然後，如果有興趣在結果集中，添加：

// bind result variables. next statement 'binds' them in the order of the select query 
$stmt->bind_result($last, $first, ....); // must have variable for EACH column 

while ($stmt->fetch()) { 
    printf ("%s (%s)\n", $first, $last); 
}