致命错误：调用非对象（mysqli）上的成员函数fetch_assoc（）

Question

我刚刚切换了webhoster，并尝试在此处部署我的PHP应用程序。不幸的是，在与此错误消息在登录过程中崩溃了：

Fatal error: Call to a member function fetch_assoc() on a non-object in /home/infobo9/www/xxx/login.php on line 41 

这是代码：

$DB = new DB; 
$mysqli = $DB->getMySQLiObject(); 

$sql = "SELECT ios_worker.username, ios_worker.prename, ios_worker.lastname, ios_worker.partnerid_fk, ios_worker.workerid, ios_worker.password, role.rolename, role.roledescription, ios_partners.accessToAdminPortal FROM ios_partners join ios_worker on ios_worker.partnerid_fk = ios_partners.partnerid join workerrole on workerrole.workerid_fk = ios_worker.workerid join role on workerrole.roleid_fk = role.roleid WHERE ios_worker.username = ? LIMIT 1"; 
$stmt = $mysqli->prepare($sql); 
$result = $stmt->bind_param('s', $username); 
$result = $stmt->execute(); 

var_dump($mysqli); 

if (!$result) { 
    echo $mysqli->error; 
} 

var_dump($result); 

while ($row = $result->fetch_assoc()) { // <-- fails here 
    ... 
} 

输出：

Notice: Undefined index: source in /home/infobo9/www/xxx/login.php on line 23 
object(mysqli)#2 (19) { ["affected_rows"]=> int(-1) ["client_info"]=> string(6) "5.5.40" ["client_version"]=> int(50540) ["connect_errno"]=> int(0) ["connect_error"]=> NULL ["errno"]=> int(0) ["error"]=> string(0) "" ["error_list"]=> array(0) { } ["field_count"]=> int(9) ["host_info"]=> string(33) "xxx via TCP/IP" ["info"]=> NULL ["insert_id"]=> int(0) ["server_info"]=> string(10) "5.5.37-log" ["server_version"]=> int(50537) ["stat"]=> string(52) "Commands out of sync; you can't run this command now" ["sqlstate"]=> string(5) "HY000" ["protocol_version"]=> int(10) ["thread_id"]=> int(82066639) ["warning_count"]=> int(0) } 
bool(true) 
    Fatal error: Call to a member function fetch_assoc() on a non-object in /home/infobo9/www/xxx/login.php on line 41 

所以$mysqli设置，并$result是true。我究竟做错了什么？ 我也复制了查询，它在PHPMyAdmin中没有任何问题。

我知道这个答案被多次询问，但我无法解决这个问题，因为小时，没有答案解决了我的问题。

Answer 1

你需要调用获取的声明不是$结果

while ($row = $stmt->fetch()) { 
    ... 
} 

http://php.net/manual/en/mysqli-stmt.fetch.php

Answer 2

你的结果var_dump是mysqli一个实例，如果它是一个结果集，这将是mysqli_result实例。

你没有得到结果的原因是因为你没有传送结果集。在你的情况下，$result只是返回值$stmt->execute()，这是一个boolean。

您需要做的是在执行语句后获取结果。

$DB = new DB; 
$mysqli = $DB->getMySQLiObject(); 

$sql = "SELECT ios_worker.username, ios_worker.prename, ios_worker.lastname, ios_worker.partnerid_fk, ios_worker.workerid, ios_worker.password, role.rolename, role.roledescription, ios_partners.accessToAdminPortal FROM ios_partners join ios_worker on ios_worker.partnerid_fk = ios_partners.partnerid join workerrole on workerrole.workerid_fk = ios_worker.workerid join role on workerrole.roleid_fk = role.roleid WHERE ios_worker.username = ? LIMIT 1"; 
$stmt = $mysqli->prepare($sql); 
$stmt->bind_param('s', $username); 
$stmt->execute(); 

//put the results in to the $result variable 
$result = $stmt->get_result(); 

while ($row = $result->fetch_assoc()) { 
    ... 
} 

Answer 3

$ stmt-> execute（）; $ result = $ stmt-> get_result（）; while（$ row = $ result-> fetch_assoc（））{...} 如果您收到错误：致命错误：调用未定义的方法mysqli_stmt :: get_result（）：它需要mysqlnd驱动程序...如果它不是没有安装在你的网站空间，你将不得不使用BIND_RESULT & FETCH！ 要通过的链接http://php.net/manual/en/mysqli-stmt.get-result.phphttp://www.php.net/manual/en/mysqli-stmt.bind-result.php http://www.php.net/manual/en/mysqli-stmt.fetch.php