Question

我刚刚切换了我的webhoster并尝试在那里部署我的PHP应用程序。不幸的是，它在登录过程中崩溃并显示以下错误消息：

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

设置$mysqli，$result为true。我究竟做错了什么？我也复制了查询，它在PHPMyAdmin中运行没有任何问题。

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

Answer 1

您需要在语句上调用fetch而不是$ result

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

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

Answer 2

var_dump结果是mysqli的实例，如果它是结果集，则它将是mysqli_result的实例。

您没有获得结果的原因是您没有转移结果集。在您的情况下，$result只是$stmt->execute()的返回值，即boolean。

您需要做的是在执行语句后获得结果。

$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-＆GT;执行（）; $ result = $ stmt-＆gt; get_result（）; while（$ row = $ result-＆gt; fetch_assoc（））{...} 如果你得到错误：致命错误：调用未定义的方法mysqli_stmt :: get_result（）：它需要mysqlnd驱动程序......如果它没有安装在你的网站空间，你将不得不使用BIND_RESULT＆amp;取！链接以完成http://php.net/manual/en/mysqli-stmt.get-result.php http://www.php.net/manual/en/mysqli-stmt.bind-result.php http://www.php.net/manual/en/mysqli-stmt.fetch.php

致命错误：在非对象（mysqli）上调用成员函数fetch_assoc（）

3 个答案: