致命錯誤：調用非對象（mysqli）上的成員函數fetch_assoc（）

Question

我剛剛切換了webhoster，並嘗試在此處部署我的PHP應用程序。不幸的是，在與此錯誤消息在登錄過程中崩潰了：

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

這是代碼：

$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 
    ... 
} 

輸出：

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$mysqli設置，並$result是true。我究竟做錯了什麼？ 我也複製了查詢，它在PHPMyAdmin中沒有任何問題。

我知道這個答案被多次詢問，但我無法解決這個問題，因爲小時，沒有答案解決了我的問題。

Answer 1

你需要調用獲取的聲明不是$結果

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

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

Answer 2

你的結果var_dump是mysqli一個實例，如果它是一個結果集，這將是mysqli_result實例。

你沒有得到結果的原因是因爲你沒有傳送結果集。在你的情況下，$result只是返回值$stmt->execute()，這是一個boolean。

您需要做的是在執行語句後獲取結果。

$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（））{...} 如果您收到錯誤：致命錯誤：調用未定義的方法mysqli_stmt :: get_result（）：它需要mysqlnd驅動程序...如果它不是沒有安裝在你的網站空間，你將不得不使用BIND_RESULT & FETCH！ 要通過的鏈接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