MySQL Select查詢可在終端中使用。 php准備好的語句失敗

Question

想要：將MySQL表獲取為PHP變量

錯誤： Fatal error: Call to a member function execute() on a non-object in /home/ubuntu/workspace/List Machines/functions.php on line 90

在瀏覽器中呈現頁面時出現此錯誤。

第90行（如錯誤所引用）是這一行： $stmt->execute();

相關代碼（為便於閱讀，略有濃縮）：

<?php
$servername = "0.0.0.0";
$username = "guest";
$password = "password";
$database = "c9";
$connection = new mysqli($servername,$username,$password,$c9);
// Check connection
if ($connection->connect_error)
{
    die("Connection failed: " . $connection->connect_error);
    echo "Problem connecting.";
} 

function retrieveMachineList(){
global $connection;

$query = "SELECT 
machine_id,
manufacturer,
model,
model_year,
type,
warranty_type,
warranty_end_date,
vendor,
purchase_date,
verified_date,
retired_date,
serial
FROM machines";

$stmt=$connection->prepare($query);

$stmt->execute();

$stmt->bind_result($machine_id,
$manufacturer,
$model,
$model_year,
$type,
$warranty_type,
$warranty_type,
$vendor,
$purchase_date,
$verified_date,
$retired_date,
$serial);

$i=0;

while ($stmt->fetch())
{
    $rows[$i]['machine_id'] = $machine_id;
    $rows[$i]['manufacturer'] = $manufacturer;
    $rows[$i]['model'] = $model;
    $rows[$i]['model_year'] = $model_year;
    $rows[$i]['type'] = $type;
    $rows[$i]['warranty_type'] = $warranty_type;
    $rows[$i]['warranty_end_date'] = $warranty_end_date;
    $rows[$i]['vendor'] = $vendor;
    $rows[$i]['purchase_date'] = $purchase_date;
    $rows[$i]['verified_date'] = $verified_date;
    $rows[$i]['retired_date'] = $retired_date;
    $rows[$i]['serial'] = $serial;
    $i++;
}
$stmt->close();
return $rows;
}

$rows = array(retrieveMachineList());
?>

SQL：

mysql> describe machines;
+-------------------+-------------+------+-----+---------+----------------+
| Field             | Type        | Null | Key | Default | Extra          |
+-------------------+-------------+------+-----+---------+----------------+
| machine_id        | int(11)     | NO   | PRI | NULL    | auto_increment |
| manufacturer      | varchar(64) | YES  |     | NULL    |                |
| model             | varchar(64) | YES  |     | NULL    |                |
| model_year        | int(11)     | YES  |     | NULL    |                |
| type              | varchar(32) | YES  |     | NULL    |                |
| warranty_type     | varchar(32) | YES  |     | NULL    |                |
| warranty_end_date | int(11)     | YES  |     | NULL    |                |
| vendor            | varchar(32) | YES  |     | NULL    |                |
| purchase_date     | int(11)     | YES  |     | NULL    |                |
| verified_date     | int(11)     | YES  |     | NULL    |                |
| retired_date      | int(11)     | YES  |     | 0       |                |
| serial            | varchar(64) | YES  |     | NULL    |                |
+-------------------+-------------+------+-----+---------+----------------+
12 rows in set (0.00 sec)

此時，SQL表machines填充了3行虛擬數據。 變量$query描述的SQL語句，后跟一個; 在終端中按預期產生這3行。

我嘗試了各種PHP方法，以在錯誤發生之前和之后的所有位置檢測$ connection和$ stmt變量中的可能錯誤。 全部變成空白。

echo var_dump($stmt);結果echo var_dump($stmt); ： /home/ubuntu/workspace/List Machines/functions.php:90: bool(false)

我還嘗試過向來賓用戶授予完全權限。

我只能說的是，此時似乎有些失敗： $stmt=$connection->prepare($query); 但我不確定失敗的確切原因。 SQL簽出，連接似乎有效。

Answer 1

您從外觀上打錯了字。

更改：

$database = "c9";
$connection = new mysqli($servername,$username,$password,$c9);

至：

$database = "c9";
$connection = new mysqli($servername,$username,$password,$database);

（您已使用$c9代替$database進行$connection ）

Answer 2

瑣碎的問題。 變量名錯誤。

“看起來您在新的mysqli（...語句– shortchng中使用了$ c9而不是$ database”