MySQL Select 查询在终端中有效。在 php 准备好的语句中失败

Question

想要：获取 MySQL table 作为 PHP 变量

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

在浏览器中呈现页面时出现此错误。

第 90 行（在错误中引用）是这一行：$stmt->execute();

相关代码（为了阅读方便，略有压缩）：

<?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)

此时，SQL table machines 填充了 3 行虚拟数据。变量 $query 中描述的 SQL 语句后跟 ; 在终端中按预期生成这 3 行。

我已经尝试了各种 PHP 方法来检测 $connection 和 $stmt 变量在错误之前和之后的代码中的所有点可能出现的错误。全部变成空白。

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

我也试过授予来宾用户完全权限。

我只能说此时似乎有什么地方失败了：$stmt=$connection->prepare($query); 但我不确定到底是什么地方失败了。 SQL 检查出来，连接似乎有效。

Answer 1

小问题。变量命名错误。

"Looks like you used $c9 instead of $database in your new mysqli(... statement – shortchng"

Answer 2

你看错字了。

变化：

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

至：

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

（您使用 $c9 而不是 $database 作为 $connection）

MySQL Select 查询在终端中有效。在 php 准备好的语句中失败

MySQL Select query works in terminal. Fails in php prepared statement

php

mysql

database-connection

prepared-statement

cloud9-ide