[英]PHP Inserting into mysql issue
I am having trouble when I run the following command: 运行以下命令时遇到麻烦:
$stmt = $this->db->prepare("INSERT INTO checked_in (attendee_id) VALUES (?)");
$stmt->bind_param("i", $id);
$stmt->execute();
$stmt->close();
This is part of a bigger page but this is the part that isn't working. 这是更大页面的一部分,但是这是行不通的。 I first pull the $id from the SQL database so I know I have a connection.
我首先从SQL数据库中提取$ id,所以我知道已经建立了连接。 I just can't seem to get this update part to work.
我似乎无法使此更新部分正常工作。
mysql > CREATE TABLE checked_in (
id int NOT NULL AUTO_INCREMENT PRIMARY KEY,
attendee_id int NOT NULL,
check_in_time TIMESTAMP DEFAULT CURRENT_TIMESTAMP);
This is the code I am using to build the table. 这是我用来构建表的代码。 Am I doing something wrong?
难道我做错了什么? I think my problem is that it can talk to the database but I don't know if I set up the table wrong where it is expecting the id and the timestamp to be manually entered.
我认为我的问题是它可以与数据库对话,但是我不知道我在期望手动输入ID和时间戳的地方是否设置了错误的表。 It works if I am on my sql prompt and use this command
如果我在sql提示符下并使用此命令,它将起作用
mysql> INSERT INTO checked_in (attendee_id) VALUES (1);
Here is the main part of the program (in the process of adapting the tutorial to my own): 这是程序的主要部分(在使教程适应我自己的过程中):
function checkIn() {
// Check for required parameters
if (isset($_POST["phone"])) {
// Put parameters into local variables
$phone = $_POST['phone'];
// Look up code in database
$user_id = 0;
$stmt = $this->db->prepare('SELECT id, first, last, email FROM attendees WHERE phone=?');
$stmt->bind_param("s", $phone);
$stmt->execute();
$stmt->bind_result($id, $first, $last, $email);
while ($stmt->fetch()) {
break;
}
$stmt->close();
// Bail if number doesn't exist
if ($id <= 0) {
sendResponse(400, 'Invalid number');
return false;
}
// Check to see if this device already redeemed
$stmt = $this->db->prepare('SELECT id FROM checked_in WHERE attendee_id=?');
$stmt->bind_param("i", $id);
$stmt->execute();
$stmt->bind_result($checkedInID);
while ($stmt->fetch()) {
break;
}
$stmt->close();
// Bail if number already checked in
if ($checkedInID > 0) {
sendResponse(403, 'Number already checked in');
return false;
}
// Add tracking of redemption
$stmt = $this->db->prepare("INSERT INTO checked_in (attendee_id) VALUES (?)");
$stmt->bind_param("i", $id);
$stmt->execute();
// Return unlock code, encoded with JSON
$result = array(
"checked_in" => "true",
);
sendResponse(200, json_encode($result));
return true;
}
sendResponse(400, 'Invalid request');
return false;
}
}
Ok so I checked my apache error logs and this is what it is saying: 好的,所以我检查了我的apache错误日志,这就是它的意思:
[Fri Jul 25 15:06:41.996107 2014] [:error] [pid 16824] [client 127.0.0.1:52699] PHP Notice: Undefined variable: db in /var/www/html/index.php on line 129
[Fri Jul 25 15:06:41.996171 2014] [:error] [pid 16824] [client 127.0.0.1:52699] PHP Fatal error: Call to a member function prepare() on a non-object in /var/www/html/index.php on line 129
If you bind your parameters while using a question mark in your queries, then you shouldn't use i
as parameter. 如果在查询中使用问号时绑定了参数,则不应使用
i
作为参数。 Use a numeric parameter instead. 请改用数字参数。
Try 尝试
$stmt->bind_param(1, $id);
For a prepared statement using named placeholders, this will be a parameter name of the form
:name
.对于使用命名占位符的准备好的语句,这将是形式为
:name
的参数:name
。 For a prepared statement using question mark placeholders, this will be the 1-indexed position of the parameter.对于使用问号占位符的准备好的语句,这将是参数的1索引位置。
That said, please describe what $id
is assigned to in this function. 也就是说,请描述在此函数中分配的
$id
。
Try this: 尝试这个:
$stmt = $this->db->prepare("INSERT INTO checked_in (attendee_id) VALUES (?)");
$stmt->execute( array( $id ) );
should work. 应该管用。
If you want to check the error, you should do: 如果要检查错误,则应该执行以下操作:
$stmt = $this->db->prepare("INSERT INTO checked_in (attendee_id) VALUES (?)");
$check = $stmt->execute( array( $id ) );
if ( $check ) {
echo 'Nice work!';
} else {
$error = $stmt->errorInfo();
echo $error[2];
}
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