 Can't use prepare() when using mysqli prepared statements
By : Iqbal hossain
Date : March 29 2020, 07:55 AM
like below fixes the issue I'm tryign to create an object orientated approach to a project I'm working on for fun, but I'm having trouble getting my head around the idea of a database class. I'm getting the following error. , You need to add the following method to your class: code :
public function prepare($query) {
return $this->connection->prepare($query);
}
public function __call($name, $arguments) {
return call_user_func_array(array($this->connection, $name), $arguments);
}
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Is it really nesseccary to PREPARE statements with PHP's mysqli?
By : nilan
Date : March 29 2020, 07:55 AM
With these it helps I think you should think about the long term benefit. For example, if in the future you are no longer the developer of that project, the habit of using non-prepared statement will be passed down to the next developer and this: code :
$id = $_POST['id'];
$result = db_query('SELECT userid, first_name, last_name FROM user
WHERE userid = ' . db_escape($id) );
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MySQLi multiple prepare statements
By : S.Y
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further I have checked everywhere thoroughly, and have gone through everything possible to find an answer to this. Besides saying "the code doesn't work" which obviously is not enough, I have yet to find anything that will even come close to this. I'm probably going to get downvotes, but let's see how this goes. , Two points:
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PHP mysqli prepare statements own class
By : vamsi chand
Date : March 29 2020, 07:55 AM
I wish this helpful for you You can't use $stmt->bind_param() that way. You would need to do argument unpacking because bind_param requires type and then each value as an argument. code :
final public function newData($sql, $placeholder, $array) {
$stmt = $this->db->prepare($sql);
$stmt->bind_param($placeholder, ...$array);
return $stmt->execute();
}
$sql = "INSERT INTO table (name, data) VALUES (?,?)";
$result = $theDbClass->newData($sql, 'ss', ['foo', 'bar']);
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how to get insert_id using mysqli prepare statements
By : Paul Laskin
Date : March 29 2020, 07:55 AM
Does that help How can I get the insert_id of the last record using mysqli prepare statement, for example following? , $new_id = $this->mysqli->insert_id; (after $stmt->execute())
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code :
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