MySQL more columns then in php query - php

I changed to another database with more columns. Nut now my register page doesn't work anymore. The tables all have default settings.
How can I let the query put all the data in the columns and use the defaults for other columns?
This is my query:
mysql_query("
INSERT INTO `users`
(`username`, `password`, `mail`, 'account_created', 'ip_last', 'ip_reg')
VALUES(
'".$naam."', '".$wachtwoord."', '".$email."',
'".$timestamp."', '".$ip."', '".$ip."'
)
");
It worked before, but now on this new database it doesn't work anymore. I didn't change my php version or something.

You can use variables in query strings without quotes.
By the way you should think about more secure -
PDO? What is this magic system
PDO Version:
$query = $db->prepare("INSERT INTO users(username, password, mail, account_created, ip_last, ip_reg) VALUES (?, ?, ?, ?, ?, ?)");
$query->execute(array($naam, $wachtwoord, $email, $timestamp, $ip, $ip));
Trash Version:
mysql_query("INSERT INTO users(username, password, mail, account_created, ip_last, ip_reg) VALUES ($naam, $wachtwoord, $email, $timestamp, $ip, $ip)");

Know the difference between back ticks and single quotes
mysql_query("
INSERT INTO `users` (`username`, `password`, `mail`, `account_created`, `ip_last`, `ip_reg`) VALUES('".$naam."', '".$wachtwoord."', '".$email."', '".$timestamp."', '".$ip."', '".$ip."')");

Related

Can a prepared statement hold multiple queries in php

I am trying to protect my queries from SQL injections, recently. I have started turning the strings I used to make the queries into statements, however, some of the strings I made need to make multiple queries simultaneously, because one insert's id will be added to the next one as a foreign key, which I'll get by using the LAST_INSERT_ID(), and I need them to be executed one after another because of it.
Can a statement hold multiple queries simultaneously and be executed at once?
Here's what the code was before, by the by.
$sql = "INSERT INTO `user_info`(`first_name`, `last_name`, `phone`, `cpf`)
VALUES ('{$firstName}', '{$lastName}', '{$phone}', '{$cpf}');";
$sql .= "SELECT LAST_INSERT_ID() INTO #mysql_variable_here;";
$sql .= "INSERT INTO `{$table}`(`email`, `password`, `active`,`user_info_id`, `created`, `role_id`" . $restaurantInsert . ")
VALUES ('{$email}','{$password}', 1, #mysql_variable_here, '{$created}', {$role}" . $restaurantValue . " );";
$sql .= "INSERT INTO `address`(number, street, city, state, zip, district, country, created, user_info_id)
VALUES ('{$number}', '{$street}', '{$city}', '{$stateCode}', '{$zip}', '{$district}', 'BR', '{$created}', #mysql_variable_here);";
$result = $conn->multi_query($sql);```
You can't execute multiple statements in a prepared query:
SQL syntax for prepared statements does not support multi-statements
(that is, multiple statements within a single string separated by ;
characters)
so you will need to prepare and execute each of the queries separately, using mysqli_stmt::insert_id to get the appropriate id value for the second and third queries:
$sql = "INSERT INTO `user_info`(`first_name`, `last_name`, `phone`, `cpf`)
VALUES (?, ?, ?, ?)";
$stmt = $conn->prepare($sql);
$stmt->bind_param('ssss', $firstName, $lastName, $phone, $cpf);
$stmt->execute();
$insert_id = $stmt->insert_id;
$stmt->close();
$sql = "INSERT INTO `{$table}`(`email`, `password`, `active`,`user_info_id`, `created`, `role_id`" . $restaurantInsert . ")
VALUES (?, ?, ?, ?, ?, ?, ?)";
$stmt = $conn->prepare($sql);
$stmt->bind_param('ssiisss', $email, $password, 1, $insert_id, $created, $role, $restaurantValue);
$stmt->execute();
$stmt->close();
$sql = "INSERT INTO `address`(number, street, city, state, zip, district, country, created, user_info_id)
VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?);";
$stmt = $conn->prepare($sql);
$country = 'BR';
$stmt->bind_param('sssssssi', $number, $street, $city, $stateCode, $zip, $district, $country, $created, $insert_id);
$stmt->execute();
$stmt->close();
Note I'm not 100% certain what you're trying to achieve with role_id" . $restaurantInsert . ", you might need to edit the second query appropriately to use that.

insert data by GET

I want insert data by GET in my sql but I can not insert data
<?php
include("config.php");
$f=$_GET["first_name"];
$l=$_GET["last_name"];
$e=$_GET["email"];
$m=$_GET["mobile"];
$b=$_GET["birthday"];
$g=$_GET["gender"];
$insert="INSERT INTO user ( `first_name`, `last_name`, `email`, `mobile`, `birthday`, `gender`)
VALUES ('$f', '$l', '$e', '$m', '$b', '$g')";
mysqli_query($insert);
?>
I try insert data by this link :
http://localhost:8888/restfull/insert.php?f=hayoo
It's been a long time since I have used mysqli the code below should most likely run though. As others have mentioned never bind unsanitized data (Even if you think you trust the data it's safe to use prepared statements still).
<?php
//Create you db connection
$conn = new mysqli('server', 'user', 'password', 'databasename');
//Create insert statement. Never concat un-sanitized data in statements
$insert="INSERT INTO user ( `first_name`, `last_name`, `email`, `mobile`, `birthday`, `gender`)
VALUES (?, ?, ?, ?, ?, ?)";
$stmt = $conn->prepare($sql);
//Values corespond to ? except the first param which represents format of expected data. "s" stands for string
$stmt->bind_param(
'ssssss',
$_GET["first_name"],
$_GET["last_name"],
$_GET["email"],
$_GET["mobile"],
$_GET["birthday"],
$_GET["gender"]
);
$stmt->execute();
Your url would look like this:
http://localhost:8888/restfull/insert.php?first_name=john&last_name=Doe&email=test#test.com&mobile=0&birthday=May&gender=male
Make sure if you are putting the url above in some type of form you correctly url encode values (I notice many of the values you are collecting will like require it slashes etc).

Why doesn't my SQL insert into work?

I know questions like this exist all over, but I can't see what's wrong with my code. The DB works and I run an update query just fine earlier on in the code.
Query 1:
mysql_query("INSERT INTO `login_history` (`memberid`, `ip`, `host`, `location`, `status`, `date`) VALUES ('".$login."', '".$ip."', '".$details->hostname."', '".$loc."', 'success', NOW()");
Query 2:
mysql_query("INSERT INTO `login_history` (`memberid`, `ip`, `host`, `location`, `status`, `date`) VALUES ('".$login."', '".$ip."', '".$details->hostname."', '".$loc."', 'failure', NOW()");
Here is an echo of the string as requested:
INSERT INTO `login_history` (`memberid`, `ip`, `host`, `location`, `status`, `date`) VALUES ('AAL', '**.60.**.**', 'c-174-**-**-**.hsd1.**.comcast.net', 'Town, US', 'success', NOW()
Looks like you are missing the final closing ). Also use prepared statements. Much cleaner and safer. Here is a quick example of what a Prepared Statement would look like (adapted from here) (you wold also need to make other changes to your PHP script to start using them)
$stmt = $mysqli->prepare("INSERT INTO `login_history` (`memberid`, `ip`, `host`, `location`, `status`, `date`) VALUES (?, ?, ?, ?, ?, NOW())")
$stmt->bind_param('sssss', $login, $ip, $details->hostname, $loc, 'failure');
$stmt->execute()

Insert auto-incremented ID using prepared statements

When inserting a new record into a table with an auto-incrementing ID column, it is normally enough to give the ID field the value NULL or omit it from the INSERT query, as explained at How to insert new auto increment ID
INSERT INTO `database`.`table` (`id`, `user`, `result`) VALUES (NULL, 'Alice', 'green')");
or
INSERT INTO `database`.`table` (`user`, `result`) VALUES ('Alice', 'green')");
My question is - how do you do the same thing when using prepared statements. I have tried the following, using NULL:
$stmt = $db->prepare("INSERT INTO `db` (id, name, password, text) VALUES (NULL, ?, ?, ?)");
$stmt->bind_param('sss', $name, $password, $text);
$stmt->execute();
and the fowllowing, omitting the ID field:
$stmt = $db->prepare("INSERT INTO `test_db` (name, password, text) VALUES (?, ?, ?)");
$stmt->bind_param('sss', $name, $password, $text);
$stmt->execute();
When I run this I get nothing inserted and no error message in the browser. I think it is because it is trying to insert a duplicate value for the ID field (stackoverflow.com/questions/12179770/…) - but why it should do that when this seems equivalent to the non-prepared-statement way of inserting data, and then give no message, I'm not sure.
Any ideas most welcome!

php prepared statement fails.

I'm guessing I'm missing something, but I can't seem to get this statement to work. When I load it into the page I get the white screen of death.
Here is what I'm trying to get to run
$statement = $db-> prepare("INSERT INTO `simplyaccomplished`.`blog_comment` (`ID`, `comment`, `date`, `ip_address`, `valid`, `name`, `blogcomment_ID`) VALUES (NULL, ?, NOW(), ?, 0, ?, ? );");
$statement -> bind_param("sssi",$comment, $ipaddress, $name , $comment_id);
$statement -> execute($statement);
$statement -> close();
The weird thing is this runs perfectly
$query = ("INSERT INTO `simplyaccomplished`.`blog_comment` (`ID`, `comment`, `date`, `ip_address`, `valid`, `name`, `blogcomment_ID`) VALUES (NULL,'$comment' , NOW(), '$ipaddress', '0', '$name', '$comment_id');");
$result =$db->query($query);
If someone could tell me where I'm going wrong I would greatly appreciate it!
The PDO method you're looking for is named bindParam, not bind_param :)
Try mysqli method,
$statement = $db-> prepare("INSERT INTO `simplyaccomplished`.`blog_comment` (`ID`,
`comment`, `date`, `ip_address`, `valid`, `name`, `blogcomment_ID`)
VALUES (?, ?, ?, ?,?, ?, ?)");
$statement -> bind_param("ssssisi",
null,$comment,NOW(),$ipaddress, 0,$name , $comment_id);
Take a look at PDO and MySqlI.

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