SQL one to many users foreign key issue - mysql

Question

Error: Cannot add or update a child row: a foreign key constraint
fails?
(testlogin.leavetable, CONSTRAINT leavetable_ibfk_1 FOREIGN KEY
(users_id) REFERENCES users (id) ON DELETE CASCADE ON UPDATE CASCADE)
Been getting this message for the last day and a half. Trying to make a one to many (foreign key) relationship between the two tables. Between users:id and leavetable:users_id. When user logs in they get a id. when they submit a form to column hours I'd like for the id to stay with the form data. It is throwing the error when someone creates a new id number and it wont update the leavetable:user_id. It throws the error above.
What's going on? Am I approaching this the right way?
Here's my code:
CREATE TABLE IF NOT EXISTS `users` (
`email` varchar(255) CHARACTER SET utf8 COLLATE utf8_unicode_ci NOT NULL,
`id` int(11) NOT NULL,
`password` char(64) CHARACTER SET utf8 COLLATE utf8_unicode_ci NOT NULL,
`salt` char(16) CHARACTER SET utf8 COLLATE utf8_unicode_ci NOT NULL,
`username` varchar(255) CHARACTER SET utf8 COLLATE utf8_unicode_ci NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
CREATE TABLE IF NOT EXISTS `leavetable` (
`hours` int(11) NOT NULL,
`users_id` int(11) DEFAULT NOT NULL,
`id` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
ALTER TABLE `users`
ADD PRIMARY KEY (`id`) USING BTREE,
ADD UNIQUE KEY `email` (`email`) USING BTREE,
ADD UNIQUE KEY `username` (`username`) USING BTREE;
ALTER TABLE `leavetable`
ADD PRIMARY KEY (`id`) USING BTREE,
ADD KEY `users_id` (`users_id`) USING BTREE;
ALTER TABLE `leavetable`
MODIFY `id` int(11) NOT NULL AUTO_INCREMENT;
-- AUTO_INCREMENT for table `users`
ALTER TABLE `users`
MODIFY `id` int(11) NOT NULL AUTO_INCREMENT;
ALTER TABLE `leavetable`
ADD CONSTRAINT `leavetable_ibfk_1` FOREIGN KEY (`users_id`) REFERENCES
`users` (`id`) ON DELETE CASCADE ON UPDATE CASCADE;
my table structure:
+---------------------+
| Tables_in_testlogin |
+---------------------+
| leavetable |
| users |
+---------------------+
+----------+--------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+----------+--------------+------+-----+---------+----------------+
| email | varchar(255) | NO | UNI | NULL | |
| id | int(11) | NO | PRI | NULL | auto_increment |
| password | char(64) | NO | | NULL | |
| salt | char(16) | NO | | NULL | |
| username | varchar(255) | NO | UNI | NULL | |
+----------+--------------+------+-----+---------+----------------+
5 rows in set (0.01 sec)
+----------+---------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+----------+---------+------+-----+---------+----------------+
| hours | int(11) | NO | | NULL | |
| users_id | int(11) | NO | MUL | NULL | |
| id | int(11) | NO | PRI | NULL | auto_increment |
+----------+---------+------+-----+---------+----------------+
3 rows in set (0.01 sec)
php code:
mysql_select_db("testlogin", $con);
$sql="INSERT INTO leavetable (hours)
VALUES
('$_POST[hours]')";
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}

Answer 1

$sql="INSERT INTO leavetable (hours)
VALUES
('$_POST[hours]')";
The above sql code does not contain any reference to user id. How should mysql guess which user this leave entry belong to? Include tge user id both in the columns and the values section of the insert statement. Probably you should store the user id in the php session.

Answer 2

Looks like I've got it going. I basically have a hidden field in my form with the users id. when the form is submitted it is submitted with the form information to leavetable. Not sure if this is the right way to go about it but it seems to work.
form code:
<input type="number" name="hours" value="" />
<label>id: </label>
<input type="text" name="id" value="<?php
echo htmlentities($_SESSION['user']['id'], ENT_QUOTES, 'UTF-8');
?>" /> <br /><br />
<input type="submit" class="btn btn-info" name="submit" value="Submit" />
sql:
$sql = "INSERT INTO leavetable ( hours, users_id )
VALUES ( '{$mysqli->real_escape_string($_POST['hours'])}',
'{$mysqli->real_escape_string($_POST['id'])}' )";
$insert = $mysqli->query($sql);
It works. Is it the best way to go about this?