PHP & MySQL Undefined variable problem
- by comma
I keep getting the following error Undefined variable: id on line 91 can some one help me correct this problem?
The error is on this line.
$query2 = "INSERT INTO users_skills (skill_id, user_id, date_created) VALUES ('$id', '$user_id', NOW())";
MySQL database tables.
CREATE TABLE tags (
id INT UNSIGNED NOT NULL AUTO_INCREMENT,
skill VARCHAR(255) NOT NULL,
experience VARCHAR(255) NOT NULL,
years VARCHAR(255) NOT NULL,
PRIMARY KEY (id)
);
CREATE TABLE users_skills (
id INT UNSIGNED NOT NULL AUTO_INCREMENT,
skill_id INT UNSIGNED NOT NULL,
user_id INT UNSIGNED NOT NULL,
date_created DATETIME UNSIGNED NOT NULL,
PRIMARY KEY (id)
);
Here is the PHP & MySQL code.
if (isset($_POST['info_submitted'])) {
$mysqli = mysqli_connect("localhost", "root", "", "sitename");
$dbc = mysqli_query($mysqli,"SELECT learned_skills.*, users_skills.*
FROM learned_skills
INNER JOIN users_skills ON learned_skills.id = users_skills.skill_id
WHERE user_id='$user_id'");
if (!$dbc) {
print mysqli_error($mysqli);
return;
}
$user_id = '5';
$skill = $_POST['skill'];
$experience = $_POST['experience'];
$years = $_POST['years'];
$mysqli = mysqli_connect("localhost", "root", "", "sitename");
$dbc = mysqli_query($mysqli,"SELECT learned_skills.*, users_skills.* FROM learned_skills INNER JOIN users_skills ON users_skills.skill_id = learned_skills.id WHERE users_skills.user_id='$user_id'");
if (mysqli_num_rows($dbc) == 0) {
if (isset($_POST['skill']) && trim($_POST['skill'])!=='') {
$mysqli = mysqli_connect("localhost", "root", "", "sitename");
$query1 = mysqli_query($mysqli,"INSERT INTO learned_skills (skill, experience, years)
VALUES ('" . $skill . "', '" . $experience . "', '" . $years . "')");
if (mysqli_query($mysqli, $query1)) {
print mysqli_error($mysqli);
return;
}
$mysqli = mysqli_connect("localhost", "root", "", "sitename");
$dbc = mysqli_query($mysqli,"SELECT id
FROM learned_skills
WHERE id='" . $skill . "'
AND experience='" . $experience . "'
AND years='" . $years . "'");
if (!$dbc) {
print mysqli_error($mysqli);
} else {
while($row = mysqli_fetch_array($dbc)){
$id = $row["id"];
}
}
$query2 = "INSERT INTO users_skills (skill_id, user_id, date_created) VALUES ('$id', '$user_id', NOW())";
}
}
