INSERT INTO multiple tables through multiple foreign keys - php

Question

I have multiple table connected together by a few different foreign keys (see attahced image).
I am trying to insert into the projects table. I've been trying to accomplish this through the following code below, but it's not working. Right now I am getting an error saying that neither client_id_fk, project_manager_id_fk have a value. That makes sense since I didn't include them in the insert, but aren't auto incrementing and I also can't just add a random int to those fields since that throws an error as well. It technically works if I set the client_id_fk and project_manager_id_fk to NULL, but then there's no data in the other tables...Please help
$sql1 = "INSERT INTO PROJECTS (Project_Name, StartDate) VALUES( '".$_POST["Project_Name"]."','".$_POST["StartDate"]."')";
$sql2 = "INSERT INTO CLIENTS(Client_Name, Client_Email, Client_Phone) VALUES ('".$_POST["Client_Name"]."','".$_POST["Client_Email"]."','".$_POST["Client_Phone"]."')";
$sql3 = "INSERT INTO PROJECT_MANAGERS(ProjectManager_Name,Project_Manager_Email, Project_Manager_Phone) VALUES ('".$_POST["ProjectManager_Name"]."','email', 'phone')";
$sql4 = "INSERT INTO TYPE_OF_WORK(TypeOfWork) VALUES ('".$_POST["TypeOfWork"]."')";

Answer 1

Pattern:
-- Insert data into slave tables
INSERT INTO slave_1 (columns_1) VALUES ('values_1');
INSERT INTO slave_2 (columns_2) VALUES ('values_2');
-- Insert data into main table,
-- query FK values from slaves
-- using values inserted above
-- as filtering conditions
INSERT INTO main (columns_main, fk_column_1, fk_column_2)
SELECT 'values_main', slave_1.id, slave_2.id
FROM ( SELECT id FROM slave_1 WHERE columns_1 = 'values_1' ) slave_1
JOIN ( SELECT id FROM slave_2 WHERE columns_2 = 'values_2' ) slave_2
If it is possible that some subquery in last INSERT may return more than 1 row then wrap output column in this subquery into MAX() function (you will receive id for the last row matched - i.e. just inserted).

Answer 2

Step back and think about the relationships you're modelling. You have told MySQL that a Project has a Client and a Project Manager; that makes sense. Then you've tried to insert a Project with only this information:
INSERT INTO PROJECTS (Project_Name, Start_Date) ...
The error you're getting is asking you "which Client is this Project for, and who is the Project Manager?" That's something only you can know.
If it's a new Client, you need to insert that before the Project; if it's a new Project Manager, you need to insert that before the Project too. Once you've inserted them, you need to get their IDs, e.g. with the LAST_INSERT_ID() function.
If it's an existing Client and/or Project Manager, you need to look up their IDs based on whatever logic makes sense in your application.
Once you have those two IDs, you can create you Project:
INSERT INTO PROJECTS (Project_Name, Start_Date, Client_Id, Project_Manager_Id) ...

Answer 3

This is wrong:
$sql1 = "INSERT INTO PROJECTS (Project_Name, StartDate) VALUES( '".$_POST["Project_Name"]."','".$_POST["StartDate"]."')";
$sql2 = "INSERT INTO CLIENTS(Client_Name, Client_Email, Client_Phone) VALUES ('".$_POST["Client_Name"]."','".$_POST["Client_Email"]."','".$_POST["Client_Phone"]."')";
$sql3 = "INSERT INTO PROJECT_MANAGERS(ProjectManager_Name,Project_Manager_Email, Project_Manager_Phone) VALUES ('".$_POST["ProjectManager_Name"]."','email', 'phone')";
$sql4 = "INSERT INTO TYPE_OF_WORK(TypeOfWork) VALUES ('".$_POST["TypeOfWork"]."')";
Because you are trying to INSERT into projects a client.id and a project_manager.id that does not exist in the database. So the above should actually be written like this (in case you use mysqli):
$link = mysqli_connect('localhost', 'my_user', 'my_password', 'my_db');
/* check connection */
if (!$link) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
// First add the client in case the table is completely empty
$stmt = mysqli_prepare($link, "INSERT INTO clients(name, email, phone) VALUES (?,?,?);");
mysqli_stmt_bind_param($stmt, 'sss', $_POST["Client_Name"], $_POST["Client_Email"], $_POST["Client_Phone"]);
/* execute prepared statement */
mysqli_stmt_execute($stmt);
/* close statement and connection */
mysqli_stmt_close($stmt);
// Then add the project_manager in case the table is completely empty
$stmt = mysqli_prepare($link, "INSERT INTO project_managers(name, email, phone) VALUES (?,?,?);");
mysqli_stmt_bind_param($stmt, 'sss', $_POST["ProjectManager_Name"], 'email', 'phone');
/* execute prepared statement */
mysqli_stmt_execute($stmt);
/* close statement and connection */
mysqli_stmt_close($stmt);
// Also add the type_of_work in case the table is completely empty
$stmt = mysqli_prepare($link, "INSERT INTO type_of_work(TypeOfWork) VALUES (?);");
mysqli_stmt_bind_param($stmt, 's', $_POST["TypeOfWork"]);
/* execute prepared statement */
mysqli_stmt_execute($stmt);
/* close statement and connection */
mysqli_stmt_close($stmt);
// Finally add your project
$stmt = mysqli_prepare($link, "INSERT INTO projects (project_name, start_date, client_id, project_manager_id) SELECT ?,?,client.id,project_managers.id FROM client,project_managers WHERE client.name = ? AND client.email = ? AND client.phone = ? AND project_managers.name = ?;");
mysqli_stmt_bind_param($stmt, 'ssssss', $_POST["Project_Name"],$_POST["StartDate"],$_POST["Client_Name"], $_POST["Client_Email"], $_POST["Client_Phone"],$_POST["ProjectManager_Name"]);
/* execute prepared statement */
mysqli_stmt_execute($stmt);
/* close statement and connection */
mysqli_stmt_close($stmt);
/* close connection */
mysqli_close($link);