how can i parse values from one php to other - php

Question

Let's start that I am newbie in php, so still I am trying to learn. I have created a form on Wordpress and I want to insert the values on one table (data_test table, i have managed that) and then take all the columns from data_test table(id that is auto increment number,name,email,product, quantity that the user enter) and insert to other table. I used this html code for the form to parse the values:
<form action="../enter_data_insert.php" method="post" onsubmit="return form_validation()" name="myForm">
Name <input id="name" name="name" type="text" />
Email <input id="email" name="email" required type="email"/>
Product<input id="prod" name="prod" required type="text" />
Quantity<input id="quant" name="quant" required type="number" min="1" / >
<input type="submit" value="Submit" />
</form>
And then this php to take the values:
<?php
if(!empty($_POST["name"]) && !empty($_POST["email"]) && !empty($_POST["prod"]) && !empty($_POST["quant"])){
//connect with database
include "database_conn.php";
//get the form elements and store them in variables
session_start();
$name=$_POST["name"];
$email=$_POST["email"];
$prod=$_POST["prod"];
$quant=$_POST["quant"];
//insert data on data_test table
$sql="INSERT INTO `site_db`.`data_test` ( `name` , `email`, `prod`,`quant`) VALUES ( '$name','$email','$prod','$quant')";
if(!mysqli_query($con,$sql)){
echo mysqli_error($con);
} else{
//retrieve data
$sql = "SELECT data_test_id FROM data_test WHERE prod='$prod'";
$result = mysqli_query($con,$sql);
if(!$result){
echo mysqli_error($con);
} else{
while($value = mysqli_fetch_object($result)){
$id = intval($value->id);
$_SESSION['myid'] = $value->id;
var_dump($value);
//insert data on data_test_ins table
$sql="INSERT INTO site_db.data_test_ins` ( id,name , email, prod,quant) VALUES ( $id,'$name','$email','$prod','$quant')";
if(!mysqli_query($con,$sql)){
echo mysqli_error($con);
} else{
//Redirects to the specified page
// header("Location: http://localhost/site/");
}
}
}
}
}
?>
Now it inserts all the values except the id on data_test table, i guess that it is null because it must close the first insert on php and then i have to call a second insert (with //insert data on data_test_ins table) on other php?
But i am not sure, can anyone help me please? or just guide me what is the right way to do.
I start to think that i have to create two php to parse the values and take on the first table and then on the other php to insert the values?
Any thoughts are helpful! :-)

Answer 1

What you are doing is not right. It is not a good approach to add value to id field to the database manually. It should be generated automatically by the database. What I would recommend is, add another field to your data_test_ins table eg: test_id which points to the id of your data_test table. This is the concept of foreign key.
Read about the concept of foreign keys here
Your code would now be:-
<?php
if(!empty($_POST["name"]) && !empty($_POST["email"]) && !empty($_POST["prod"]) && !empty($_POST["quant"])){
//connect with database
include "database_conn.php";
//get the form elements and store them in variables
session_start();
$name=$_POST["name"];
$email=$_POST["email"];
$prod=$_POST["prod"];
$quant=$_POST["quant"];
//insert data on data_test table
$sql="INSERT INTO `site_db`.`data_test` ( `name` , `email`, `prod`,`quant`) VALUES ( '$name','$email','$prod','$quant')";
if(!mysqli_query($con,$sql)){
echo mysqli_error($con);
} else{
//retrieve data
$sql = "SELECT data_test_id FROM data_test WHERE prod='$prod'";
$result = mysqli_query($con,$sql);
if(!$result){
echo mysqli_error($con);
} else{
while($value = mysqli_fetch_object($result)){
$id = $value->id;
//insert data on data_test_ins table
$sql="INSERT INTO `site_db`.`data_test_ins` ( `id`,`name` , `email`, `prod`,`quant`, `test_id`) VALUES ('$name','$email','$prod','$quant', '$id')";
if(!mysqli_query($con,$sql)){
echo mysqli_error($con);
} else{
//Redirects to the specified page
header("Location: http://localhost/site/");
}
}
}
}
}
?>

Answer 2

have you tried passing $value->id into the query instead of $value?
its an object which has the current row of a result set, so you should only pass the id attribute of this object.
$sql="INSERT INTO `site_db`.`data_test_ins` ( `id`,`name` , `email`, `prod`,`quant`) VALUES ( '$value->id','$name','$email','$prod','$quant')";
Addition:
stop using the mysql deprecated library.
you should check the posted data if its isset or not
EDIT:
your code should looks like:
<?php
if(!empty($_POST["name"]) && !empty($_POST["email"]) && !empty($_POST["prod"]) && !empty($_POST["quant"])){
//connect with database
include "database_conn.php";
//get the form elements and store them in variables
session_start();
$name=$_POST["name"];
$email=$_POST["email"];
$prod=$_POST["prod"];
$quant=$_POST["quant"];
//insert data on data_test table
$sql="INSERT INTO `site_db`.`data_test` ( `name` , `email`, `prod`,`quant`) VALUES ( '$name','$email','$prod','$quant')";
if(!mysqli_query($con,$sql)){
echo mysqli_error($con);
} else{
//retrieve data
$sql = "SELECT data_test_id FROM data_test WHERE prod='$prod'";
$result = mysqli_query($con,$sql);
if(!$result){
echo mysqli_error($con);
} else{
while($value = mysqli_fetch_object($result)){
$_SESSION['myid'] = $value->data_test_id;
$id = intval($value->data_test_id);
//insert data on data_test_ins table
$sql="INSERT INTO `site_db`.`data_test_ins` ( id,name , email, prod,quant) VALUES ( '$id','$name','$email','$prod','$quant')";
if(!mysqli_query($con,$sql)){
echo mysqli_error($con);
} else{
//Redirects to the specified page
header("Location: http://localhost/site/");
}
}
}
}
}
?>