Error inserting two query in once by php - php

Question

I'm trying to insert data in two tables, with exploiting the the primary key of the first one and add it as foreign key in the second table,i think the problem is the two query are inserting each info in the same time and it give us a conflict between foreign key and primary one,any ideas ? maybe making something like a delay between each query !!
<?php
$InputCIN = filter_input(INPUT_POST,"CIN");
$InputID_PAYEMENT = filter_input(INPUT_POST,"ID_PAYEMENT");
$InputDATE_PAYEMENT = filter_input(INPUT_POST,"DATE_PAYEMENT");
$InputTYPE_DE_PAYEMENT = filter_input(INPUT_POST,"TYPE_DE_PAYEMENT");
$InputPRIX = filter_input(INPUT_POST,"PRIX");
$InputMOIS_PAYEMENT = filter_input(INPUT_POST,"MOIS_PAYEMENT");
$queryAj2_1 = "insert into payement values($InputID_PAYEMENT,'$InputDATE_PAYEMENT','$InputTYPE_DE_PAYEMENT',$InputPRIX,'$InputMOIS_PAYEMENT')";
$queryAj2_2 = "insert into payer values('','$InputCIN',$InputID_PAYEMENT)";
if(isset($_POST['butAj2']))//submit button
{
if($InputID_PAYEMENT && $InputCIN)
{
if(mysqli_query($con,$queryAj2_1) && mysqli_query($con,$queryAj2_2))
{
$msg = "<div class='alert alert-success'><span class='fa fa-square-check'></span>Adding successfull</div>";
}
else
{
$msg = "<div class='alert alert-danger'><span class='fa fa-square-check'></span>Erorr adding</div>".mysqli_error($con);
}
}
else if(!$InputID_PAYEMENT && !$InputCIN)
{
$msg = "<div class='alert alert-danger'><span class='fa fa-square-check'></span>ADD CIN/ID PAYEMENT</div>".mysqli_error($con);
}
else{
$msg = "<div class='alert alert-danger'><span class='fa fa-square-check'></span>erorrrrrrrrrrrrr</div>".mysqli_error($con);
}//it show me this msg
}
>

Answer 1

The mysqli_multi_query() function performs one or more queries against the database.
The queries must be separated with a semicolon.
from php.net
MySQL optionally allows having multiple statements in one statement
string. Sending multiple statements at once reduces client-server
round trips but requires special handling.
Multiple statements or multi queries must be executed with
mysqli_multi_query(). The individual statements of the statement
string are separated by semicolon. Then, all result sets returned by
the executed statements must be fetched.
The MySQL server allows having statements that do return result sets
and statements that do not return result sets in one multiple
statement.
$queryAj2_1 = "insert into payement values($InputID_PAYEMENT,'$InputDATE_PAYEMENT','$InputTYPE_DE_PAYEMENT',$InputPRIX,'$InputMOIS_PAYEMENT');";
$queryAj2_2 = "insert into payer values('','$InputCIN',$InputID_PAYEMENT)";
// Execute multi query
if (mysqli_multi_query($con, $queryAj2_1. $queryAj2_2))
{
// print message on success
Use of the multiple statement with prepared statements is not supported.

Answer 2

There is been conflict when you are trying to execute both the query at the same time. The 2nd table does not able to take the reference of the 1st tables primary key.
Try this out....
if(mysqli_query($con,$queryAj2_1))
{
if(mysqli_query($con,$queryAj2_2))
{
$msg = "<div class='alert alert-success'><span class='fa fa-square-check'></span>Adding successfull</div>";
}
}
So, here there will be no chance of conflict and primary key will be inserted first into the table and then it's foreign key.

Answer 3

You can Try the code below to fire multiple query in PHP.
$sql = "INSERT INTO requestdetails (`userid`, `dayte`) VALUES ('$userid','$date');INSERT INTO managerrepo (`userid`, `dayte`) VALUES ('$userid', '$date') ";
if($conn->multi_query($sql) === TRUE)
{
echo "You have requested successfully!";
unset($_POST);
//header("location: index.php?id=".$id);
}
else
{
echo "Problem in request. Try Again!";
}
OR
you use the below code too, for insertion in same table:
$sql = "INSERT INTO requestdetails (`userid`, `dayte`) VALUES ('$userid','$date')";
$sql_new = "INSERT INTO requestdetails (`userid`, `dayte`) VALUES ('$userid', '$date') ";
if($conn->multi_query($sql) === TRUE)
{
if($conn->multi_query($sql_new) === TRUE)
{
}
else
{
echo "error";
}
}
else
{
echo "Problem in request. Try Again!";
}