store data to 2 tables with foreign key in MySql php

Question

I have two table in MySql, which are table 1 and table 2. As I want to link table 1 to 2 via userID. However, the funciton I have come out is not working.

MySQl tables are below:

In my case, userId will be the foreign key to link these 2 tables. However, my functions are not working. This is my function below:

function insert() {
    function adduserdetails($con, $accountId, $name, $contact) {

        $query = "insert into userdetails(accountId,name,contact) 
            values('$accountId','$name','$contact')";
        //echo "{$sqlString}";



        $insertResult = mysqli_query($con, $query);


        if ($insertResult) {
            echo " Applicant Detail Added !<br />";
            echo "<a href='index.php'>Back to Home</a>";
        } else {
            echo " Error !";
            echo "{$query}";
            //header('Location: post.php');
        }
    }

}

if ($con->query($query) === TRUE) {
    $last_id = $con->insert_id;
    echo "New record created successfully. Last inserted ID is: " . $last_id;
} else {
    echo "Error: " . $query . "<br>" . $con->error;
}

function adduseremployment($con,$last_id,$occupationMain,$comapny){

    $query1 = "insert into useremployment(userId,Occupation,company) 
            values('$last_id',$occupationMain','$comapny')";            
                 //echo "{$sqlString}";


                 $insertResult = mysqli_query($con, $query1);


                 if($insertResult){
                     echo " Applicant Detail Added !<br />";
                     echo "<a href='index.php'>Back to Home</a>";
                 }
                 else {
                     echo " Error !";
                     echo "{$query1}";
                     //header('Location: post.php');
                 }


}

you can not create`function` inside `function`. `function insert() { function` — urfusion, Jul 08 '17 at 05:42

ScaisEdge · Answer 1 · 2017-07-08T06:13:33.520

0

In second insert you are using wrong column name (userID and not accountID)

   $query1 = "insert into useremployment(userID,Occupation,company) 
                                         ^^^^^^ here
          values('$last_id',$occupationMain','$comapny')";

and in function signature and body you have comapny (could be you want company)

edited Jul 08 '17 at 06:13

answered Jul 08 '17 at 05:47

ScaisEdge

131,976
10
91
107

is the if function is correct? I tried to use 'if' to link $last_id to the userId in the userdetails table – xhinvis Jul 08 '17 at 06:10
If you have not the userID column in you insert statement you can't assign the correspondant var to the column .. the missing userID is inside the column name declaration ... asnwer updated with ^^^^here for position – ScaisEdge Jul 08 '17 at 06:13

score 0 · Answer 2 · answered Jul 08 '17 at 05:53

You have vertical table.

Check below code,

Be remember to prepare statement.

<?php
    function addUser(){
        // parent table.
        $queryParent = ''; // Your userdetails table insert query
        $insertResult = mysqli_query($con, $queryParent);
        $userId = mysqli_insert_id($con); // This is your last inserted userId.

        // child table.
        $queryChild = ''; // Your useremployment table insert query with userId.
        $insertResult = mysqli_query($con, $queryChild);
    }
?>

score 0 · Answer 3 · answered Jul 08 '17 at 05:53

0

your Query is wrong, please check this,

$query1 = "insert into useremployment(userId,occupation,company) 
        values('$last_id',$occupationMain','$comapny')";

you have to store record on userId as a refrence not to "accountID".

hope it helps,

answered Jul 08 '17 at 05:53

Ritesh Khatri

1,253
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29

store data to 2 tables with foreign key in MySql php

3 Answers3