CODEWITHSUNDEEP
phphtmlmysqldatabase
xhinvis
xhinvis

Reputation: 211

Insert data to MySql with 2 input value

I have 1 table and I want to insert 2 values into MySql. However, there is only one value goes into my database and the other remains empty.

The design of my mySql table: enter image description here

My Html table code is below:

<table>

   <tr>
       <th></th>
       <th>Main Applicant</th>
       <th>Joint Applicant1</th>

  </tr>
  <tr>
       <td>Name</td>
       <td><input type="text"   id="nameMain" name="nameMain" class="form-control" autocomplete="off" required></td>
      <td><input type="text"   id="nameJoint1" name="nameJoint1" class="form-control" autocomplete="off" required></td>


  </tr>

  <tr>
      <td>Nationality(Country)</td>
      <td><input type="text"  id="nationMain" name="nationMain" class="form-control" autocomplete="off" required></td>
      <td><input type="text"  id="nationJoint1" name="nationJoint1" class="form-control" autocomplete="off" required></td>

 </tr>
 <tr>
    <td>Age</td>
    <td><input type="number"  id="ageMain" name="ageMain" class="form-control" autocomplete="off" required></td>
    <td><input type="number"  id="ageJoint1" name="ageJoint1" class="form-control" autocomplete="off" required></td>


</tr>

My post.php code is below:

       <?php 


    require_once 'db/dbfunction.php';
    require_once 'db/dbCreditAssessment.php';

    session_start();
    $con = open_connection();



    $name = $_POST['nameMain'];
    $nationality = $_POST['nationMain'];
    $age = $_POST['ageMain'];


    $nameJoint1 = $_POST['nameJoint1'];
    $nationJoint1 = $_POST['nationJoint1'];
    $ageJoint1 = $_POST['ageJoint1'];

    addApplicantPersonalDetails($con,$name,$nationality,$age);

    addApplicantPersonalDetails2($con,$name,$nationality,$age);

    close_connection($con);


    ?>

My addCreditAssessment.php code is below:

    <?php

function addApplicantPersonalDetails($con,$name,$nationality,$age){

    $query = "insert into zzz(name,nationality,age) 
            values('$name','$nationality','$age')";            
                 //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');
                 }


}

function addApplicantPersonalDetails2($con,$name,$nationality,$age){



    $query2 = "insert into zzz(name,nationality,age) 
            values('$nameJoint1','$nationJoint1','$ageJoint1')"; 


                 $insertResult2 = mysqli_query($con, $query2);



                 if($insertResult2){
                     echo " Applicant Detail Added !<br />";
                     echo "<a href='index.php'>Back to Home</a>";
                 }
                 else {
                     echo " Error !";
                     echo "{$query2}";
                     //header('Location: post.php');
                 }
}


?>

Upvotes: 0

Views: 969

Answers (6)

Haj Mohamed A
Haj Mohamed A

Reputation: 83

POST.PHP

Change the variable name

addApplicantPersonalDetails2($con,$nameJoint1,$nationJoint1,$ageJoint1);

addCreditAssessment.php

function addApplicantPersonalDetails2($con,$nameJoint1,$nationJoint1,$ageJoint1)

Upvotes: 0

Murad Hasan
Murad Hasan

Reputation: 9583

You are receiving $name,$nationality,$age and insert '$nameJoint1','$nationJoint1','$ageJoint1' at second function addApplicantPersonalDetails2().

You need to use the same query for both the function. Your second query should like: 

$query2 = "insert into zzz(name,nationality,age) values('$name','$nationality','$age')";

Or you can change your function parameters with '$nameJoint1','$nationJoint1','$ageJoint1'

Upvotes: 0

Huy Trịnh
Huy Trịnh

Reputation: 753

Instead of using nameMain and nameJoint1, you can use name[] for both field

On post.php, the name user input will belong to $_POST['name'] array.

So that you can use for-loop to submit all these data.

So your html file will looks like this: 

<table>

   <tr>
       <th></th>
       <th>Main Applicant</th>
       <th>Joint Applicant1</th>

  </tr>
  <tr>
       <td>Name</td>
       <td><input type="text"   id="nameMain" name="name[]" class="form-control" autocomplete="off" required></td>
      <td><input type="text"   id="nameJoint1" name="name[]" class="form-control" autocomplete="off" required></td>


  </tr>

  <tr>
      <td>Nationality(Country)</td>
      <td><input type="text"  id="nationMain" name="nation[]" class="form-control" autocomplete="off" required></td>
      <td><input type="text"  id="nationJoint1" name="nation[]" class="form-control" autocomplete="off" required></td>

 </tr>
 <tr>
    <td>Age</td>
    <td><input type="number"  id="ageMain" name="age[]" class="form-control" autocomplete="off" required></td>
    <td><input type="number"  id="ageJoint1" name="age[]" class="form-control" autocomplete="off" required></td>


</tr>

then your function will looks like:

function addApplicantPersonalDetails($con,$name,$nationality,$age){
if(count($name) == count($nationality) == count($age) = $count){
    for($i = 0; $i < $count; $i++){
       $query = "insert into zzz(name,nationality,age) 
            values('$name[$i]','$nationality[$i]','$age[$i]')";            
                 //echo "{$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');
                 }


}

Upvotes: 0

Deepak Kishore
Deepak Kishore

Reputation: 138

You are using variable that are not define i the function that why empty

Do like this 

function addApplicantPersonalDetails2($con,$nameJoint1,$nationJoint1,$ageJoint1){
    $query2 = "insert into zzz(name,nationality,age) 
            values('$nameJoint1','$nationJoint1','$ageJoint1')"; 


                 $insertResult2 = mysqli_query($con, $query2);



                 if($insertResult2){
                     echo " Applicant Detail Added !<br />";
                     echo "<a href='index.php'>Back to Home</a>";
                 }
                 else {
                     echo " Error !";
                     echo "{$query2}";
                     //header('Location: post.php');
                 }
}

It will work for you.

Upvotes: 1

Cynical
Cynical

Reputation: 9568

I think there are two errors here. The first lies in this call, where you use the same parameters as the one above:

addApplicantPersonalDetails2($con,$name,$nationality,$age);

I guess what you mean is:

addApplicantPersonalDetails2($con,$nameJoint1,$nationJoint1,$ageJoint1);

Secondly, as @FrayneKonok and @Richard have pointed out, you are using the wrong formal parameters in the query inside the function's body.

Upvotes: 0

Richard
Richard

Reputation: 628

Your second function is wrong:

Your function is:

addApplicantPersonalDetails2($con,$name,$nationality,$age)

Write your SQL like this:

$query2 = "insert into zzz(name,nationality,age) 
            values('$name','$nationality','$age')";

You set the old variables from your $_POST request, not from your function

Upvotes: 0

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