how to insert data in database when I have two different names in my html for one column?

Question

I know my title is a bit confusing so I'll explain it further on here. I know how to insert data in database, but I do not know how to do it when I have two different names in my HTML that should be inserted in my database.

I have a dropdown list for country, and the select name is country1. but when the user selects "Other" , a textbox will appear on the side that has a name country2and he should input his country. And I want that when he chooses "Other" and input his data, the data will insert in the database OR when he chooses from the dropdown list, it will do insert also.

When the user selects "Other", I want the data that he input in the textbox will be the one to be inserted in the database and not the word "Other", and when he selects from the list except "Other", I want the data that he selected will be the one to be inserted in the db.

Please please please do help me. If my question isn't clear enough please tell. Thank you.

My javascript code

function ValidateCountry(val){
 var element=document.getElementById('country');
 if(val==='Other')
   element.style.display='block';
 else  
   element.style.display='none';
}

HTML code

<form action="send.php" method="post">
    <select name="country1" onchange='ValidateCountry(this.value);'>
        <option> Select Country</option>
        <option value="USA"> Lawyer </option>
        <option value="Japan"> Nurse </option>
        <option value="Other"> Other </option>
    </select>
    <input type="text" name="country2" id="country" style='display:none;'/>
    <input type="submit" name="submit" />
</form>

And my PHP code:

include 'dbcontroller.php';

if(isset($_POST['submit']))
{
    $country = $conn->real_escape_string(trim($_POST['country1']));
}
$query = "INSERT INTO info(occupation) VALUES ('$occupation')";
mysqli_query($conn,$query);

Name them the same name like `country`. The input that comes last will be the value received. — AbraCadaver, Apr 27 '16 at 15:38
add an if condition in your controller and check the country1 is not equal to other then set another variable country = country1 else set country = country2 — Ibrahim, Apr 27 '16 at 15:42
First thing first, alter your `$query` to actually insert the country data. — Prusse, Apr 27 '16 at 15:50

score 3 · Accepted Answer · answered Apr 27 '16 at 15:41

3

Very simply, use an if/else

if($_POST['country1'] == 'Other')
{
     $country = $_POST['country2'];
} else {
     $country = $_POST['country1'];
}

Or with one line:

$country = $_POST['country1'] == 'Other' ? $_POST['country2'] : $_POST['country1'];

Then you can use $country in your query.

answered Apr 27 '16 at 15:41

aynber

22,380
8
50
63

i will try this one. :) – Felix Apr 27 '16 at 15:42
No problem, glad I could help! – aynber Apr 27 '16 at 15:54

score 1 · Answer 2 · answered Apr 27 '16 at 15:41

1

You would just need to do a simple comparison:

if ($_POST['country1'] == 'Other') {
    $country = $conn->real_escape_string(trim($_POST['country2']));
} else {
    $country = $conn->real_escape_string(trim($_POST['country1']));
}

answered Apr 27 '16 at 15:41

mister martin

6,197
4
30
63

score 0 · Answer 3 · answered Apr 27 '16 at 15:44

include 'dbcontroller.php';

if(isset($_POST['submit']))
{
    $country = $conn->real_escape_string(trim($_POST['country1'] != 'Other' ? $_POST['country1'] : $_POST['country2']));
}
$query = "INSERT INTO info(occupation) VALUES ('$occupation')";
mysqli_query($conn,$query);

score 0 · Answer 4 · answered Apr 27 '16 at 15:44

0

Just check on your php if the user input something into country2

if($_POST["country2"] != "")
{
  $country = $conn->real_escape_string(trim($_POST['country2']));
}
else
{
  $country = $conn->real_escape_string(trim($_POST['country1']));
}

answered Apr 27 '16 at 15:44

DIEGO CARRASCAL

1,999
14
16

Farhad Rakib · Answer 5 · 2016-04-27T15:55:40.720

You can use If else condition on your php code.Like the following

    include 'dbcontroller.php';

if(isset($_POST['submit']))
{
    $occupation = "";
    $country = $conn->real_escape_string(trim($_POST['country1']);
    if($country == "other"){
        $occupation = $_POST['country2'];
    }else{
        $occupation = $_POST['country1'];
    }
}
$query = "INSERT INTO info(occupation) VALUES ('$occupation')";
mysqli_query($conn,$query);

how to insert data in database when I have two different names in my html for one column?

5 Answers5