Inserting values of JS-created input textboxes into a database

Question

I am using jQuery to create as many input textboxes as the user needs like so:

<script type="text/javascript">
$(document).ready(function() {
$('#names').on({
    blur: function() {
        var name = $("<p><input class='input' type='text' /></p>")
        var nullFields = 0;
        $(this).closest('div#names').find('input.input').each(function(){
            if($(this).val() == ""){
                nullFields++;
            }
        });
        console.log(nullFields);
        if(nullFields <= 1){
         $('#names').append(name.fadeIn(500));
        }
    }
}, 'input');
 });
</script>

Inserting a static textbox into a database isn't a problem using $_POST['blah'] andmysql_query("INSERT INTO ..."), but how do I insert the values of the dynamically created textboxes? I know I'll have to give the textboxes different names as they're created and I presume the MySQL query will be by way of some sort of loop.

EDIT

The website in question is here, specifically at step 4. As mentioned above, step 3 was quite straightforward.

Daniel Hunter · Accepted Answer · 2012-06-12 21:39:55Z

1

This is an example to get you started, not the complete solution. You create an array for the names then have the php insert each array item

var currentArrayNum = 1;    
$('#someClickable').click(function(){

        $('#td').append('<input name="arrayName['+currentArrayNum+']" value="" />');
        currentArrayNum += 1;
    });

php:

foreach ($_POST as $key){
        if (is_array($key)){
                foreach ($key as $key2 => $value){
                //$key2 will equal arrayName[currentArrayNum]
               //$value will equal the user input for the text field
do some stuff
}

edited Jun 12, 2012 at 21:39

answered Jun 12, 2012 at 20:29

Daniel Hunter

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3 Comments

scriptin Over a year ago

You can just use name[] without any index.

Sebastian Over a year ago

Thanks very much for your reply. A couple of questions: 1. is there a way to test the javascript? The dynamically created textboxes don't show up in the source code. 2. What exactly is the PHP doing? I don't quite understand what $key is doing...

Daniel Hunter Over a year ago

When you POST your form to the PHP page that enters into the database the POST is an array. $_POST[key] = value. The dynamically created text fields are now an array inside that array. This script will find the array (is_array) and take each value as $key2 and assign $value. now you can access this inside your foreach loop. might not be the best way for what you are doing but i would need to see how you want to write it to the database.. use print_r($_POST); die(); in your php script to see the array

Nurickan · Accepted Answer · 2012-06-12 20:38:37Z

0

You can create arrays out of name, simply try this:

<input type="text" name="post_input[input_1]" value="">
<input type="text" name="post_input[input_2]" value="">

After Submit, you would get an Array out of $_POST["post_input"] with the keys input_1 and input_2 and their assigned values. Then you just loop them as a normal array, for example

$aTextFields = $_POST["post_input"];
foreach( $aTextFields as $sValue ) {
  ...
}

answered Jun 12, 2012 at 20:38

Nurickan

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