Popular input with data from the same table via dropdown

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Popular input with data from the same table via dropdown

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#1 by (3 votes)
#2 by (1 votes)

2

I have a select that pulls the name of a product.

I want you to select the product name via dropdown and automatically fill in an input with product code. The information is in a single table, in the id, id_produto, nome ,serial, log case.

I can pull the data into the dropdown but I do not know how to bind an input with id_produto .

I hope I have been able to pass on my doubt.

<select name="produto" id="produto">
<?php $sql = mysql_query("select * from produto");
    while ($row = mysql_fetch_array($sql)) {
       print '<option value="'.$row['nome'].'">'.$row['nome'].'</option>';
    }
?>
</select><br/>

php input dropdown

asked by anonymous 08.09.2014 / 20:21

2 answers

1

<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script>
$(document).ready(function()
{
    $('#produto').change(function() {
        $('#recebe').val( $( this ).val() );
    });
});
</script>

<select name="produto" id="produto">
    <option value="ID-A">a</option>
    <option value="ID-B">b</option>
    <option value="ID-B">c</option>
</select>

<input id="recebe" />

You can check out online here at jsfiddle .

08.09.2014 / 21:03

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score 3 · Accepted Answer

<input type="hidden" id="inputdesejado">
<select name="produto" id="produto">
<!-- Isso ira evitar que dispare o change quando carregar. -->
<option value="" disable>Escolha um produto:</option>
<?php $sql = mysql_query("select * from produto");
    while ($row = mysql_fetch_array($sql)) {
       print '<option value="'.$row['id_produto'].'">'.$row['nome'].'</option>';
    }
?>
</select><br/>

And do a javascript with the following code:

$("#produto").change(function(){
  $("#inputdesejado").val() = $(this).val();
})'