Good morning,
I have a small problem to show data results of a project.
On the one hand I have a table in Mysql called WORKS, with the fields CODE and DESCRIPTION.
In HTML, I have a dropdown list with search engine, in which all the codes of works that are open are shown, and you can also search them by code. I do this with Ajax and JSON.
And finally I have a locked textbox (disabled) where I want to show the description of the selected work in the dropdown list.
Tests.html: Dropdown:
<select class="itemName form-control" style="width:500px" name="itemName" id="itemName"></select>
Textbox:
<p><input class="form-control" type="text" disabled="disabled" id="obranom" name="obranom"></p>
script:
<script type="text/javascript">
$('.itemName').select2({
placeholder: 'Selecciona una obra',
ajax: {
url: 'ajaxpro.php',
dataType: 'json',
delay: 250,
processResults: function (data) {
return {
results: data
};
},
cache: true
}
});
</script>
ajaxpro.php:
<?php
define (DB_USER, "*****");
define (DB_PASSWORD, "*******");
define (DB_DATABASE, "******");
define (DB_HOST, "localhost");
$mysqli = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_DATABASE);
$sql = "SELECT CODIGO, DESCRIPCION FROM obras
WHERE CODIGO LIKE '%".$_GET['q']."%' AND ESTADO='Abierto'";
$result = $mysqli->query($sql);
$json = [];
while($row = $result->fetch_assoc()){
$json[] = ['id'=>$row['CODIGO'], 'text'=>$row['CODIGO']];
}
echo json_encode($json);
?>
I need to show the description in the textbox "obranom" of the selected code in the dropdown "itemName"
Thanks in advance and greetings