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D Oliveira
- AppGini Super Hero
- Posts: 348
- Joined: 2018-03-04 09:30
- Location: David
Post
by D Oliveira » 2019-09-10 22:56
Code: Select all
$j('#ID').change(function(){
p3 = $j('#ID').val();
$j.ajax({
url: 'hooks/ajax-copiadieta.php',
data: { p: p3 },
success: function(data){
if (data) {
console.log(data);
$j('#ID2').val(data);
}else{
console.log('nodata');
}
}
});
});
Code: Select all
$p = makesafe($_REQUEST['p']);
$sql = "SELECT ID FROM dietasuplib_new WHERE ID = '{$p}'";
$data1 = sqlValue($sql);
echo $data1;
how to return the entire array * from dietasuplib_new instead of single value using sqlVal() function?
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D Oliveira
- AppGini Super Hero
- Posts: 348
- Joined: 2018-03-04 09:30
- Location: David
Post
by D Oliveira » 2019-09-11 00:11
found the solution.
Code: Select all
$j('#imp-container').change(function(){
valor_entidad3 = $j('#imp-container').val();
$j.ajax({
url: 'hooks/ajax-copiadieta.php',
data: { entidad_id: valor_entidad3 },
success: function(data){
if (data) {
var tbl = jQuery.parseJSON(data)
console.log(tbl);
AppGini.current_id1_1.value=tbl.id1;
id1_1_reload();
$j('#desc1_1').val(tbl.desc1);
$j('#cal1_1').val(tbl.cal1);
$j('#fat1_1').val(tbl.fat1);
}else{
//console.log('nodata');
}
}
});
});
Code: Select all
<?php
$currDir = dirname(__FILE__) . '/..';
include("$currDir/defaultLang.php");
include("$currDir/language.php");
include("$currDir/lib.php");
/* grant access to all users who have acess to the orders table */
//db connection
require ('../config.php');
$entidad = makesafe($_REQUEST['entidad_id']);
$con = mysqli_connect($dbServer, $dbUsername, $dbPassword);
mysqli_select_db($con,$dbDatabase);
$query = mysqli_query($con,"select * from dietasuplib_new where ID= '{$entidad}'");
$dietasuplib_new = mysqli_fetch_array($query);
echo json_encode($dietasuplib_new);
?>