Passing variable - Ajax and Php

Question

I have this simple form wich allows to search and I want to show results into a DIV, so I am using ajax for it.

<script type="text/javascript">  
$(document).ready(function(){  
    $('#boton_cargar').click(function() {   

    var nombre = $("#nombre").val(); 
        $.ajax({ 
        type: "GET",           
        url: 'resultados.php?nombre='+nombre, 
            success: function(data) {  
                $('#resultados').html(data);  
                $('#resultados div').slideDown(1000);  
            }  
        });  
    });  

});  
</script> 




<form>
<input id="nombre" name="nombre" type="text" />

<input name="boton_cargar" id="boton_cargar" type="button" value="buscar" />
</form>

<div id="resultados">
   // I want to show results here
</div>

and this is resultados.php

<?php
include('loader.php'); //call db

$conn = new conection();
$rs = new RecordSet($conn);

if(isset($_GET['nombre']))

$sql="SELECT * FROM clientes INNER JOIN alquiler ON clientes.id_cliente = alquiler.id_cliente INNER JOIN insumos ON  insumos.id_insumo = alquiler.id_insumo WHERE `clientes`.`nombre` = {$_GET['nombre']}";
else
die('error');


unset($rs);
unset($conn);
?>

<?php foreach($resultados as $res){ ?> 
    <?php echo $res->nombre ?>
<?php }?>

I don't know what it's wrong, for example if I replace {$_GET['nombre']} for "jhon" I can get the results.

Hope can help me, thank u so much in advance!

warning your code is vulnerable to sql injection attacks! — Daniel A. White
– Daniel A. White, Commented Aug 27, 2013 at 23:00
where is the execution of $sql ? and where is the definition of $resultados? — MaveRick
– MaveRick, Commented Aug 27, 2013 at 23:02

Khawer Zeshan · Accepted Answer · 2013-08-27 23:04:19Z

1

You need to put quotes around {$_GET['nombre']}

$sql="SELECT * FROM clientes INNER JOIN alquiler ON clientes.id_cliente = alquiler.id_cliente INNER JOIN insumos ON  insumos.id_insumo = alquiler.id_insumo WHERE `clientes`.`nombre` = '{$_GET['nombre']}'";

answered Aug 27, 2013 at 23:04

Khawer Zeshan

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Seether · Accepted Answer · 2013-08-27 23:05:08Z

0

Try by replacing the sql line with this:

$sql="SELECT * FROM clientes INNER JOIN alquiler ON clientes.id_cliente = alquiler.id_cliente INNER JOIN insumos ON  insumos.id_insumo = alquiler.id_insumo WHERE `clientes`.`nombre` = '" . $_GET['nombre'] . "'";

answered Aug 27, 2013 at 23:05

Seether

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Comments

Darwing · Accepted Answer · 2013-08-27 23:05:54Z

0

Your problem is the query, that is not returning nothing because its format, try this:

$sql = "SELECT * FROM clientes INNER JOIN alquiler ON clientes.id_cliente = alquiler.id_cliente INNER JOIN insumos ON  insumos.id_insumo = alquiler.id_insumo WHERE `clientes`.`nombre` = '".$_GET['nombre']."'";

I hope that can help you.

answered Aug 27, 2013 at 23:05

Darwing

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1 Comment

Maru Over a year ago

thanks for the answears! I am trying with 2 variables, maybe I am doing it wrong becouse it's not working url: 'resultados.php?nombre='+nombre'&insumos='+insumos, <select id="insumos" name="insumos"> <?php foreach($insumos as $in){ ?> <option id="<?php $in->id_insumo ?>"><?php echo $in->insumo ?></option> <?php }?> </select> if(isset($_GET['nombre']) && ($_GET['insumos'])) $sql="SELECT * FROM ....... WHERE clientes.nombre = '".$_GET['nombre']."' AND alquiler.id_insumo = '".$_GET['insumos']."'";

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