AJAX won't call the php file

Question

so I have a php file that brings information from a database about an user. My idea is that when an button is pressed I send the id to the php file and the get the user that corresponds. However I can't seem to execute the php file, and the ajax response is empty.

How can I fix this? Thank you for your help

EDIT: I removed the die function, and add the echo the $usuario array. Now i get it in the response but i have a dumb question: how do i access the data of that array? Thank you again

and can u plz share the code inside this file `editar_persona.php` — devpro, Mar 06 '19 at 14:48
What do you mean by "the response is empty"? Does that mean the AJAX code is successfully calling the PHP file and successfully getting a blank response? If so then it sounds like the problem is server-side, not here. — David, Mar 06 '19 at 14:48
@JulieRuchel: Then the code you're showing us appears to be working. Time to start debugging your PHP code. — David, Mar 06 '19 at 14:50
Impossible to say what's wrong without seeing your php file. — Jared Smith, Mar 06 '19 at 14:51
Why do you call `die()` in your PHP code? Looks obvious to me why you see nothing in your response — Nico Haase, Mar 06 '19 at 14:55
@NicoHaase to see if it would enter there. I get the same result with it or without it — Juliette Ruchel, Mar 06 '19 at 14:58
@JulieRuchel you're unconditionally calling `die`, so your script will stop executing at that point and you'll get an empty response. I also don't see you `echo`ing anything in your script anywhere. — Jared Smith, Mar 06 '19 at 15:00
So, what happens there? What do you expect from that `die()` call? — Nico Haase, Mar 06 '19 at 15:01
Additionally, if you want to check what `$_GET` contains, try looking it up in your browser's network console first to see which data is **really** sent to the backend — Nico Haase, Mar 06 '19 at 15:02
use `die($usuario);` before `return $usuario;` and share the result — devpro, Mar 06 '19 at 15:05
`"SELECT * FROM persona WHERE nro = $nro"` Don't do this. You're vulnerable to SQL injection. [Use parameterized queries.](https://stackoverflow.com/q/60174/9029328) — AuxTaco, Mar 07 '19 at 05:58

score 0 · Answer 1 · answered Mar 06 '19 at 16:27

<?php
 require_once 'constantes.php';
 require_once 'smarty.php';
 require_once 'verificar_rol.php';

 $nro = $_GET['nro'];
try{
   $mysqli = new mysqli(HOST,USUARIO,PASSWORD,DB);
   if($mysqli->connect_errno){
      throw new Exception("Error al conectarse a servidor Mysql");
   }
  $sql = "SELECT * FROM persona WHERE nro = $nro";
   // here is some change
  $resultado = $mysqli->query($sql)  
  if(!$resultado){
  // end change
    throw new Exception("Error al obtener los datos");
  }
  $usuario = $resultado->fetch_assoc();
  echo $usuario;

}catch(Exception $excepcion){
  $smarty->assign("error", 1);
  $smarty->assign("mensaje", $excepcion->getMessage());
 }

and this

$("#editar").click(function(){
    var nro = $(this).data('id');
    $.ajax({
        url: "editar_persona.php",
        type: 'GET',
        data: {nro: nro},
    }).done(function(response){
        console.log(response);
        alert(response);
    });

Try this

score -1 · Accepted Answer · answered Mar 06 '19 at 14:52

-1

What about

$("#editar").click(function(){
    var nro = $(this).data('id');
    $.ajax({
        url: "editar_persona.php?nro="+nro,
        type: 'GET'
    }).done(function(response){
        console.log(response);
        alert(response);
    });

GET request uses URL to pass Data

answered Mar 06 '19 at 14:52

tanawat yakit

62
5

3

Yeah, what about that? Can you explain how this solves the problem? – Nico Haase Mar 06 '19 at 14:55
if i do that in the response i get the value of nro but not the user, it doesn't get to the php – Juliette Ruchel Mar 06 '19 at 14:57
@JulieRuchel: because u r using `die($nro);` – devpro Mar 06 '19 at 14:58
@devpro I still get nothing without it – Juliette Ruchel Mar 06 '19 at 15:03
If you want SQL result, you need to print out the result on screen. In my oppinion, I might use echo json_encode($array); to print out the sql result. So, the code will change to $usuario = $resultado->fetch_assoc(); echo json_encode($usuario); And you will get those result in json format in respone variable. – tanawat yakit Mar 06 '19 at 15:07

AJAX won't call the php file

2 Answers2