Return of PHP Query in Ajax

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How do I get the PHP query values in AJAX?

AJAX

$.ajax({
   type: 'post',
   dataType: 'json',
   url: 'listaGabaritosSelecionados.php',
   success: function(url){
     $('.retorno').html(url);
   }
 });

PHP

$mysqli = new mysqli("localhost", "root", "", "site"); 
$idProduto = 42;
$query = "SELECT * FROM produto where produto_id = '$idProduto'";
$consulta = $mysqli->query($query);

while($row = $consulta->fetch_assoc()){

    $item = $row['item'];
    $ids = explode(',', $item);
    foreach ($ids as $valores) {

        $selecionaGabaritos = "SELECT * FROM gabaritos where id = '$valores' ";
        $banco = $mysqli->query($selecionaGabaritos);
        while ($resultado = $banco->fetch_assoc()) {
            $url[] = utf8_encode($resultado['url']);
        }
    }
}


json_encode($url);

Because I want to be sent a product ID to PHP, it makes a query in the database in 2 tables, if I run only the PHP file it runs, the problem is in AJAX for return!

RETURN PHP.FILE

<pre>array(2) {
  [0]=>
  string(14) "urlDoArquivoAI"
  [1]=>
  string(23) "http://www.teste.com.br"
}
</pre>

In AJAX is BLANK does not return me anything!

php ajax mysqli

  • Try to give a: echo json_encode($url); in your AJAX file.

  • Getting like this? $. ajax({ type: 'post', date: {id = Prod}, dataType: 'json', url: 'listingGabattribs.php', Success: Function(url){ //$('. return'). html(url); //console.log(url); echo json_encode($url); } });

  • No, this would be inside the AJAX file you call, in case no listaGabaritosSelecionados.php.

  • Oh yes, thank you. But he continues to return nothing!

3 answers

1

The error was due to want to run the.php file inside another folder, fixed with the help of @Israel Merljak

1


Friend, you can send the parameter (id) directly by ajax script. using "date".

$.ajax({
    url: "listaGabaritosSelecionados.php",
    method: "POST",
    data: { id : seuIdAqui }, //{[nome_do_parametro_post]: [valor_do_parametro]}
    dataType: "json",
    success: function(data) {
        // Transforma json em string pra renderizar no html
        $(".retorno").html(JSON.stringify(data));
    }
});

Don’t forget to recover this value in the PHP script using $_POST

A good tip is to read the documentation: jQuery.ajax().

To return the value of the PHP script you can do this way: 

echo json_encode($valor_retorno)

Example code (tested)

PHP: 
<?php

echo json_encode([
    "testando" => "retorno",
    "json" => "de",
    "um" => "script",
    "php" => "!!!!"
]);
Javascript: 
<div id="here"></div>

<script src="https://code.jquery.com/jquery-3.3.1.min.js" integrity="sha256-FgpCb/KJQlLNfOu91ta32o/NMZxltwRo8QtmkMRdAu8="
    crossorigin="anonymous"></script>


<script>
    $(document).ready(function () {
        $.ajax({
            url: "rest.php",
            method: "POST",
            dataType: "json",
            success: function(data) {
               $("#here").html(JSON.stringify(data)); 
            }
        });
    });
</script>

Exit: {"testing":"return","json":"of","a":"script","php":"!!!!"}

  • Buddy, I’m testing with the ID right in the query, but thanks for the tip, the problem is in the AJAX bring me this return, it can not return me anything!

  • tried a echo json_encode($valor) ?

  • Yes I am trying with echo json_encode($url); and it does not return me!

  • 1

    Amigo.. it’s simple. you’ve tried giving a console.log() in the return ?? You’re trying to print a javascript object in html. will not be possible, try to do $(".retorno").html(JSON.stringify(url))

  • Buddy, as amazing as it sounds, it didn’t work!

  • I’ll post the code I tested in the answer for you to see..

  • and it doesn’t work yet! Nothing returns to me!

  • So do us a favor, run your php script and paste us the full result. one makes a echo "<pre>"; var_dump($url); echo "</pre>"; and glue what return to see.

  • This is the return directly in the.php <pre>array(2) { [0]=> string(14) "urlDoAquivoAI" [1]=> string(23) "http://www.teste.com.br" } </pre> But AJAX still returns nothing!

  • cola on your question please.. are you sure this return is correct? It does not seem to me a valid json (or php).

  • Answered there in question!

  • It makes no sense not to run it. You’re making a echo json_encode($url); at the end of the PHP script? Update the code in your question to the current version. Have you tried to copy my example and run?? See if it works.

  • Doing here by your example worked! I will now identify in mine which error!

  • Amazing how my files don’t work! Your script works!

  • Let’s go continue this discussion in chat.

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On the last line of your file .php replaces json_encode($url); for print_r($url) and checks the result. Because you are not giving any instruction to PHP to return the query data.

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