Error reading dynamically generated JSON with file_get_contents

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I have this code that generates me a json file

header("Content-Type: application/json; charset=utf-8;"); 
$codigo = $_GET['cod']; //variável para parametro que será passado ao servidor via    
URL
$sql1 = mysql_query("Select nome, valor from produtos where id_produtos = '21' "); // 
comando SQL para buscar informações do banco de dados

$jsonObj= array(); // cria uma variável que receberá um array (JSON)
while($result=mysql_fetch_object($sql1))
{
$jsonObj[] = $result;
}

$final_res =json_encode($jsonObj); // "transforma" o array recebido em JSON
echo $final_res;
exit;

And that you’re returning this code to me

[{"nome":"SAO PAULO CENTRO X COPACABANA","valor":"200,00"}]

It’s just that you’re giving me the error of reading it with this reading code Warning: Invalid argument supplied for foreach() in /home/mrangelc/public_html/mpitech.com.br/transport/test/Webler.php on line 13

$json = file_get_contents('webserv.php');
$lista = json_decode($json, true);
// Veja como fica o resultado
var_dump($lista);

// Manipulando o resultado como PHP.
foreach($lista as $objeto) {
print "nome: {$objeto['nome']} ,a valor: {$objeto['valor']}";
}

$objeto = json_decode($json);

echo 'Nome: ' . $objeto->nome;
echo 'valor: ' . $objeto->valor;

What must be wrong?

php mysql json
  • 1

    Check the value of $lista and of $json some of them are wrong.

  • 2

    When you give a file_get_contents('webserv.php'), means you’re taking the content from it - exactly a PHP file. Replace this line with include('webserv.php'), put the query result into a variable and use this variable to generate JSON.

  • I tried to use include plus ai only repeated the result of webserv.php

  • value gives variable list probably not coming as an array, check why

  • I think that’s almost what @Rodrigorigotti said, but a simple include `

2 answers

1


I simulated your case with the following code:

$json = file_get_contents('arquivo.json');
$variable = json_decode($json);
foreach ($variable as $key) {
    echo "nome: " . $key->nome;
    echo "valor: " . $key->valor;
}

And with a.json file:

[{"nome":"SAO PAULO CENTRO X COPACABANA","valor":"200,00"}]

Your case var_dump of the already decoded variable (in this example is the $variable) be the below, you should get the result accessed as an object of the default php class stdClass.

array(1) {
  [0]=>
  object(stdClass)#1 (2) {
    ["nome"]=>
    string(29) "SAO PAULO CENTRO X COPACABANA"
    ["valor"]=>
    string(6) "200,00"
    }
}

I know you use php to dynamically generate your values, revise your result would also be a good tip.

0

As @Rodrigorigotti said in his comment, do file_get_contents of a PHP file brings its source code, not the processed file.

According to this reply in English you have two options to solve this:

  1. Make the included file return a value (with return, nay echo), and then include it like this:

    $json = include 'webserv.php';

  2. If you cannot change the file to return a value, you need to use output buffering:

    ob_start();
include 'webserv.php';
$json = ob_get_clean();

    It is also possible to create a function for this, as in the accepted answer of the question in English. Highly recommended if you are going to use this kind of thing more than once.

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