JSON encode with SQL and PHP query

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JSON encode with SQL and PHP query

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#1 by (1 votes)
#2 by (0 votes)
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#4 by (0 votes)

2

Good morning, everyone. This question I have already seen in several issues but I do not understand very well what is the problem of why it does not enter the results in an array. I know that the query returns results.

I leave code:

function jsonEcho(){
include('conn.php');
$servername = "localhost";


$ttenQ="SELECT 'id', 'nombre_cock', 'categoria_fk', 'descripcion','Preparacion', 'img_path', 'puntuacion' FROM 'def_cockteles'  ORDER BY 'puntuacion' DESC LIMIT 10";
$resultado = mysqli_query($conexion,$ttenQ);

$topTen = array();

$rowcount=mysqli_num_rows($resultado);
printf("Result set has %d rows.\n",$rowcount);
while ($row=mysqli_fetch_array($resultado)) {

    $topTen[]=array('ruta'=> $row["img_path"],'descripcion'=>$row["descripcion"]);
}

 return json_encode($topTen);
}

print_r(jsonEcho());

?>

Thanks in advance

php json

asked by RafaelM 11.10.2016 в 17:36

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4 answers

JSON.parse error how to get the item from a list in java and pass it to a variable?

score 1 · Answer 1

1

Replace the return with echo and voila, that will show you the json.

echo json_encode($topTen);

Greetings

answered by 11.10.2016 в 21:31

score 0 · Answer 2

If you are doing it with ajax you have to comment on this line printf("Result set has %d rows.\n",$rowcount); and if you want to see the result of that line of code it would be good if you also get the fix. When you make a ajax request, the correct thing is that only the json_encode is printed. Try this code:

$topTen = array();

$rowcount=mysqli_num_rows($resultado);
$topTen['rowcount'] = "Result set has $rowcount rows.\n";
while ($row=mysqli_fetch_array($resultado)) {    
    $topTen['data'][]=array('ruta'=> $row["img_path"],'descripcion'=>$row["descripcion"]);
}
return json_encode($topTen);

score 0 · Answer 3

0

Test by setting the Content-Type header:

<?php
$data = /** lo que sea que necesitas serializar **/;
header('Content-Type: application/json');
echo json_encode($data);
?>

Original answer: link

answered by 11.10.2016 в 21:15

score 0 · Answer 4

0

Test:

$topTen = array();
while($row = mysqli_fetch_assoc($resultado)) {
  $topTen[] = $row;
}
print json_encode($topTen);

and remove the printf

answered by 14.11.2016 в 10:45