CODEWITHSUNDEEP
phpjquery
Misters
Misters

Reputation: 1347

Uncaught SyntaxError: Unexpected token ; BUT WHERE

I'm getting this error but i can't see what is wrong. I need a little help. I've tried everything: looked for illegal characters, erasing the semi colon but i can't fix it anyway.

(function(){


$.ajax({
dataType:           "json",
url:                "prueba.php",
type:               "get",
data:               "t=javascript",
success:function(inf){

    var div += "<span class='carpeta'>";//<-- here is where the error comes but i dont see anything!
    div =+ "<span class='titulo'>"+inf["categoria"]+"<h2>"+inf["titulo"]+"</h2></span>";//cerrando titulo
    div =+"<span class='mascara'>"; 
    div =+"<h2>"+inf["titulo"]+"</h2>"; 
    div =+ "<p>"+info["descripcion"]+"</p>";
    div =+ "<a onclick('tutorial_numero("+1+")') href='javascript:void(0)'>Leer mas</a>";
    div =+"</span>"; //cerrando el span de la class mascara
    div =+ "</span>";//cerrando el span de la class carpeta


    document.getElementById("contenido").innerHTML = "porque tio";
    alert("ola k ase")
}
});

})();

the php code

<?php


if(isset($_GET["t"]))

{



require 'conexion.php';

$getConexion = new conexion();

$conexion_mysql  = $getConexion->mysql();

$id_tutorial = $_GET["t"];$query = "SELECT titulo,contenido_tutorial,descripcion,categoria FROM tituriales WHERE tags LIKE '%$id_tutorial%'";

    $ejecutar_query = $conexion_mysql->prepare($query);

    $ejecutar_query->execute();

        $enviar = $ejecutar_query->fetchAll(PDO::FETCH_ASSOC);

        foreach($enviar as $resultado)
        {

            $f['titulo'] = $resultado['titulo'];
            $f['contenido'] = $resultado['contenido_tutorial'];
            $f['descripcion'] = $resultado['descripcion'];

            if($resultado['categoria']=="web");
            {
                $f['categoria'] = "<img src='../pc.png' />";
            }
            if($resultado['categoria']=="desktop")
            {
                $f['categoria'] = "@";
            }
            if($resultado['categoria']=="ambos")
            {
                $f['categoria'] = "klk";
            }

            //echo $f["categoria"];

            echo json_encode($f);

        }



}



?>

Upvotes: 0

Views: 476

Answers (4)

Medo42
Medo42

Reputation: 3821

Shane Andrade already gave you one part of your answer, but that doesn't actually fix the error you get first:

var div += "<span class='carpeta'>";

This line is invalid all on its own, no other code necessary to see that. div is not defined yet, but you try to add something to it, since div += x is just shorthand for div = div + x. Just assign the string here:

var div = "<span class='carpeta'>";

Upvotes: 2

Bergi
Bergi

Reputation: 664538

Your assignment operators are wrong:

  • The first one, in the declaration and initialisation of the variable, should be a single =
  • The following ones, shorthand assignment and concatenation, should be += instead of =+

Upvotes: 2

Shane Andrade
Shane Andrade

Reputation: 2675

Your string concatenation operators aren't correct. Replace =+ with +=

(function(){


$.ajax({
dataType:           "json",
url:                "prueba.php",
type:               "get",
data:               "t=javascript",
success:function(inf){

    var div += "<span class='carpeta'>";//<-- here is where the error comes but i dont see anything!
    div += "<span class='titulo'>"+inf["categoria"]+"<h2>"+inf["titulo"]+"</h2></span>";//cerrando titulo
    div +="<span class='mascara'>"; 
    div +="<h2>"+inf["titulo"]+"</h2>"; 
    div += "<p>"+info["descripcion"]+"</p>";
    div += "<a onclick('tutorial_numero("+1+")') href='javascript:void(0)'>Leer mas</a>";
    div +="</span>"; //cerrando el span de la class mascara
    div += "</span>";//cerrando el span de la class carpeta


    document.getElementById("contenido").innerHTML = div;

}
});
})();

Upvotes: 3

Yevgen
Yevgen

Reputation: 1300

It looks like PHP doesn't return json. I think you should debug PHP script.

Upvotes: 0

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