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I need to create div’s with data from two urls with json via ajax, and the data will be returned on the same page, but I need to pass two urls as parameter, as I can do this?
var url_aparecida = 'https://upvagasweb.000webhostapp.com/gerenciamento/php_access/access_json_aparecida.php';
var url_pinda = 'https://upvagasweb.000webhostapp.com/gerenciamento/php_access/access_json_pinda.php';
//Carrega dados de um arquivo json
$(document).on('click', '#bt_prosseguir', function() {
//preciso passar também a url_pinda
$.getJSON(url_aparecida, function(result) {
elemento = "<div class='list radius white'>";
$.each(result, function(i, valor) {
elemento += "<div class='item'>";
elemento += "<div class='left'>";
elemento += "<img class='avatar radius' src='" + valor.logo_empresa + "'>";
elemento += "</div>";
elemento += "<h2 style='margin-left:40%;'><strong>" + valor.setor + "</strong></h2>";
elemento += "<p style='margin-left:40%;'>" + valor.empresa + "</p>";
elemento += "<p style='margin-left:40%;'> Por " + valor.vinculo + "</p>";
elemento += "<p class='text-grey-500' style='margin-left:40%;'>" + valor.cidade + " - SP</p>";
elemento += "<p class='bt_ver_mais' style='margin-left:40%;color:#00f;'>Ver mais</p>";
elemento += "<div class='more_info'>";
elemento += "<p class='ver_mais' style='margin-left:40%;'><strong>Benefícios:</strong>" + valor.beneficios + "</p>";
elemento += "<p class='ver_mais' style='margin-left:40%;'><strong>Remuneração</strong>" + valor.remuneracao + "</p>";
elemento += "<p class='ver_mais' style='margin-left:40%;'><strong>Nível de estágio:</strong>" + valor.nivel_estagio + "</p>";
elemento += "<p class='ver_mais' style='margin-left:40%;'><strong>Inscrição:</strong>" + valor.processo_seletivo + "</p>";
// elemento += "<a href='"+valor.contato+"' class='ver_mais' style='margin-left:40%;'><strong>Link:</strong>"+valor.contato+"</p>";
elemento += "</div>";
elemento += "</div>";
});
elemento += "</div>";
$('#feed_vagas').html(elemento);
//esconde a div more_info
$('.more_info').hide();
$('.bt_ver_mais').click(function() {
$(this).closest("div.item").find(".more_info").toggle();
//https://api.jquery.com/closest/
});
});
});
It did not generate an error, but it also did not load the data of any of the urls, the json of each one is the same, only changes the values, because it is a json of each city, however the information needs to load in the same page as a feed
– Luan Rodrigues
Did not load the data or mounted your html?
– Leandro Angelo
Did not load the data, my html is already done
– Luan Rodrigues