jquery json PHP HTML 不显示结果


jquery json php html does not show results

我正在尝试将 Jquery 与 json 一起使用,并通过用类警报替换 DIV 来返回结果。 我也知道JS使用警报,所以我稍后会改变它。
但是我有点卡在这里,它确实发布到秒文件(offerte2.php),但没有显示任何结果。
我不确定这是java还是php的错误。有人有一些想法吗?

脚本.js:

$(document).ready(function() {
var form = $('#form');
var submit = $('#submit1');
var page1 = $('.page1');
form.on('submit', function(e) {
e.preventDefault();
$.ajax({
  url: '******/shop/templates/Euphoria-Art/offerte2.php', // form action url
  type: 'post', // form submit method get/post
  dataType: 'json', // request type html/json/xml
  data: form.serialize(), // serialize form data 
  beforeSend: function() {
    page1.fadeOut();
  },
success: function(result) {
    if(result.error){
        page1.html(result.html).fadeIn();
        console.log(e)
    }else{
        page1.html(result.html).fadeIn();
        form.trigger('reset');   
    }
}
});
});
});

捕获值

<?php
header('Content-Type: application/json');
if( isset( $_SERVER['HTTP_X_REQUESTED_WITH'] ) ){            
$class1 = filter_var($_POST['class1'], FILTER_SANITIZE_STRING);
$class2 = filter_var($_POST['class2'], FILTER_SANITIZE_STRING);
$class3 = filter_var($_POST['class3'], FILTER_SANITIZE_STRING);
$class4 = filter_var($_POST['class4'], FILTER_SANITIZE_STRING);
$class5 = filter_var($_POST['class5'], FILTER_SANITIZE_STRING);
$result = array("error" => false, "html" => null);
$totaal1 = $class1 + $class2 + $class3 + $class4 + $class5 
$result["error"] = false;
$result["html"] = "<h3>Het totaal is .$totaal1.</h3>";
echo json_encode($result);
exit;
}
?>
一切

似乎都很好,除非你在$class5之后错过了一个分号

$totaal1 = $class1 + $class2 + $class3 + $class4 + $class5;

if( isset( $_SERVER['HTTP_X_REQUESTED_WITH'] ) ){            
    $class1 = filter_var($_POST['class1'], FILTER_SANITIZE_STRING);
    $class2 = filter_var($_POST['class2'], FILTER_SANITIZE_STRING);
    $class3 = filter_var($_POST['class3'], FILTER_SANITIZE_STRING);
    $class4 = filter_var($_POST['class4'], FILTER_SANITIZE_STRING);
    $class5 = filter_var($_POST['class5'], FILTER_SANITIZE_STRING);
    $result = array("error" => false, "html" => null);
    $totaal1 = $class1 + $class2 + $class3 + $class4 + $class5;
    $result["error"] = false;
    $result["html"] = "<h3>Het totaal is .$totaal1.</h3>";
} else {
    $result["error"] = true;
    $result["html"] = "<h3>Error</h3>";
}
echo json_encode($result);
exit;