Question

我想向PHP文件发送ajax调用，其中我有Javascript代码将数据发送到SurveyJS（https://surveyjs.io/）。

使用＆＃34; localhost / testSurvey / ajaxOriginator.html＆＃34;调用ajaxOriginator.html成功调用receiver.php并在receiver.php中运行JavaScript但不向SurveyJS发送请求。

如果我直接使用＆＃34; localhost / testSurvey / receiver.php＆＃34;来调用receiver.php，它会成功将请求发送给SurveyJS，我可以看到收到的数据。

这是ajaxOriginator.html

<!DOCTYPE HTML>
<html lang="en">
<head>
<meta charset="utf-8"/>
<title>Ajax originator</title>

<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

<script>
$.get('receiver.php', { "name": "Julia", "car": "Kia" });
</script>

</head>

这是receiver.php

<?php
$name = $_GET["name"];
$car = $_GET["car"];
?>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<script src="https://surveyjs.azureedge.net/1.0.16/survey.jquery.min.js"></script>
<script>
var json = {};

window.survey = new Survey.Model(json);

//To test if ajax call works and it does
var values = ["<?php echo $name; ?>", "<?php echo $car; ?>"];

//To send data directly to SurveyJS
values = ["John", "somecar"];

var deepSurveyCopy = $.extend(true, {}, survey);
console.log(deepSurveyCopy);
deepSurveyCopy.data.name = values[0];   
deepSurveyCopy.data.car = values[1];    
deepSurveyCopy.sendResult(my access key to surveyJS);
</script>

<?php
echo "Welcome ". $name. "<br />";
echo "You want ". $car;
?>

以下是Chrome和SurveyJS的屏幕截图： AjaxOriginator.html to receiver.php Calling receiver.php directly Data received from calling receiver.php directly

我不是非常精通PHP，并且非常感谢您帮助理解所有这些工作（而不是工作）。

基本上ajaxOriginator.html是模拟一个插件，它将把数据发送到receiver.php，我试图用SurveyJS接受的格式包装收到的数据并重新发送到SurveyJS。

非常感谢任何帮助。谢谢！

直接从PHP发送请求，但从AJAX调用PHP发送请求不起作用

0 个答案: