处理针对JSON数据的PHP URL的AJAX请求。(Handle AJAX request to PHP url for JSON data. I cant pass the data)

 我试图将我的数据从mySQL数据库传递给returnData.PHP作为一个AJAX请求作为JSON数据传递到.js文件中的javascript。 到目前为止，我可以从mysql查询中检索我的数据并将其编码为json就好了。 以下是returnData.php 
 
<?php
  include('../sqlFunctions/sqlFunctions.php');

//Establish connection to database
  $link = linkDB();

//set up a MySQL query
  $sql = "SELECT * FROM shelters;";

  if(!$results = $link->query($sql)){
    die("Query Unsuccessful");
  }

  $rows = array();

  while ($data = $results->fetch_assoc()) {
    $rows[] = $data;
  }
  $JSONRows = json_encode($rows);
  return $JSONRows;
?>  
 
 我使用AJAX从javascript函数调用此页面，需要传递这些数据。 
 
  <script type="text/javascript">

    function getData(){
      var dataXMLhttp = new XMLHttpRequest();
      dataXMLhttp.open("GET", "./js/returnData.php", true);
      dataXMLhttp.send();
      if(dataXMLhttp.readyState == 4 && dataXMLhttp.status == 200){
        var XMLdataResult = dataXMLhttp.responseText;
        window.alert("XMLdataResult Contains something.");
      }else{
        window.alert("404");
      }
    }getData();
</script>
 
 我收到了404消息。 显然我的AJAX请求被破坏了。 返回$ JSONow显然显示了我的noobness。 我已经查看并阅读了有关AJAX与PHP的交互，但是看不到我想念的那些拼图。 
 
 我的AJAX请求有什么问题？ 如何将我的JSON转换为PARSing的javascript变量？ 
 
 谢谢阅读。 

Im trying to pass my data from mySQL database to returnData.PHP to an AJAX request as JSON data to javascript in a .js file. So far I can retrieve my data from a mysql query and encode it into json just fine. Following is returnData.php 
 
<?php
  include('../sqlFunctions/sqlFunctions.php');

//Establish connection to database
  $link = linkDB();

//set up a MySQL query
  $sql = "SELECT * FROM shelters;";

  if(!$results = $link->query($sql)){
    die("Query Unsuccessful");
  }

  $rows = array();

  while ($data = $results->fetch_assoc()) {
    $rows[] = $data;
  }
  $JSONRows = json_encode($rows);
  return $JSONRows;
?>  
 
I call this page from a javascript function using AJAX and need to hand off this data. 
 
  <script type="text/javascript">

    function getData(){
      var dataXMLhttp = new XMLHttpRequest();
      dataXMLhttp.open("GET", "./js/returnData.php", true);
      dataXMLhttp.send();
      if(dataXMLhttp.readyState == 4 && dataXMLhttp.status == 200){
        var XMLdataResult = dataXMLhttp.responseText;
        window.alert("XMLdataResult Contains something.");
      }else{
        window.alert("404");
      }
    }getData();
</script>
 
I am recieving the 404 message. Apparently my AJAX request is broken. The return $JSONow obviously shows my noobness. I have looked and read about AJAX intereacting with PHP, but can't see the pieces of the puzzle I am missing. 
 
What is wrong with my AJAX request? How can I get my JSON into a javascript variable for PARSing?
 
Thanks for reading.

原文：https://stackoverflow.com/questions/44743957