xmlhttprequest没有responseText

Question

我尝试使用xmlhttprequest向php响应器发送对象，运行搜索查询，然后将结果作为对象返回，但由于某种原因没有结果。我可以在network选项卡下看到响应者生成了所需的记录，但该记录未被处理。我无论如何也看不出这个问题。

请求：

​

function returnJSON(variable, URL, callback) {
  var ajaxObj = new XMLHttpRequest();
  ajaxObj.open("POST", URL, true);
  console.log("posting");
  ajaxObj.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
  ajaxObj.onreadystatechange = function() {
    if (ajaxObj.status === 200)
      if (ajaxObj.readyState === 4)
        callback(JSON.parse(ajaxObj.responseText));
        console.log(ajaxObj.responseText);
      };
  ajaxObj.send(variable);
}

​

使用所述回调的示例函数。

​

function getSearch(e){
  e.preventDefault();
  var prodCode = document.getElementById("productCode").value;

  var productDetails = "productCode="+prodCode;
  returnJSON(
    productDetails,
    'api/database/returnSearch.php',
    function(data) {
      getSearchResult(data);
      console.log("working");

    }
  );

  if (searchResults.length > 1) {
      alert("There are too many results to display.");
  }else if (searchResults.length = 0){
     alert("There are no results for "+prodCode);

  }
  else if (searchResults.length > 0){
    document.getElementById("pCode").value = searchResults[0][0];
    document.getElementById("productN").value = searchResults[0][1];
    document.getElementById("description").value = searchResults[0][2];
    document.getElementById("productType").value = searchResults[0][3];
    document.getElementById("price").value = searchResults[0][4];
    document.getElementById("quantity").value = searchResults[0][5];

  }
  // setAdmin();

}

​

有问题的php响应器。

​

<?php
global $range;
$range = [];
$hostname = 'localhost';

/*** mysql username ***/
$username = 'root';

/*** mysql password ***/
$password = '';
if(isset($_POST['productCode'])){
    $prodCode= $_POST['productCode'];
    $productCode = NULL;
    $prodName = NULL;
    $desc = NULL;
    $prodType = NULL;
    $price = NULL;
    $quantity = NULL;
    $db = new PDO("mysql:host=$hostname;dbname=webcw", $username, $password);
    $sql = "SELECT productCode, productName, productType, description, price, quantity
            FROM product
            WHERE productCode = '$prodCode';";
    foreach ($db->query($sql) as $row) :

      $productCode = $row ['productCode'];
      $prodName = $row ['productName'];
      $desc = $row ['description'];
      $prodType = $row ['productType'];
      $price = $row ['price'];
      $quantity = $row ['quantity'];
      $product = array($prodCode, $prodName, $desc, $prodType, $price, $quantity);
      $range[] = $product;
    endforeach;

    echo json_encode($range);
    $db = null;
  }
  ?>

​

你们有人能帮上忙吗？

提前感谢您的帮助。

Answer 1

替换以下代码行：

if (ajaxObj.status === 200)
  if (ajaxObj.readyState === 4)
    callback(JSON.parse(ajaxObj.responseText));
    console.log(ajaxObj.responseText);

有了这些：

if (ajaxObj.readyState === 4) {
   if (ajaxObj.status === 200){
      callback(JSON.parse(ajaxObj.responseText));
      console.log(ajaxObj.responseText);
   }
}

Answer 2

我设法让它工作起来

​

function returnJSON(variable, URL, callback) {
  var ajaxObj = new XMLHttpRequest();
  ajaxObj.open("POST", URL, true);
  console.log("posting");
  ajaxObj.setRequestHeader("Content-type", "application/x-www-form-urlencoded");

  ajaxObj.addEventListener("load", 
    function() {
        console.log("received");
        callback(JSON.parse(ajaxObj.responseText));
    }
  );
  ajaxObj.send(variable);
}

​