[英]AJAX POST request is failing
Apologies for the generic title. 对通用标题表示歉意。
Essentially, when the script runs 'error' is alerted as per the jQuery below. 本质上,当脚本运行时,将按照以下jQuery发出“错误”警报。 I have a feeling this is being caused by the structuring of my JSON, but I'm not sure how I should change it.
我感觉这是由JSON的结构引起的,但是我不确定如何更改它。
The general idea is that there are several individual items, each with their own attributes: product_url
, shop_name
, photo_url
, was_price
and now_price
. 通常的想法是,有几个单独的项目,每个项目都有自己的属性:
product_url
, shop_name
, photo_url
, was_price
和now_price
。
Here's my AJAX request: 这是我的AJAX请求:
$.ajax(
{
url : 'http://www.comfyshoulderrest.com/shopaholic/rss/asos_f_uk.php?id=1',
type : 'POST',
data : 'data',
dataType : 'json',
success : function (result)
{
var result = result['product_url'];
$('#container').append(result);
},
error : function ()
{
alert("error");
}
})
Here's the PHP that generates the JSON : 这是生成JSON的PHP:
<?php
function scrape($list_url, $shop_name, $photo_location, $photo_url_root, $product_location, $product_url_root, $was_price_location, $now_price_location, $gender, $country)
{
header("Access-Control-Allow-Origin: *");
$html = file_get_contents($list_url);
$doc = new DOMDocument();
libxml_use_internal_errors(TRUE);
if(!empty($html))
{
$doc->loadHTML($html);
libxml_clear_errors(); // remove errors for yucky html
$xpath = new DOMXPath($doc);
/* FIND LINK TO PRODUCT PAGE */
$products = array();
$row = $xpath->query($product_location);
/* Create an array containing products */
if ($row->length > 0)
{
foreach ($row as $location)
{
$product_urls[] = $product_url_root . $location->getAttribute('href');
}
}
$imgs = $xpath->query($photo_location);
/* Create an array containing the image links */
if ($imgs->length > 0)
{
foreach ($imgs as $img)
{
$photo_url[] = $photo_url_root . $img->getAttribute('src');
}
}
$was = $xpath->query($was_price_location);
/* Create an array containing the was price */
if ($was->length > 0)
{
foreach ($was as $price)
{
$stripped = preg_replace("/[^0-9,.]/", "", $price->nodeValue);
$was_price[] = "£".$stripped;
}
}
$now = $xpath->query($now_price_location);
/* Create an array containing the sale price */
if ($now->length > 0)
{
foreach ($now as $price)
{
$stripped = preg_replace("/[^0-9,.]/", "", $price->nodeValue);
$now_price[] = "£".$stripped;
}
}
$result = array();
/* Create an associative array containing all the above values */
foreach ($product_urls as $i => $product_url)
{
$result = array(
'product_url' => $product_url,
'shop_name' => $shop_name,
'photo_url' => $photo_url[$i],
'was_price' => $was_price[$i],
'now_price' => $now_price[$i]
);
echo json_encode($result);
}
}
else
{
echo "this is empty";
}
}
/* CONNECT TO DATABASE */
$dbhost = "xxx";
$dbname = "xxx";
$dbuser = "xxx";
$dbpass = "xxx";
$con = mysqli_connect("$dbhost", "$dbuser", "$dbpass", "$dbname");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$id = $_GET['id'];
/* GET FIELDS FROM DATABASE */
$result = mysqli_query($con, "SELECT * FROM scrape WHERE id = '$id'");
while($row = mysqli_fetch_array($result))
{
$list_url = $row['list_url'];
$shop_name = $row['shop_name'];
$photo_location = $row['photo_location'];
$photo_url_root = $row['photo_url_root'];
$product_location = $row['product_location'];
$product_url_root = $row['product_url_root'];
$was_price_location = $row['was_price_location'];
$now_price_location = $row['now_price_location'];
$gender = $row['gender'];
$country = $row['country'];
}
scrape($list_url, $shop_name, $photo_location, $photo_url_root, $product_location, $product_url_root, $was_price_location, $now_price_location, $gender, $country);
mysqli_close($con);
?>
The script works fine with this much simpler JSON: 该脚本可以使用以下简单得多的JSON正常运行:
{"ajax":"Hello world!","advert":null}
You are looping over an array and generating a JSON text each time you go around it . 您正在遍历数组, 每次遍历数组时都会生成JSON文本。
If you concatenate two (or more) JSON texts, you do not have valid JSON. 如果串联两个(或更多)JSON文本,则没有有效的JSON。
Build a data structure inside the loop. 在循环内部构建数据结构。
json_encode
that data structure after the loop. json_encode
循环后的数据结构。
If i have to guess you are echoing multiple json strings which is invalid. 如果我不得不猜测您正在回显无效的多个json字符串。 Here is how it should work:
这是应该如何工作的:
$result = array();
/* Create an associative array containing all the above values */
foreach ($product_urls as $i => $product_url)
{
// Append value to array
$result[] = array(
'product_url' => $product_url,
'shop_name' => $shop_name,
'photo_url' => $photo_url[$i],
'was_price' => $was_price[$i],
'now_price' => $now_price[$i]
);
}
echo json_encode($result);
In this example I am echoing the final results only once. 在此示例中,我仅回显最终结果一次。
You are sending post request but not sending post data using data 您正在发送发布请求,但未使用data发送发布数据
$.ajax(
{
url : 'http://www.comfyshoulderrest.com/shopaholic/rss/asos_f_uk.php?id=1',
type : 'POST',
data : {anything:"anything"}, // this line is mistaken
dataType : 'json',
success : function (result)
{
var result = result['product_url'];
$('#container').append(result);
},
error : function ()
{
alert("error");
}
})
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