当我在我的笔记本电脑上使用我的本地主机(MAMP)时,请求工作得很好但是当我将我的Web应用程序移动到另一台服务器时,它似乎无法返回error: function() { console.log("oh no");
。代码如下:
var backendURL = "http://dayout.uphero.com/dayout-app/php/";
.factory('AttractionService', function() {
// Returns a JSON array
var attractions;
return {
all: function() {
jQuery.ajax( backendURL + "attractions_summary.php", {
async: false,
cache: false,
error: function() { console.log("oh no"); },
success: function(oData) {
attractions = oData;
}
});
return attractions;
}
}
})
这是与服务器连接并返回JSON数组的attractions_summary.php文件:
<?php
include "../../config.php";
header('Content-type: application/json');
header('Access-Control-Allow-Origin: *');
$summary = mysql_query ('SELECT attraction.attraction_name AS NAME, attraction.image_name AS IMAGE,
attraction.attraction_price AS PRICE, attraction.id AS ID, bus.b_name AS BUSINESS FROM Attractions AS attraction
INNER JOIN BusinessInfo AS bus ON attraction.user_id = bus.user_id ORDER BY attraction.id DESC');
$result_formatted = array();
$i = 0;
while($result = mysql_fetch_array($summary)) {
$result_formatted[$i] = array();
$result_formatted[$i]['id'] = mysql_real_escape_string($result["ID"]);
$result_formatted[$i]['attraction_name'] = mysql_real_escape_string($result["NAME"]);
$result_formatted[$i]['attraction_price'] = mysql_real_escape_string($result["PRICE"]);
$result_formatted[$i]['bname'] = mysql_real_escape_string($result["BUSINESS"]);
$result_formatted[$i]['image_name'] = mysql_real_escape_string($result["IMAGE"]);
$i++;
}
echo json_encode($result_formatted);
?>
答案 0 :(得分:1)
[{"id":"4","attraction_name":"gerfwedq","attraction_price":"324","bname":"Texaco Garage","image_name":"icon-6013953941166125.png"},{"id":"1","attraction_name":"Starbucks Attraction Day!","attraction_price":"175.99","bname":"Starbucks","image_name":"footprints.png"}]
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这是你网址的输出......
删除以下
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我认为这是通过自动托管添加的