PHP文件仅发送"parsererror"消息，通过AJAX [英] PHP files sending only a "parsererror" through AJAX
问题描述
解决方案:在与@skobaljic进行teamviewer会话后，他发现我实际上不是在localhost中打开html，而是使用文件系统(如file://...).我为浪费大家的时间而道歉.
SOLUTION: After a teamviewer session with @skobaljic he figured out that I was not actually opening the html in localhost but using the file system (as in file://...). I apologize for wasting everyone's time like that.
我正在尝试通过Ajax发送一些php数组并打印出来，尽管得到200 OK响应，但是接收到的html中没有实际效果.状态文本显示为"parsererror".
I am trying to send some php arrays through Ajax and print them out and despite getting a 200 OK response, there is no actual effect in the received html. The status text says "parsererror".
文件如下:
<!DOCTYPE html>
<html>
<head>
<title>Page Title Woo!</title>
<script type='text/javascript' src='http://ajax.googleapis.com/ajax/libs/jquery/2.2.4/jquery.min.js'></script>
</head>
<body>
<h1>Heading</h1>
<p>Paragraph.</p>
<ul></ul>
<script type='text/javascript'>
$(document).ready(function(){
$.getJSON('DbGetter.php', function(data) {
console.log(data);
$.each(data, function(key, array) {
$('ul').append('<li id="' + key + '">'
+ array.longitude + ' '
+ array.latitude + '</li>');
});
})
.fail(function(error) {
console.log(error);
});
});
</script>
</body>
</html>
和php:
<?php
$servername = "localhost";
$username = "testuser";
$password = "password";
$dbname = "Locations";
header('Content-Type: application/json');
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM places";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
//declare associative array
$array = array();
$num = 0;
// output data of each row
while($row = $result->fetch_assoc()) {
//store them in an array
$place = array(
'id' => $row["id"],
'latitude'=> $row["latitude"] ,
'longitude'=> $row["longitude"],
'place_name' => $row["place_name"],
'country_code'=> $row["country_code"],
'postal_code'=> $row["postal_code"]);
/*
echo "Coordinates: " . $row["latitude"]. " " . $row["longitude"]. " - Name: " . $row["place_name"]. " " . "<br>";
*/
//building the second associative array
$array[$num] = $place;
$num += 1;
}
echo json_encode($array);
} else {
echo json_encode("0 results");
}
$conn->close();
?>
我尝试通过Firebug来查看值，但是我是盲目的还是没有将其存储在DOM中的任何位置.一般来说，我对Web应用程序还很陌生，所以我不知道如何调试它.
I've tried looking at the value through firebug, but either I'm blind or it's just not stored anywhere in the DOM. I'm pretty new to web application in general so I don't know how to go about debugging it.
我自己运行php，我得到:[{"id":"1","latitude":"57.0502","longitude":"9.9173","place_name":"1000fryd","country_code":"DK","postal_code":"9000"},{"id":"2","latitude":"58.0502","longitude":"10.9173","place_name":"same_place","country_code":"DK","postal_code":"9000"}]
我期望的是两行.
Running the php by itself I get: [{"id":"1","latitude":"57.0502","longitude":"9.9173","place_name":"1000fryd","country_code":"DK","postal_code":"9000"},{"id":"2","latitude":"58.0502","longitude":"10.9173","place_name":"same_place","country_code":"DK","postal_code":"9000"}]
Which are the 2 rows I expect.
也没有Firebug标记的XHR请求.
There is also no XHR request marked by Firebug.
推荐答案
您在此处未对有效的JSON进行编码.如果尝试将其与JQuery的$ getJSON结合使用，则会给您"parseerror".请注意第275-281行中的 JQuery Ajax源.
You're not encoding valid JSON here.. This will give you a "parseerror" if you try using JQuery's $getJSON with it. Notice the JQuery Ajax Source from lines 275-281.
echo json_encode("0 results");
您可能可以将其更改为:
You could probably change it to something like:
echo json_encode(["0 results"]);
这篇关于PHP文件仅发送"parsererror"消息，通过AJAX的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
