JSON未在PHP的javascript中解析
我是PHP新手,我是PHP的ajax,我从服务器获得结果,但我的JSON不可解析 以下是我的PHP代码:JSON未在PHP的javascript中解析,javascript,php,json,Javascript,Php,Json,我是PHP新手,我是PHP的ajax,我从服务器获得结果,但我的JSON不可解析 以下是我的PHP代码: <?php header("Access-Control-Allow-Origin: *"); header('Content-type: application/json'); require 'connection.php'; $ReturnObject = (object) [ 'error' => false, 'errorMessage' =>
<?php
header("Access-Control-Allow-Origin: *");
header('Content-type: application/json');
require 'connection.php';
$ReturnObject = (object) [
'error' => false,
'errorMessage' => "",
'data' => "1"
];
echo json_encode($ReturnObject);
?>
我得到这个错误:
未捕获的语法错误:意外标记 在JSON中的位置0
但数据是:
"{"error":false,"errorMessage":"","data":"1"}"
现在过了一段时间,我搬走了
require 'connection.php';
并将代码粘贴为:
<?php
header("Access-Control-Allow-Origin: *");
header('Content-type: application/json');
$servername = "localhost";
$database = "testMe";
$username = "Mika";
$password = "123123";
$conn = mysqli_connect($servername, $username, $password, $database); // Establishing Connection with Server..
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$ReturnObject = (object) [
'error' => false,
'errorMessage' => "",
'data' => "1"
];
echo json_encode($ReturnObject);
?>
我的要求有问题吗??这不是连接的问题,因为它成功了,我可以查询数据库,应该没有问题,除了出于某种原因需要'connection.php'打印一些错误,尝试console.log(data),并确保连接错误立即以JSON格式打印(“连接失败:…将只打印一个字符串。因此javascript无法将其解析为JSON对象。在开头
之前是否有任何内容?这不是一个空白,在帖子中,代码中和“token”之后有\ufeff
字符(零宽度无中断空间)。您可以在使用适当的文本编辑器时看到它,或者确保将文件保存为“UTF-8无BOM”编码。\ufeff
…这就是从visual studio粘贴时的BOM
-感谢MS的帮助:|删除它,您应该会很好。@CD001是的,您是正确的,当从visual studio粘贴到服务器上的文件编辑器时,它会在文件开头自动添加\ufeff
字符。现在我移除它,一切正常。谢谢
<?php
header("Access-Control-Allow-Origin: *");
header('Content-type: application/json');
$servername = "localhost";
$database = "testMe";
$username = "Mika";
$password = "123123";
$conn = mysqli_connect($servername, $username, $password, $database); // Establishing Connection with Server..
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$ReturnObject = (object) [
'error' => false,
'errorMessage' => "",
'data' => "1"
];
echo json_encode($ReturnObject);
?>
"{"error":false,"errorMessage":"","data":"1"}"