我想在Chrome扩展程序中使用ajax将数据插入服务器,以便我的用户可以注册使用我的扩展程序。 我拼命寻找解决方案而不是找不到例子
当我在popup.html上编码ajax时,我收到了这个错误 “拒绝执行内联脚本,因为它违反了以下内容安全策略指令:”script-src'self''unsafe-eval'“。'unsafe-inline'关键字,一个哈希('sha256-lXzxt12nj + 7ATE1j5ucGnYo1VkLZpNS / cGA9SL9nCv0 ='),或启用内联执行所需的随机数('nonce -...')。“
popup.html代码
<!DOCTYPE html>
<html>
<head>
<style>
body {
width:300px;
font-size:12px;
}
</style>
</head>
<body>
<form action="" method="POST">
<input type="text" id="name" name="uname">
<input type="text" id="pass" name="upass">
<input type="submit" id="submit" name="submit" id="btn">
</form>
</body>
</html>
<script type="text/javascript" src="check.js"></script>
<script type="text/javascript">
$("#btn").click(function(){
var btn =$('#btn').val();
var name= $('#name').val();
var pas = $('#pass').val();
$.ajax({
data :{btn: btn, name : name, pas: pas},
method="POST",
url:"http://localhost/status/index.php",
success : function(data){
lb.val(data);
}
});
});
}
</script>
服务器上的index.php代码
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "test";
$conn = mysqli_connect($servername, $username, $password, $dbname);
if(isset($_POST['byn'])){
$localIP = getHostByName(getHostName());
$name = $_POST['name'];
$pass = $_POST['pass'];
$insert = mysqli_query($conn,"INSERT INTO usercheck VALUES('','$name','$pass','1111,'$localIP')");
}
?>