我认为 CI 返回所有页面,我需要字符串“GOOD”。
这是一张图片。http://d.pr/i/h4oi 请帮帮我)这是我对 Codeigniter 的看法:
<html>
<head>
<script src="<?php echo base_url(); ?>js/jquery.js" type="text/javascript"></script>
</head>
<body>
<input type="text" id="username" />
<input type="button" id="check" value="Check" />
<div id="name_feedback"></div>
<script>
$("#check").click(function() {
$.post("<?php echo base_url(); ?>main/about", { name : $("#username").val() }, function(data) {
$("#name_feedback").html(data);
console.log(data.length);
} );
} );
</script>
</body>
</html>
控制器:
function about()
{
if (isset($_POST['name'])) {
$name = $_POST['name'];
$this->db->where('username', $name);
$users = $this->db->get('users');
if ($users->num_rows() > 0) {
echo "Bad";
} else
{
echo "Good";
}
}
$this->load->view('about_view');
}
Codeigniter 返回#name_feedback 中的所有代码页,并回答“好或坏”
这是控制台日志:
Good<script src="http://localhost/project/js/jquery.js" type="text/javascript"></script>
<html>
<head>
<script src="http://localhost/project/js/jquery.js" type="text/javascript"></script>
</head>
<body>
<input type="text" id="username" />
<input type="button" id="check" value="Check" />
<div id="name_feedback"></div>
<script>
$("#check").click(function() {
$.post("http://localhost/project/main/about", { name : $("#username").val() }, function(data) {
$("#name_feedback").html(data);
console.log(data);
} );
} );
</script>
</body>
</html>