我是jQuery的新手,正在尝试从我发现的使用AJAX的网站上复制一个示例。我需要获取PHP变量并将其传递到另一个PHP页面,而不刷新整个页面。我已经想出了下面的代码,但我得到了一个语法错误
Uncaught SyntaxError: Unexpected token )
这是我的代码:
$("Button").click(function(){
var textBox = $("textBox").val();
$.post("script.php",
{textBox: textBox},
function(data,status)
)};
)};
我的PHP页面:
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<head>
<script type='text/javascript' src='script.js'></script>
<script type='text/javascript' src='http://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js'></$
<title>LCD Display</title>
</head>
<body>
<form id="lcd" action="script.php" method="post">
<input name='textBox' type='text' />
<input type='submit' id= 'Button' value='Press to see something cool'/>
</form>
<div id='ResponseDiv'>
</div>
<iframe id="video" src="http://192.168.0.11:8081" height="120" width="160"></iframe>
</body>
</html>
我也得到了以下错误,但我认为它们可能与语法错误有关:
Resource interpreted as Document but transferred with MIME type multipart/x-mixed-replace: "http://192.168.0.11:8081/".
Resource interpreted as Document but transferred with MIME type image/jpeg: "http://192.168.0.11:8081/".
您的结束标签:
)};
应该是这样的:
$("Button").click(function(){
var textBox = $("textBox").val();
$.post("script.php",
{textBox: textBox},
function(data,status){
});
});
脚本顺序是这样的,注意关闭</$
:
<head>
<script src='http://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js'>
</script>
<script type='text/javascript' src='script.js'></script>
<title>LCD Display</title>
$.post("script.php",
{textBox: textBox},
function(data,status){}
);
另请阅读:http://api.jquery.com/jQuery.post/