如果我理解正确,您希望稍后在代码中重用 ajax 响应。如果是这种情况,您当前的代码将无法工作,因为默认情况下,javascript 引擎不会等待 ajax 请求的响应。换句话说,下面的代码将不起作用:
<script type="text/javascript">
$(document).ready(function(){
var x;
var y;
$.ajax({
url: 'ajaxload.php',
dataType: "json",
success: function(data) {
x= data.posX;
y= data.posX;
alert (x+" "+y); // I can se data but I need outside ajax call
}
});
alert(x+" "+y); // You won't see anything, because this data isn't yet populated. The reason for this is, the "success" function is called when the ajax request has finished (it has received a response).
})
</script>
您需要等待 ajax 响应。要使用 jQuery 做到这一点,您需要稍微修改您的代码:
<script type="text/javascript">
$(document).ready(function(){
var data = $.parseJSON($.ajax({
url: 'ajaxload.php',
dataType: "json",
async: false
}).responseText); // This will wait until you get a response from the ajax request.
// Now you can use data.posX, data.posY later in your code and it will work.
var x = data.posX;
var y = data.posY;
alert(x+" "+y);
alert(data.posX+" "+data.posY);
});
</script>