我相信我没有完全掌握jQuery中的变量范围,并且需要一些帮助来理解为什么变量“final_shasign”在AJAX内部表现良好,但在AJAX外部未定义,尽管它已被声明在AJAX之外。Javascript变量不通过ajax更新
var final_shasign; // initial declaration outside of ajax block below
$.ajax({
type: "POST",
url: "sha.php",
data: "amount="+amount+"&cn="+cn+"¤cy="+currency+"&language="+language+"&orderid="+orderid+"&pspid="+pspid,
success: function(response_data, final_shasign){
var parsed_data = $.parseJSON(response_data);
final_shasign = parsed_data.shasign;
console.log ("inside nested ajax: " + final_shasign); //comes out fine
}
});
console.log ("outside of nested ajax: " + final_shasign); //comes out as undefined!
AJAX异步...的 –
可能重复[如何返回从AJAX调用的响应?](http://stackoverflow.com/questions/14220321 /如何对返回的响应-从-AN-Ajax的呼叫) –