Question

ajax : function(typ,url,callback) {

            if(window.XMLHttpRequest) {
                var xml = new XMLHttpRequest();
            }

            if(window.ActiveXObject) {
                var xml = new ActiveXObject("Microsoft.XMLHTTP");
            }
            xml.onreadystatechange = function(callback) {

            if(xml.readyState == 4 && xml.status == 200) {
                    callback();
            }
        }
            xml.open(typ,url,true);
            xml.send();
    }

  }

 //Function being called
window.onload = function() {    
    JS.ajax("GET","/server/chkEmail.php?email=email@email.com",function()                       { 
        alert(xml.responseText);
    });
}

它抛出错误说：

未捕获TypeError：回调不是functionxml.onreadystatechange @ global.js：30

有什么想法吗？

我通过onreadystatechange函数传递数据变量并在回调函数中调用它来修复它。

ajax : function(typ,url,callback) {

            if(window.XMLHttpRequest) {
                var xml = new XMLHttpRequest();
            }

            if(window.ActiveXObject) {
                var xml = new ActiveXObject("Microsoft.XMLHTTP");
            }

            xml.onreadystatechange = function(data) {

            if(xml.readyState == 4 && xml.status == 200) {
                    var data = xml.responseText;
                    callback(data);
            }
        };
            xml.open(typ,url,true);
            xml.send();
    }

}

window.onload = function() {    
    JS.ajax("GET","/server/chkEmail.php?  email=jonwcode@gmail.com",function(data){
        alert(data);
    });
}

Answer 1

试试这样：

xml.onreadystatechange = function() {
    if (xml.readyState == 4 && xml.status == 200) {
        callback();
    }
};

请注意，onreadystatechange函数不带任何参数，而在代码中，您传递了一个名为callback的参数，该参数将覆盖callback变量。外部范围。

更新：

看起来你没有正确确定xml变量的范围，并且它在你的AJAX回调中不可用。我强烈建议您阅读有关javascript变量范围的更多信息。以下是如何使xml变量可见：

var xml = null;
if (window.XMLHttpRequest) {
    xml = new XMLHttpRequest();
} else if (window.ActiveXObject) {
    xml = new ActiveXObject("Microsoft.XMLHTTP");
}

if (xml == null) {
    alert('Sorry, your browser doesn\'t seem to support AJAX - please upgrade to a modern browser');
} else {
    xml.onreadystatechange = function() {
        if(xml.readyState == 4 && xml.status == 200) {
            var data = xml.responseText;
            callback(data);
        }
    };

    xml.open(typ,url,true);
    xml.send();
}

回调函数在调用时不会运行

1 个答案: