javascript - 在函数和对象之外设置变量

Question

这个问题已经有答案了:

How do I return the response from an asynchronous call? (42 个回答)

已关闭 10 年前。

我正在尝试设置变量 result但它仍然未定义。

        if (window.XMLHttpRequest)
        {// code for IE7+, Firefox, Chrome, Opera, Safari
            xmlhttp=new XMLHttpRequest();
        }
        else
        {// code for IE6, IE5
            xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
        }
        xmlhttp.onreadystatechange=function()
        {
            if(xmlhttp.readyState == 4 && xmlhttp.status == 200)
            {
                if(xmlhttp.responseText)
                {
                    result = true
                    alert(xmlhttp.responseText+" here1")
                }
                else
                {
                    document.getElementById("report").innerHTML = "wrong password/username"
                    alert(xmlhttp.responseText+" here2")
                    result = false
                }
            }
        }
        xmlhttp.open("POST", "process.php", true);
        xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
        xmlhttp.send("name="+encodeURIComponent(name.value));
        alert("final value: "+result)
        return result//always undefined

如何让对象影响其所在函数之外的变量？这个问题比我说的要复杂一些。当用户尝试提交表单时调用此函数。如果返回 true，则应提交表单，但如果返回 false，则不应提交表单。我现在有了代码(谢谢甜淀粉 enzyme )

   var result
   var checkResult = function(result) {
         alert("final value: " + result);
         return result
   };

    xmlhttp.onreadystatechange=function()
    {
        var result = null;
        if(xmlhttp.readyState == 4 && xmlhttp.status == 200)
        {
            if(xmlhttp.responseText)
            {
                result = true
                alert(xmlhttp.responseText+" here1")
            }
            else
            {
                document.getElementById("report").innerHTML = "wrong password/username"
                alert(xmlhttp.responseText+" here2")
                result = false
            }
            checkResult(result);     // pass result to another function
        }
    }
    xmlhttp.open("POST", "process.php", true);
    xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    xmlhttp.send("name="+encodeURIComponent(name.value));

但表单始终会提交，甚至在显示“最终值...”之前也是如此。如果我在代码末尾添加 return false ，则表单永远不会提交。

Answer 1

AJAX 是异步的，因此您需要修改 onreadystatechange() 函数，您可以将 result 传递给另一个函数来处理您的逻辑:

   var checkResult = function(result) {
         alert("final value: " + result);
         /* handle your result here */
   };
   ...

    xmlhttp.onreadystatechange=function()
    {
        var result = null;
        if(xmlhttp.readyState == 4 && xmlhttp.status == 200)
        {
            if(xmlhttp.responseText)
            {
                result = true
                alert(xmlhttp.responseText+" here1")
            }
            else
            {
                document.getElementById("report").innerHTML = "wrong password/username"
                alert(xmlhttp.responseText+" here2")
                result = false
            }
            checkResult(result);     // pass result to another function
        }
    }
    xmlhttp.open("POST", "process.php", true);
    xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    xmlhttp.send("name="+encodeURIComponent(name.value));