如何将四个下拉列表中的四个值或四个参数传递给ajax?

时间:2018-11-08 09:54:46

标签: javascript ajax

我试图将4个参数从下拉列表传递给Ajax,但出现以下错误:

  

未捕获的TypeError:无法在以下位置设置null的属性“ innerHTML”   XMLHttpRequest.xmlhttp.onreadystatechange

function showreports() {
    var str = document.getElementById("p").value;
    var str1 = document.getElementById("t").value;
    var str2 = document.getElementById("sem").value;
    var str3 = document.getElementById("ses").value;


    if (str == 0) {
        document.getElementById("txtHint").innerHTML = "";
        return;
    } else {
        var xmlhttp = new XMLHttpRequest();
        xmlhttp.onreadystatechange = function () {
            if (this.readyState == 4 && this.status == 200) {
                document.getElementById("txtHint").innerHTML = this.responseText;
            }
        };
        xmlhttp.open("GET", "get_reports.php?p=" + str, true);
        xmlhttp.send();
    }
}

0 个答案:

没有答案