Question

嘿伙计们，当我试图通过ajax发送两个变量时，我遇到了问题我做了与此链接中使用的相同的代码w3schools http://www.w3schools.com/ajax/tryit.asp ... 但它只适用于我只使用一个变量!! 我的代码：

<script>
function showHintname() {
    var ln = document.getElementById("lname").value ;
    var fn= document.getElementById("fname").value ;
    if (fn.length == 0) { 
        document.getElementById("txtHint-name").innerHTML = "";
        return;
    }

    else {
        var xmlhttp = new XMLHttpRequest();
        xmlhttp.onreadystatechange = function() {
            if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
                document.getElementById("txtHint-name").innerHTML = xmlhttp.responseText;
            }
        }
        //xmlhttp.open("GET", "ini/name.php?f=" + fn , true); // working

        // xmlhttp.open("GET", "ini/name.php?f=" + fn &'l=' + ln , true); // not working
        xmlhttp.open("GET", "ini/name.php?f=1&l=1", true); // not working
        xmlhttp.send();
    }
}
</script>

Answer 1

$.ajax({
  method: ,
  url: ,
  data: { var1: data1, var2: data2 }
}, done(function(data) {
  //Do some actions
});

Answer 2

使用它会起作用

 xmlhttp.open("GET", "ini/name.php?f=" + fn+ "&l=" + ln , true);

Answer 3

xmlhttp.open("GET", "ini/name.php?f=" + fn &'l=' + ln , true);

这句话不起作用，因为＆amp;符号不在引号内。您还应该作为一般惯例，尝试与单/双引号保持一致。这应该工作

xmlhttp.open("GET", "ini/name.php?f=" + fn+ "&l=" + ln , true);

Answer 4

<script>
 function showHintname() {
 var ln = document.getElementById("lname").value 
 var fn= document.getElementById("fname").value ;
 if (fn.length == 0) { 
   document.getElementById("txtHint-name").innerHTML = "";
   return;
 }

 else {
  var xmlhttp = new XMLHttpRequest();
  xmlhttp.onreadystatechange = function() {
  if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
   document.getElementById("txtHint-name").innerHTML = xmlhttp.responseText;
  }
 }
 //the url
 var url = "ini/name.php?f=" + fn + "&l=" + ln;

 xmlhttp.open("GET", url, true);

 xmlhttp.send(null);
}
}

通过ajax发送两个变量

4 个答案: