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Question

我正在尝试提交表单而不进行硬刷新，但我的页面很新鲜。我尝试将action=“javascript:void(0);"放在表单标记中，但这不起作用。我也不能使用jquery。

导致错误的工作流程：

提交一份应该将某些信息发送到php文件的表单。

页面被定向到另一个不存在的页面。

    function toggle_visibility(id){
        var e = document.getElementById(id);

        if(e.style.visibility == 'hidden'){
            e.style.visibility = 'visible';
        }
        else{
            e.style.visibility = 'hidden';
        }

    }
    function createAjaxRequestObject() {
        var httpRequest;
        if (window.XMLHttpRequest) { // Mozilla, Safari, ...
            httpRequest = new XMLHttpRequest();
            if (httpRequest.overrideMimeType) {
                httpRequest.overrideMimeType('text/xml');
            }
        }
        else if (window.ActiveXObject) { // IE
            try {
                httpRequest = new ActiveXObject("Msxml2.XMLHTTP");
                }
            catch (e) {
                try {
                    httpRequest = new ActiveXObject("Microsoft.XMLHTTP");
                }
                catch (e) {}
            }
        }
        if (!httpRequest) {
            alert('Giving up :( Cannot create an XMLHTTP instance');
            return false;
        }
        // Create the object
        return httpRequest;
    }

    function sendTicket() {
        var firstName = document.getElementById("firstName").value;
        var lastName = document.getElementById("lastName").value;
        var email = document.getElementById("email").value;
        var subject = document.getElementById("subject").value;
        var msg = document.getElementById("msg").value;

        var encFirst = encodeURIComponent(firstName);
        var encLast  = encodeURIComponent(lastName);
        var encEmail = encodeURIComponent(email);
        var encsubject = encodeURIComponent(subject);
        var encmsg = encodeURIComponent(msg);

        var info = "firstName="+encFirst+"&lastName="+encLast+"&email="+encEmail+"&subject="+encsubject+"&msg="+encmsg;

        var http3 = createAjaxRequestObject();

        if (http3.readyState == 4) {
            if (http3.status == 200){
                var result = JSON.parse(http3.responseText);
                console.log(result);
            }
        }

        http3.open("POST", "send_mail.php", true);
        http3.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
        http3.send(info);

    }

</script>

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Answer 1

您应该将type的{{1}}从submit更改为button，以便触发sendTicket()，否则表单会在到达{{1}之前被提交函数应该是：

sendTicket()

而不是：

<input type="button" value="Submit" onclick="sendTicket();" />

希望这有帮助。

＆＃13;

<input type="submit" value="Submit" onclick="sendTicket();" />

＆＃13;

function sendTicket() {
  var firstName = document.getElementById("firstName").value;
  var lastName = document.getElementById("lastName").value;
  var email = document.getElementById("email").value;
  var subject = document.getElementById("subject").value;
  var msg = document.getElementById("msg").value;

  var encFirst = encodeURIComponent(firstName);
  var encLast  = encodeURIComponent(lastName);
  var encEmail = encodeURIComponent(email);
  var encsubject = encodeURIComponent(subject);
  var encmsg = encodeURIComponent(msg);

  var info = "firstName="+encFirst+"&lastName="+encLast+"&email="+encEmail+"&subject="+encsubject+"&msg="+encmsg;
  console.log(info);

}

＆＃13;

表单提交没有刷新不起作用

欢迎您！

1 个答案: