使用 javascript 没有 jQuery 的简单 ajax 表单

Question

我正在使用我无法更改标记且无法使用 jQuery 的表单。 目前，表单将结果发布到新窗口。是否可以将其更改为 ajax 表单，以便在提交时显示结果而不更改任何标记？ 将结果（标记）从结果页面拉回表单页面。

这是表单的标记。

<form class="form-poll" id="poll-1225962377536" action="/cs/Satellite" target="_blank">
<div class="form-item">
    <fieldset class="form-radio-group">
        <legend><span class="legend-text">What mobile phone is the best?</span></legend>
                <div class="form-radio-item">
                    <input type="radio" class="radio" value="1225962377541" name="option" id="form-item-1225962377541">
                    <label class="radio" for="form-item-1225962377541">
                        <span class="label-text">iPhone</span>
                    </label>
                </div><!-- // .form-radio-item -->
                <div class="form-radio-item">
                    <input type="radio" class="radio" value="1225962377542" name="option" id="form-item-1225962377542">
                    <label class="radio" for="form-item-1225962377542">
                        <span class="label-text">Android</span>
                    </label>
                </div><!-- // .form-radio-item -->
                <div class="form-radio-item">
                    <input type="radio" class="radio" value="1225962377543" name="option" id="form-item-1225962377543">
                    <label class="radio" for="form-item-1225962377543">
                        <span class="label-text">Symbian</span>
                    </label>
                </div><!-- // .form-radio-item -->
                <div class="form-radio-item">
                    <input type="radio" class="radio" value="1225962377544" name="option" id="form-item-1225962377544">
                    <label class="radio" for="form-item-1225962377544">
                        <span class="label-text">Other</span>
                    </label>
                </div><!-- // .form-radio-item -->
    </fieldset>
</div><!-- // .form-item -->
<div class="form-item form-item-submit">
    <button class="button-submit" type="submit"><span>Vote now</span></button>
</div><!-- // .form-item -->
<input type="hidden" name="c" value="News_Poll">
<input type="hidden" class="pollId" name="cid" value="1225962377536">
<input type="hidden" name="pagename" value="Foundation/News_Poll/saveResult">
<input type="hidden" name="site" value="themouth">

非常感谢任何提示/教程。 :)

Answer 1

以下是另一个答案的更优雅的解决方案，更适合现代浏览器。

我的理由是，如果您需要对旧版浏览器的支持您很可能已经使用了像 jQuery 这样的库，因此这个问题毫无意义。

/**
 * Takes a form node and sends it over AJAX.
 * @param {HTMLFormElement} form - Form node to send
 * @param {function} callback - Function to handle onload. 
 *                              this variable will be bound correctly.
 */

function ajaxPost (form, callback) {
    var url = form.action,
        xhr = new XMLHttpRequest();

    //This is a bit tricky, [].fn.call(form.elements, ...) allows us to call .fn
    //on the form's elements, even though it's not an array. Effectively
    //Filtering all of the fields on the form
    var params = [].filter.call(form.elements, function(el) {
        //Allow only elements that don't have the 'checked' property
        //Or those who have it, and it's checked for them.
        return typeof(el.checked) === 'undefined' || el.checked;
        //Practically, filter out checkboxes/radios which aren't checekd.
    })
    .filter(function(el) { return !!el.name; }) //Nameless elements die.
    .filter(function(el) { return el.disabled; }) //Disabled elements die.
    .map(function(el) {
        //Map each field into a name=value string, make sure to properly escape!
        return encodeURIComponent(el.name) + '=' + encodeURIComponent(el.value);
    }).join('&'); //Then join all the strings by &

    xhr.open("POST", url);
    // Changed from application/x-form-urlencoded to application/x-form-urlencoded
    xhr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");

    //.bind ensures that this inside of the function is the XHR object.
    xhr.onload = callback.bind(xhr); 

    //All preperations are clear, send the request!
    xhr.send(params);
}

所有主流浏览器都支持以上，IE9及以上。

Answer 2

这是我用来做你想做的事情的一个漂亮的函数：

HTML：

<form action="/cs/Satellite">...</form>
<input type="button" value="Vote now" onclick="javascript:AJAXPost(this)">

JS：

function AJAXPost(myself) {
    var elem   = myself.form.elements;
    var url    = myself.form.action;    
    var params = "";
    var value;

    for (var i = 0; i < elem.length; i++) {
        if (elem[i].tagName == "SELECT") {
            value = elem[i].options[elem[i].selectedIndex].value;
        } else {
            value = elem[i].value;                
        }
        params += elem[i].name + "=" + encodeURIComponent(value) + "&";
    }

    if (window.XMLHttpRequest) {
        // code for IE7+, Firefox, Chrome, Opera, Safari
        xmlhttp=new XMLHttpRequest();
    } else { 
        // code for IE6, IE5
        xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
    }

    xmlhttp.open("POST",url,false);
    xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    xmlhttp.setRequestHeader("Content-length", params.length);
    xmlhttp.setRequestHeader("Connection", "close");
    xmlhttp.send(params);

    return xmlhttp.responseText;
}

Answer 3

现在使用FormData 是最简单的方法。您使用对 Form 元素的引用来构造它，它会为您序列化所有内容。

MDN 有一个 here 的示例——大致如下：

const form = document.querySelector("#debarcode-form");
form.addEventListener("submit", e => {
    e.preventDefault();
    const fd = new FormData(form);
    const xhr = new XMLHttpRequest();
    xhr.addEventListener("load", e => {
        console.log(e.target.responseText);
    });
    xhr.addEventListener("error", e => {
        console.log(e);
    });
    xhr.open("POST", form.action);
    xhr.send(fd);
});

如果你想要它作为一个对象（JSON）：

const obj = {};
[...fd.entries()].forEach(entry => obj[entry[0]] = entry[1]);

Answer 4

扩展 Madara 的答案：我必须进行一些更改才能使其在 Chrome 47.0.2526.80 上运行（未在其他任何设备上进行测试）。希望这可以节省一些时间。

此 sn-p 是对该答案的修改，具有以下更改：

• 过滤!el.disabled,
• 在排除!checked之前检查输入类型
• 请求类型为x-www-form-urlencoded

结果如下：

function ajaxSubmit(form, callback) {
    var xhr = new XMLHttpRequest();
    var params = [].filter.call(form.elements, function (el) {return !(el.type in ['checkbox', 'radio']) || el.checked;})
    .filter(function(el) { return !!el.name; }) //Nameless elements die.
    .filter(function(el) { return !el.disabled; }) //Disabled elements die.
    .map(function(el) {
        return encodeURIComponent(el.name) + '=' + encodeURIComponent(el.value);
    }).join('&'); //Then join all the strings by &
    xhr.open("POST", form.action);
    xhr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    xhr.onload = callback.bind(xhr);
    xhr.send(params);
};

Answer 5

策略是序列化表单并使用 XHR 发送数据，然后对响应执行您想要的操作。 Matt Krus 的Ajax Toolbox 和相关的Javascript Toolbox 提供了一套很好的实用程序和帮助。

如果您只是对发布的表单进行序列化，那么以下内容就可以解决问题。它可以轻松扩展以包含其他表单控件类型：

var serialiseForm = (function() {

  // Checkboxes that have already been dealt with
  var cbNames;

  // Return the value of a checkbox group if any are checked
  // Otherwise return empty string
  function getCheckboxValue(cb) {
    var buttons = cb.form[cb.name];
    if (buttons.length) {
      for (var i=0, iLen=buttons.length; i<iLen; i++) {
        if (buttons[i].checked) {
          return buttons[i].value;
        }
      }
    } else {
      if (buttons.checked) {
        return buttons.value;
      }
    }
    return '';
  }

  return function (form) {
    var element, elements = form.elements;
    var result = [];
    var type;
    var value = '';
    cbNames = {};

    for (var i=0, iLen=elements.length; i<iLen; i++) {
      element = elements[i];
      type = element.type;

      // Only named, enabled controls are successful
      // Only get radio buttons once
      if (element.name && !element.disabled && !(element.name in cbNames)) {

         if (type == 'text' || type == 'hidden') {
          value = element.value;

        } else if (type == 'radio') {
          cbNames[element.name] = element.name;
          value = getCheckboxValue(element);

        }
      }

      if (value) {
        result.push(element.name + '=' + encodeURIComponent(value));
      }
      value = '';

    }
    return '?' + result.join('&');
  }
}());

Answer 6

使用fetch 的现代方式是：

const formData = new FormData(form);
fetch(form.action, {
  method: 'POST',
  body: formData
});

如果需要 IE 支持，请注意 browser support 并使用 this polyfil

Answer 7

function ajaxSubmit(form, callback) {
var xhr = new XMLHttpRequest();
var params = [].filter.call(form.elements, function (el) {return !(el.type in ['checkbox', 'radio']) || el.checked;})
.filter(function(el) { return !!el.name; }) //Nameless elements die.
.filter(function(el) { return !el.disabled; }) //Disabled elements die.
.map(function(el) {
    if (el.type=='checkbox') return encodeURIComponent(el.name) + '=' + encodeURIComponent(el.checked);
   else return encodeURIComponent(el.name) + '=' + encodeURIComponent(el.value);
}).join('&'); //Then join all the strings by &
xhr.open("POST", form.action);
xhr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xhr.onload = callback.bind(xhr);
xhr.send(params);

};

Answer 8

我刚刚接受了上面 Coomie 的回答，并使其适用于 Radio/Checkboxes。我不敢相信这是多么简单明了。除了少数例外，我已经用完了框架。

    var params = "";
    var form_elements = form.elements;
    for (var i = 0; i < form_elements.length; i++) 
    {
        switch(form_elements[i].type)
        {
            case "select-one":
            {
                value = form_elements[i].options[form_elements[i].selectedIndex].value;
            }break;
            case "checkbox":
            case "radio": 
            {
                if (!form_elements[i].checked)
                {
                    continue; // we don't want unchecked data
                }
                value = form_elements[i].value;
            }break;
            case "text" :
            {
                value = form_elements[i].value;                
            }break;

        }

        params += encodeURIComponent(form_elements[i].name) + "=" + encodeURIComponent(value) + "&";
    }



    var xhr = new XMLHttpRequest();    
    xhr.open('POST', "/api/some_url");
    xhr.setRequestHeader('X-Requested-With', 'XMLHttpRequest');
    xhr.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
    xhr.onreadystatechange = function() {
        if (xhr.readyState == 4) {
            if (xhr.status == 200)
            {

                console.log("xhr.responseText");
            }
            else
            {
                console.log("Error! Status: ", xhr.status, "Text:", xhr.responseText);
            } 
        }
    };

    console.log(params);
    xhr.send(params);

Answer 9

这是我想出的最简单的方法。我还没有找到使用这种确切方法的示例。代码使用非提交类型按钮提交表单并将结果放入 div，如果表单无效（未填写所有必填字段），它将忽略提交操作，浏览器本身将显示哪些字段未填写填写正确。

此代码仅适用于支持“FormData”对象的现代浏览器。

<script>
function ajaxSubmitForm() {
  const form = document.getElementById( "MyForm" );
  if (form.reportValidity()) {
    const FD = new FormData( form );
    var xhttp = new XMLHttpRequest();
    xhttp.onreadystatechange = function() { if (this.readyState == 4 && this.status == 200) { document.getElementById("content_area").innerHTML = this.responseText; } };
    xhttp.open("POST","https://example.com/whatever.php",true);
    xhttp.send( FD );
  }
}
</script>

<div id="content_area">
<form id="MyForm">
  <input type="hidden" name="Token" Value="abcdefg">
  <input type="text" name="UserName" Value="John Smith" required>
  <input type="file" accept="image/jpeg" id="image_uploads" name="ImageUpload" required>
  <button type="button" onclick="ajaxSubmitForm()">
</form>
</div>