是否有一种简单的单行方式来获取表单的数据,就像以经典的仅html方式提交表单一样?

例如:

<form>
    <input type="radio" name="foo" value="1" checked="checked" />
    <input type="radio" name="foo" value="0" />
    <input name="bar" value="xxx" />
    <select name="this">
        <option value="hi" selected="selected">Hi</option>
        <option value="ho">Ho</option>
</form>

输出:

{
    "foo": "1",
    "bar": "xxx",
    "this": "hi"
}

像这样的东西太简单了,因为它没有(正确地)包括文本区域,选择,单选按钮和复选框:

$("#form input").each(function () {
    data[theFieldName] = theFieldValue;
});

当前回答

function getFormData($form){
    var unindexed_array = $form.serializeArray();
    var indexed_array = {};

    $.map(unindexed_array, function(n, i){
        if(indexed_array[n['name']] == undefined){
            indexed_array[n['name']] = [n['value']];
        }else{
            indexed_array[n['name']].push(n['value']);
        }
    });

    return indexed_array;
}

其他回答

显示表单输入元素字段和输入文件提交您的表单没有页面刷新和抓取所有值与文件包括在这里

<form id="imageUploadForm" action="" method="post" enctype="multipart/form-data"> <input type="text" class="form-control" id="fname" name='fname' placeholder="First Name" > <input type="text" class="form-control" name='lname' id="lname" placeholder="Last Name"> <input type="number" name='phoneno' class="form-control" id="phoneno" placeholder="Phone Number"> <textarea class="form-control" name='address' id="address" rows="5" cols="5" placeholder="Your Address"></textarea> <input type="file" name="file" id="file" > <input type="submit" id="sub" value="Registration"> </form> on Submit button page will send ajax request to your php file. $('#imageUploadForm').on('submit',(function(e) { fname = $('#fname').val(); lname = $('#lname').val(); address = $('#address').val(); phoneno = $('#phoneno').val(); file = $('#file').val(); e.preventDefault(); var formData = new FormData(this); formData.append('file', $('#file')[0]); formData.append('fname',$('#fname').val()); formData.append('lname',$('#lname').val()); formData.append('phoneno',$('#phoneno').val()); formData.append('address',$('#address').val()); $.ajax({ type:'POST', url: "test.php", //url: '<?php echo base_url().'edit_profile/edit_profile2';?>', data:formData, cache:false, contentType: false, processData: false, success:function(data) { alert('Data with file are submitted !'); } }); }))

$("#form input, #form select, #form textarea").each(function() {
 data[theFieldName] = theFieldValue;
});

除此之外,你可能想看看serialize();

这是一个很好的香草JS函数,我写来提取表单数据作为一个对象。它还具有向对象中插入附加内容和清除表单输入字段的选项。

const extractFormData = ({ form, clear, add }) => {
  return [].slice.call(form.children).filter(node => node.nodeName === 'INPUT')
  .reduce((formData, input) => {
    const value = input.value
    if (clear) { input.value = '' }
    return {
      ...formData,
      [input.name]: value
    }
  }, add)
}

下面是一个使用post请求的例子:

submitGrudge(e) {
  e.preventDefault()

  const form = e.target
  const add = { id: Date.now(), forgiven: false }
  const grudge = extractFormData({ form, add, clear: true })

  // grudge = {
  //  "name": "Example name",
  //  "offense": "Example string",
  //  "date": "2017-02-16",
  //  "id": 1487877281983,
  //  "forgiven": false
  // }

  fetch('http://localhost:3001/api/grudge', {
    method: 'post',
    headers: { 'Content-Type': 'application/json' },
    body: JSON.stringify(grudge)
  })
    .then(response => response.json())
    .then(grudges => this.setState({ grudges }))
    .catch(err => console.log('error: ', err))
}

使用.serializeArray()获取数组格式的数据,然后将其转换为对象:

function getFormObj(formId) {
    var formObj = {};
    var inputs = $('#'+formId).serializeArray();
    $.each(inputs, function (i, input) {
        formObj[input.name] = input.value;
    });
    return formObj;
}

你们都不完全正确。你不能写:

formObj[input.name] = input.value;

因为这样,如果你有一个多选列表,它的值将被最后一个覆盖,因为它被传输为:"param1": "value1", "param1": "value2"。

所以,正确的方法是:

if (formData[input.name] === undefined) {
    formData[input.name] = input.value;
}
else {
    var inputFieldArray = $.merge([], $.isArray(formData[input.name]) ? formData[input.name] : [formData[input.name]]);
    $.merge(inputFieldArray, [input.value]);
    formData[input.name] = $.merge([], inputFieldArray);
}