如何使用JavaScript/jQuery获取表单数据?

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

这里是一个工作的JavaScript实现，它正确地处理复选框、单选按钮和滑块(也可能是其他输入类型，但我只测试了这些)。

function setOrPush(target, val) {
    var result = val;
    if (target) {
        result = [target];
        result.push(val);
    }
    return result;
}

function getFormResults(formElement) {
    var formElements = formElement.elements;
    var formParams = {};
    var i = 0;
    var elem = null;
    for (i = 0; i < formElements.length; i += 1) {
        elem = formElements[i];
        switch (elem.type) {
            case 'submit':
                break;
            case 'radio':
                if (elem.checked) {
                    formParams[elem.name] = elem.value;
                }
                break;
            case 'checkbox':
                if (elem.checked) {
                    formParams[elem.name] = setOrPush(formParams[elem.name], elem.value);
                }
                break;
            default:
                formParams[elem.name] = setOrPush(formParams[elem.name], elem.value);
        }
    }
    return formParams;
}

工作的例子:

编辑:

如果您正在寻找一个更完整的实现，那么请查看我为其制作的项目的这一部分。我最终会用我想出的完整解决方案来更新这个问题，但也许这对某人会有帮助。

2016-02-17 22:19:38

其他回答

我编写了一个库来解决这个问题:JSONForms。它采用一个表单，遍历每个输入并构建一个易于阅读的JSON对象。

假设你有以下表单:

<form enctype='application/json'>
  <input name='places[0][city]' value='New York City'>
  <input type='number' name='places[0][population]' value='8175133'>
  <input name='places[1][city]' value='Los Angeles'>
  <input type='number' name='places[1][population]' value='3792621'>
  <input name='places[2][city]' value='Chicago'>
  <input type='number' name='places[2][population]' value='2695598'>
</form>

将表单传递给JSONForms的encode方法将返回以下对象:

{
  "places": [
    {
      "city": "New York City",
      "population": 8175133
    },
    {
      "city": "Los Angeles",
      "population": 3792621
    },
    {
      "city": "Chicago",
      "population": 2695598
    }
  ]
}

下面是表单的演示。

2015-01-28 23:01:41

var formData = new FormData($('#form-id'));
params   = $('#form-id').serializeArray();

$.each(params, function(i, val) {
    formData.append(val.name, val.value);
});

2016-02-01 13:22:21

这是我的香草JS版本(在Chrome上测试)

适用于:

name = "输入" Name ="form[Name]"(创建一个对象) Name ="checkbox[]"(创建一个数组对象) Name ="form[checkbox][]"(创建一个数组) Name ="form[select][Name]"(创建一个只包含所选值的对象)

/**
 * Get the values from a form
 * @param formId ( ID without the # )
 * @returns {object}
 */
function getFormValues( formId )
{
    let postData = {};
    let form = document.forms[formId];
    let formData = new FormData( form );

    for ( const value of formData.entries() )
    {
        let container = postData;
        let key = value[0];
        let arrayKeys = key.match( /\[[\w\-]*\]/g ); // Check for any arrays

        if ( arrayKeys !== null )
        {
            arrayKeys.unshift( key.substr( 0, key.search( /\[/ ) ) );  // prepend the first key to the list
            for ( let i = 0, count = arrayKeys.length, lastRun = count - 1; i < count; i++ )
            {
                let _key = arrayKeys[i];
                _key = _key.replace( "[", '' ).replace( "]", '' ); // Remove the brackets []
                if ( _key === '' )
                {
                    if ( ! Array.isArray( container ) )
                    {
                        container = [];
                    }

                    _key = container.length;
                }

                if ( ! (_key in container) ) // Create an object for the key if it doesn't exist
                {
                    if ( i !== lastRun && arrayKeys[i + 1] === '[]' )
                    {
                        container[_key] = [];
                    }
                    else
                    {
                        container[_key] = {};
                    }
                }

                if ( i !== lastRun ) // Until we're the last item, swap container with it's child
                {
                    container = container[_key];
                }

                key = _key;
            }
        }
        container[key] = value[1]; // finally assign the value
    }

    return postData;
}

2021-02-19 09:17:12

你可以使用这个函数从表单中获取一个对象或JSON。

使用它:

var object = formService.getObjectFormFields（“#idform”）;

2016-07-07 14:37:55

$( "form" ).bind( "submit", function(e) { e.preventDefault(); console.log( $(this).serializeObject() ); //console.log( $(this).serialize() ); //console.log( $(this).serializeArray() ); }); $.fn.serializeObject = function() { var o = {}; var a = this.serializeArray(); $.each( a, function() { if ( o[this.name] !== undefined) { if ( ! o[this.name].push ) { o[this.name] = [o[this.name]]; } o[this.name].push(this.value || ''); } else { o[this.name] = this.value || ''; } }); return o; }; <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js"></script> <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> </select> <input type="submit" value="Submit" /> </form>

Codepen

2019-10-29 03:58:49

如何使用JavaScript/jQuery获取表单数据?

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