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;
});
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”);
function getObjectFormFields(formSelector)
{
/// <summary>Função que retorna objeto com base nas propriedades name dos elementos do formulário.</summary>
/// <param name="formSelector" type="String">Seletor do formulário</param>
var form = $(formSelector);
var result = {};
var arrayAuxiliar = [];
form.find(":input:text").each(function (index, element)
{
var name = $(element).attr('name');
var value = $(element).val();
result[name] = value;
});
form.find(":input[type=hidden]").each(function (index, element)
{
var name = $(element).attr('name');
var value = $(element).val();
result[name] = value;
});
form.find(":input:checked").each(function (index, element)
{
var name;
var value;
if ($(this).attr("type") == "radio")
{
name = $(element).attr('name');
value = $(element).val();
result[name] = value;
}
else if ($(this).attr("type") == "checkbox")
{
name = $(element).attr('name');
value = $(element).val();
if (result[name])
{
if (Array.isArray(result[name]))
{
result[name].push(value);
} else
{
var aux = result[name];
result[name] = [];
result[name].push(aux);
result[name].push(value);
}
} else
{
result[name] = [];
result[name].push(value);
}
}
});
form.find("select option:selected").each(function (index, element)
{
var name = $(element).parent().attr('name');
var value = $(element).val();
result[name] = value;
});
arrayAuxiliar = [];
form.find("checkbox:checked").each(function (index, element)
{
var name = $(element).attr('name');
var value = $(element).val();
result[name] = arrayAuxiliar.push(value);
});
form.find("textarea").each(function (index, element)
{
var name = $(element).attr('name');
var value = $(element).val();
result[name] = value;
});
return result;
}
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。