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javascript - 通过表单发布 Json 使用 jQuery 提交

转载 作者:行者123 更新时间:2023-12-03 12:44:02 24 4
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HTML

<form id="join">
<div class="content">
<h2>Join the Dodilio Exchange</h2>
<ul>
<li><input id="firstname" type="text" placeholder="Fast Name" name="firstname" required></li>
</ul>
</div>
</form>

JS

    var register = function (firstname) {

var obj = {
"firstname": firstname
}

$.ajax({
type: "POST",
dataType: 'json',
data: obj,
url: "/rest/register/?format=json",
success: function (data) {
console.log('success')
//window.location.href = '/ideas'
}
});
}


var joinForm = $('#join');
var firstname = $('#firstname').val()

joinForm.submit(function(e){
console.log('submit')
e.preventDefault();

register(firstname)


return false;
})

但它是发布传统形式:

POST http://127.0.0.1:8000/rest/register/?format=json HTTP/1.1
Host: 127.0.0.1:8000
Connection: keep-alive
Content-Length: 233
Accept: application/json, text/javascript, */*; q=0.01
Origin: http://127.0.0.1:8000
X-Requested-With: XMLHttpRequest
User-Agent: Mozilla/5.0 (Windows NT 6.3; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/34.0.1847.131 Safari/537.36
Content-Type: application/x-www-form-urlencoded; charset=UTF-8
Referer: http://127.0.0.1:8000/jointheexchange.html
Accept-Encoding: gzip,deflate,sdch
Accept-Language: en-US,en;q=0.8

firstname=&lastname=&

我对 jQuery 不是很擅长,我在网络和堆栈上看过一些这样的例子,老实说,我看不出我哪里错了/不同。请帮忙 - 谢谢。

最佳答案

在我看来,var firstname = $('#firstname').val() 在页面加载时设置。您希望将其移至提交处理程序内,以便在提交表单时分配其值:

var joinForm = $('#join');

joinForm.submit(function(e) {
console.log('submit');
e.preventDefault();

var firstname = $('#firstname').val();

register(firstname);

return false;
})

此外,您在上面的代码中缺少一些分号,因此我会仔细检查其余代码是否存在语法错误。

关于javascript - 通过表单发布 Json 使用 jQuery 提交,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/23408225/

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