我正在使用Django 1.5和Dojo 1.8。我试图让Dojo在单击按钮时将表单提交回Django视图。
这是我的Django视图:
def report(request, report_id, report_url=None, template='report_parameters.html'):
if request.method == 'POST':
form = ReportParametersForm(request.POST)
if form.is_valid():
report_params = form.save()
html = "Success!"
return HttpResponse(html)
else:
form = ReportParametersForm()
return render(request,template, {
'form': form,
'report_url': report_url,
'report_id': report_id,
})
这是html页面:
<div id="report_body">
<form data-dojo-type="dijit/form/Form" id="parameters_form" data-dojo-id="parameters_form">
{% csrf_token %}
<table>
{{ form.as_table }}
</table>
<p><button id="submit_parameters" dojoType="dijit.form.Button" type="submit">Submit</button></p>
</form>
</div>
<script type="dojo/on" data-dojo-event="submit" data-dojo-args="e">
e.preventDefault();
require(["dojo/dom", "dojo/request", "dojo/dom-form"], function(dom, request, domForm){
on(dom.byId("submit_parameters"), "click", function() {
console.log("Dojo Post");
request.xhr("/report_parameters/report_id/report_url/", {
method: "post",
handleAs: "json",
data: domForm.toJson("parameters_form"),
}).then(
function(response){
alert(response);
dom.byId("report_body").innerHTML = "Report!";
},
function(error){
dom.byId("report_body").innerHTML = "<div class=\"error\">"+error+"<div>";
}
);
});
});
</script>
当我点击Submit
按钮时,我想向网址发送POST
请求,传递我在表单中的数据。但是,现在当我点击Submit
时,页面会重新加载一个类似于以下内容的网址:/?csrfmiddlewaretoken=Y9gaNMFRWZNXMbJ2L3Ev7A5iKPGTuWeF¶m_1=0¶m2=0/report_parameters/report_id/report_url/
。
我没有看到应该出现在我的控制台中的Dojo Post
。
如何提交表单?
答案 0 :(得分:0)
This fiddle似乎做你想做的事。
主要差异似乎是:
<form>
实际上是<div>
。 Dojo documentation for Form的reasons why this is done for IE链接。{/ li>
<div>
。on(dom.byId("submit_parameters")...
代码,因为已经存在声明性提交事件处理程序。HTML code:
<div id="report_body"></div>
<div data-dojo-type="dijit/form/Form" id="parameters_form" data-dojo-id="parameters_form" encType="multipart/form-data" action="" method="">
<input name="dummy" value="dummy">
<script type="dojo/on" data-dojo-event="submit" data-dojo-args="e">
console.log("submit");
e.preventDefault();
require(["dojo/dom", "dojo/request/xhr", "dojo/dom-form"], function(dom, xhr, domForm) {
console.log("Dojo Post");
var url = "/report_parameters/report_id/report_url/";
var data = domForm.toJson("parameters_form");
// overwrite url and data for jsfiddle
url = "/echo/json/";
data = {
json: data
};
xhr(url, {
method: "post",
handleAs: "json",
data: data,
}).then(function(response) {
alert(JSON.stringify(response, null, 2));
dom.byId("report_body").innerHTML = "Report!";
}, function(error) {
dom.byId("report_body").innerHTML = "<div class=\"error\">" + error + "<div>";
});
});
</script>
<button data-dojo-type="dijit/form/Button" id="submit_button" type="submit" name="submitButton" value="Submit">Submit</button>
</div>
JS代码:
require(["dojo/parser", "dijit/registry", "dijit/form/Form", "dijit/form/Button", "dijit/form/ValidationTextBox", "dijit/form/DateTextBox", "dojo/domReady!"], function (parser, registry) {
parser.parse().then(function () {
console.log("parsed");
console.log(registry.byId("parameters_form"));
console.log(registry.byId("submit_button"));
});
});
答案 1 :(得分:0)
我不得不稍微修改上面的内容。这最终对我有用:
<div id="report_body"></div>
<form data-dojo-type="dijit/form/Form" id="parameters_form" data-dojo-id="parameters_form" encType="multipart/form-data" action="" method="POST">
{% csrf_token %}
<table>
{{ form.as_table }}
</table>
<script type="dojo/on" data-dojo-event="submit" data-dojo-args="e">
e.preventDefault();
require(["dojo/dom", "dojo/request/xhr", "dojo/dom-form"], function(dom, xhr, domForm){
var url = "/report_parameters/report_id/report_url/"
var data = domForm.toObject("parameters_form")
xhr(url, {
method: "post",
data: data,
}).then(
function(response){
alert(response);
dom.byId("report_body").innerHTML = "Report!";
},
function(error){
dom.byId("report_body").innerHTML = error;
}
);
});
</script>
<p><button id="submit_parameters" dojoType="dijit/form/Button" type="submit" name="submitButton" value="Submit">Submit</button></p>
</form>
使用<div>
或<form>
标记将整个内容包装好。