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Javascript Django使用ajax更新表数据_Javascript_Html_Django_Ajax_Forms - Fatal编程技术网

Javascript Django使用ajax更新表数据

Javascript Django使用ajax更新表数据,javascript,html,django,ajax,forms,Javascript,Html,Django,Ajax,Forms,我尝试过滤表数据,然后使用ajax更新表。我有: 过滤形式: <form id="search" method="POST" action="{% url 'article-filter' %}"> <input type="text" class="form-control" id="tokenfield1" name="txtSearch"> <button type="submit" class="btn btn-default js-filt

我尝试过滤表数据,然后使用ajax更新表。我有:

过滤形式:

<form id="search" method="POST" action="{% url 'article-filter' %}">
    <input type="text" class="form-control" id="tokenfield1" name="txtSearch">
    <button type="submit" class="btn btn-default js-filter">Submit</button>
</form>
ajax处理程序:

<script>
    var filter =
    function() {

        $.ajax({
            type: "POST",
            url: "blog/article-filter", 
            dataType: 'json',
        success: function(response) {
            $('#article-table').html(response.html_table);
        }
        });
    };

    $("#search").on("submit", ".js-filter", filter);
</script>

无功滤波器=
函数(){
$.ajax({
类型:“POST”,
url:“博客/文章过滤器”,
数据类型:“json”,
成功:功能(响应){
$('#article table').html(response.html#u table);
}
});
};
$(“#搜索”)。在(“提交”,“.js过滤器”,过滤器)上;

但是当我提交表单时,它只打印出原始html文本(表),我如何在页面中替换表并呈现它呢

所以问题是ajax甚至没有出现。我将ajax编辑为表单提交,而不是“提交”按钮:

$("#search").submit(function(e) {
    e.preventDefault();
    $.ajax({
        type: "POST",
        url: "blog/article-filter",
        dataType: 'json',
    }).done(function(data) {
        $('#article-table').html(data.html_table);
    });
});
它工作得很好

$("#search").submit(function(e) {
    e.preventDefault();
    $.ajax({
        type: "POST",
        url: "blog/article-filter",
        dataType: 'json',
    }).done(function(data) {
        $('#article-table').html(data.html_table);
    });
});