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Ajax不能处理POST表单提交,而是使用默认POST

ajax jquery html

colt.exe · 技术社区 · 3 年前

我在模态中确实有形式。我试图实现的是重写默认的POST方法并使用AJAX。我这样做是为了防止页面切换。然而,似乎脚本并没有被触发。如果我删除脚本,就会发生正常的POST。代码如下:

<div class="modal" id="modal1" tabindex="-1" role="dialog" aria-labelledby="exampleModalLabel" aria-hidden="true">
<div class="modal-dialog" role="document">
    <div class="modal-content">
        <div class="modal-header">
            <h5 class="modal-title" id="exampleModalLabel">Modal title</h5>
            <button type="button" class="close" data-dismiss="modal" aria-label="Close">
                <span aria-hidden="true">&times;</span>
            </button>
        </div>
        <form action="process.php" method="POST" id="submitform">
            <div class="modal-body">
                <div class="form-group">
                    <label for="source">source:</label>
                    <textarea class="form-control" id="kaynak" name="kaynak" rows="6"></textarea>
                </div>
            </div>
            <div class="modal-footer">
                <button type="button" class="btn btn-secondary" data-dismiss="modal">Ä°ptal</button>
                <button type="submit" class="btn btn-primary ">Kaydet</button>
            </div>
        </form>
    </div>
</div>

    $(document).ready(function() {

        $('#submitform').submit(function(event) {
            event.preventDefault();
            var post_url = $(this).attr("action");
            var request_method = $(this).attr("method");
            var formData = $(this).serialize();
            var username = "<?php echo $username ?>";
            formData.push({
                name: 'username',
                value: username
            });

            $.ajax({
                    type: request_method,
                    url: post_url,
                    data: formData,
                    dataType: 'json',
                    encode: true
                })
                .done(function(data) {
                    console.log(data);
                });
        })

0 回复 | 直到 3 年前

Musa Kishore Kumar 3 年前

你想用 .serializeArray()

        var formData = $(this).serializeArray();
        var username = "<?php echo $username ?>";
        formData.push({
            name: 'username',
            value: username
        });

-1

Or Assayag 3 年前

尝试:

$(document).ready(function() {

    $('#submitform').submit(function(event) {
        event.preventDefault();
        var post_url = $(this).attr("action");
        var request_method = $(this).attr("method");
        var formData = $(this).serialize();
        var username = "<?php echo $username ?>";
        formData['name'] = username;
        $.ajax({
                type: request_method,
                url: post_url,
                data: formData,
                dataType: 'json',
                encode: true
            })
            .done(function(data) {
                console.log(data);
            });
    })