Question

我有一个我发送的表单，目前它使用类似.button类的p标记。出于某种原因，我没有得到任何回复而没有在我的chrome dev工具的网络标签中提交？可以帮忙吗？

$(".button-like").each(function () {
    $(this).click(function () {
        var $thisItem = $(this);
        var $parent = $thisItem.parent(".forms");
        $parent.submit(function () {
            var data = {
                "action": "like"
            };
            data = $parent.serialize() + "&" + $.param(data);
            var itemId = $parent.find('input.id').val();
            $.ajax({
                type: "POST",
                url: "/actions/",
                data: data,
                success: function (data) {
                    console.log('Like submitted successfully sent');
                    $('body').addClass('liked');
                }
            });
            return false;
        });
    });
});

谢谢，标记

Answer 1

$(".button-like").each(function () {
    $(this).click(function () {
        var $thisItem = $(this);
        var $parent = $thisItem.parent(".forms");
        $parent.submit(function () {
            var data = {
                "action": "like"
            };
            data = $parent.serialize() + "&" + $.param(data);
            var itemId = $parent.find('input.id').val();
            $.ajax({
                type: "POST",
                url: "/actions/",
                data: data,
                success: function (data) {
                    console.log('Like submitted successfully sent');
                    $('body').addClass('liked');
                }
            });
            return false;
        }).submit();
    });
});

这将在设置后立即触发提交。但我认为你想要做的是：

$(".button-like").each(function () {
    var $thisItem = $(this);
    var $parent = $thisItem.parent(".forms");
    $parent.submit(function () {
        var data = {
            "action": "like"
        };
        data = $parent.serialize() + "&" + $.param(data);
        var itemId = $parent.find('input.id').val();
        $.ajax({
            type: "POST",
            url: "/actions/",
            data: data,
            success: function (data) {
                console.log('Like submitted successfully sent');
                $('body').addClass('liked');
            }
        });
        return false;
    });
    $thisItem.click(function () {
        $parent.submit();
    });
});

为什么我的表单没有在点击时从p标签发送？

1 个答案: