Question

我有一个“邮政”纽芬兰语,在文本被打入文字区时,可以激活(从蓝色到颜色,从残疾到残疾)。

This works fine but I would like the text area to detect when there is nothing in the text area. So let s say I typed "abc" into the text area.. the submit button would become active but let s say I now delete all the text.. The button is still active.. well not fully as I have this:

commentButton.prop({ disabled: commentBox.val() == 0 });

不能在案文中提交任何内容。问题是,提交国仍然有颜色,没有回到表面上不活跃的国家。

因此,我要谈谈我的问题。

Can I set up something that will activate and start listening as soon as i focus in the text area and type and know when I ve deleted everything from the text area?

这样,我就可以说,如果案文全部删除,那么就删除了彩色邦纽顿的地图集,并添加模糊邦纽顿的地图集。

这也适用于白人空间。

我目前的法典是:

www.un.org/Depts/DGACM/index_french.htm

.activeCommentButton = coloured submit button

.commentButton = blurred submission button

$(".microposts").off().on("focus", ".comment_box", function () {
    $(this).prop("rows", 7);
    $(this).parent().find(".activeCommentButton").removeClass("activeCommentButton").addClass("commentButton");
    var commentBox = $(this),
        form = $(this).parent(),
        cancelButton = form.children(".commentButtons").children(".cancelButton"),
        commentButton = form.children(".commentButtons").children(".commentButton");
    $(this).removeClass("comment_box").addClass("comment_box_focused").autoResize();
    form.children(".commentButtons").addClass("displayButtons");
    commentButton.prop({
        disabled: true
    });
    commentBox.off().on("keyup keypress input change", function () {
        commentButton.prop({
            disabled: commentBox.val() == 0
        });
        commentButton.removeClass("commentButton").addClass("activeCommentButton");
    });
    cancelButton.click(function () {
        commentBox.removeClass("comment_box_focused").addClass("comment_box");
        form.children(".commentButtons").removeClass("displayButtons");
        commentBox.val("");
    });
});

幼儿园

Answer 1

是的,你可以修改:

$(".commentButton").prop({ disabled: true });

$(".comment_box").keypress(function() {

    $(".commentButton").prop({ disabled: !$(this).val().length >= 1 })           

});

Answer 2

这对我来说是行之有效的,但是否采取了更清洁的方法?

 commentButton.prop({ disabled: true });

                commentBox.off().on("keyup keypress input change", function () {
                  commentButton.prop({ disabled: commentBox.val().length == 0 });
                  if ( commentBox.val().length > 1 ) {
                  commentButton.removeClass("commentButton").addClass("activeCommentButton");
              } else {
                commentButton.removeClass("activeCommentButton").addClass("commentButton");
              }
                });

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