Jquery ，获取元素的位置（）？
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Jquery , get position().right of an element?
提问by gregdevs
So I am working on having a draggable image.
所以我正在努力制作一个可拖动的图像。
very simply, When you drag the image to the left or right side at a specific position, I want the image to snap back to the left or right side.
很简单，当您将图像在特定位置的左侧或右侧拖动时，我希望图像重新回到左侧或右侧。
In the fiddle below, I have the functionality for the left side..but Im not quite sure what the best approach is for determining the right side since Jquery only allows top and left properties.
在下面的小提琴中，我有左侧的功能......但我不太确定确定右侧的最佳方法是什么，因为 Jquery 只允许 top 和 left 属性。
consider this code:
考虑这个代码：
$(function() {
$(".headerimage").css('cursor', 's-resize');
var y1 = $('.picturecontainer').height();
var y2 = $('.headerimage').height();
$(".headerimage").draggable({
scroll: false,
drag: function(event, ui) {
if (ui.position.left >= 200) {
ui.position.left = 0;
}
},
stop: function(event, ui) {
//####
}
});
采纳答案by Jonathan Marzullo
what about this
那这个呢
drag: function(event, ui) {
if (ui.position.left >= 200 || ui.position.left <= -300) {
ui.position.left = 0;
}
},
$(function() {
$(".headerimage").css('cursor', 's-resize');
var y1 = $('.picturecontainer').height();
var y2 = $('.headerimage').height();
$(".headerimage").draggable({
scroll: false,
drag: function(event, ui) {
if (ui.position.left >= 200 || ui.position.left <= -300) {
ui.position.left = 0;
}
},
stop: function(event, ui) {
//####
}
});
});
checking against negative value of position.left
检查 position.left 的负值
回答by Chad Kuehn
To get the position to the right of an element I use this jQuery:
要获得元素右侧的位置，我使用了这个 jQuery：
var target = $("selector");
var result = target.position().left + target.width();
Make sure the element is fully rendered when you are getting this value. Otherwise the results could be off.
确保在获取此值时完全呈现元素。否则结果可能会关闭。
回答by Nikola Mitic
What about this one:
这个如何：
var rightOffset = $('.parent').width() - ( $(target).position().left + $(target).width() )
note: I believe that this shoudn't work if you have margins and border around target element, not sure if that could be fixed with box-sizing: border-box
注意：我相信如果目标元素周围有边距和边框，这应该不起作用，不确定是否可以使用 box-sizing: border-box
回答by Huy Nguy?n
to get position right=0. Set left = leftpositionwithout margin css is set
获得位置right=0。Set left = leftposition没有边距 css 被设置
var leftposition = $(parent).width()-$(child).width();
回答by Mohamad Hamouday
This work for me:
这对我有用：
(function($) {
$.fn.Right = function()
{
return this.parent().outerWidth() - this.position().left + this.outerWidth();
};
})(jQuery);
To use it:
要使用它：
$("selector").Right();
回答by Matt
Do position left + width.
做左+宽的位置。
$(ui).position().left + $(ui).width();
