js或者jquery如何实现拖拽表格单元格内容交换

Question

如题所述

Answer 1

<script type="text/javascript" src="https://cdn.bootcss.com/jquery/1.9.1/jquery.min.js"></script>
<style>
td.hover{border:dotted 3px #f00 }
td.selected{background:#f00;color:#fff;}
td.disabled{background:#eee;color:#000}
#tdMove{position:absolute;display:none;border:dotted 3px #f00;background:#fff}
table{user-select:none;-moz-user-select:none}</style>
<table border="1">
<script>
var s = ''
for (var i = 0; i < 10; i++) {
s += '<tr>'
for (var j = 0; j < 10; j++) s += '<td' + (Math.random() > 0.8 ? ' class="disabled"' : '') + '>' + i + '-' + j + '</td>'
s += '</tr>'
}
document.write(s)
</script>
</table>
<div id="tdMove"></div>
<script>
var tdMove = $('#tdMove'),px=5;//浮动块偏移量，不偏移鼠标移动过程会出现在这个div移动的情况，移动到td上无法响应高亮样式
$('td:not(.disabled)').mousedown(function (e) {
this.className = 'selected'
tdMove.html(this.innerHTML).css({ left: e.clientX + px, top: e.clientY + px }).show();;
$(document).mousemove(function (e) {
var el = e.target;
tdMove.css({ left: e.clientX + px, top: e.clientY + px });
if (el.tagName == 'TD' && el.className != 'disabled' && el.className != 'selected') {
$('td').removeClass('hover'); el.className = 'hover';
}
}).mouseup(function () {
$(document).unbind();

tdMove.hide();
var td = $('td.hover,td.selected')
if (td.length == 2) {
var s = td[1].innerHTML;
td[1].innerHTML = td[0].innerHTML;
td[0].innerHTML = s;
}
td.removeClass('hover selected')
});
})
</script>本回答被网友采纳