$(function() {
$("tr").click(function() {
$(this).addClass("selected");
});
//************** for left working***************//
$("button").click(function() {
var CurrentTrId = $("tr.selected").attr("data-id");
var ParrentTrId = (CurrentTrId) - 1;
if (ParrentTrId == 0) {
return
}
var CurrentTrLeftValue = parseInt($(".selected > td:first-child ").css("padding-left"));
console.log(CurrentTrLeftValue);
var ParrentTrLeftValue = parseInt($("tr[data-id='" + parrentid + "']td:first-child").css("padding-left"));
if (NewCurrentTrLeftValue == ParrentTrLeftValue) {
if (CurrentTrLeftValue <= 40) {
CurrentTrLeftValue = (CurrentTrLeftValue + 20);
$(".selected > td:first-child").css({ "padding-left": CurrentTrLeftValue });
$(".selected > td:first-child").addClass("normal");
$('#', ParrentTrId).addClass('bold');
}
} });
})
.table {
position:static;
width:100%;
border-collapse:collapse;
}
.table td {
border: 1px solid;
border-color: lightgray;
height: 17px;
}
.selected{
background-color:lightskyblue;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<table class="table">
<tr>
<td>hi</td> <td>hi</td>
</tr>
<tr>
<td>me</td> <td>me</td>
</tr>
<tr>
<td>you</td> <td>you</td>
</tr>
</table>
<button>clickme</button>
我想選擇TD沒有采取有沒有任何方法呢? 「parrentid是一個變量,存儲parrent data-id」
我有一張表和一個按鈕,我想將第三次點擊按鈕時將我的可選tr內容移動到20 px左側。第二次點擊我想再次移動它20px(總計40px)。我在這裏寫一些代碼
您可以重現stacksnippet中的問題...? –
嗨,我在這裏添加完整的代碼 –