Zebra striping a table with hidden rows [duplicate] - css

Question

I've got a table
<table id="mytable">
<tr style="display: none;"><td> </td></tr>
<tr><td> </td></tr>
<tr style="display: none;"><td> </td></tr>
<tr><td> </td></tr>
<tr><td> </td></tr>
<tr><td> </td></tr>
</table>
I'm trying to set the table striping to use nth-child selectors but just can't seem to crack it.
table #mytable tr[#display=block]:nth-child(odd) {
background-color: #000;
}
table #mytable tr[#display=block]:nth-child(odd) {
background-color: #FFF;
}
I'm pretty sure I'm close ... can't quite seem to crack it.
anyone pass along a clue?

Answer 1

Here's as close as you're going to get. Note that you can't make the nth-child count only displayed rows; nth-child will take the nth child element no matter what, not the nth child that matches a given selector. If you want some rows to be missing and not affect the zebra-striping, you will have to remove them from the table entirely, either through the DOM or on the server side.
#mytable tr:nth-child(odd) {
background-color: #000;
}
#mytable tr:nth-child(even) {
background-color: #FFF;
}
<table id="mytable">
<tr><td> </td></tr>
<tr><td> </td></tr>
<tr><td> </td></tr>
<tr><td> </td></tr>
<tr><td> </td></tr>
<tr><td> </td></tr>
</table>
Here are the fixes that I made:
table #mytable tr[#display=block]:nth-child(odd) {
background-color: #000;
}
There's no need to specify an ancestor selector for an id based selector; there is only ever one element that will match #table, so you're just adding extra code by adding the table in.
#mytable tr[#display=block]:nth-child(odd) {
background-color: #000;
}
Now, [#display=block] would match elements which had an attribute display set to block, such as <tr display=block>. Display isn't a valid HTML attribute; what you seem to be trying to do is to have the selector match on the style of the element, but you can't do that in CSS, since the browser needs to apply the styles from the CSS before it can figure that out, which it's in the process of doing when it's applying this selector. So, you won't be able to select on whether table rows are displayed. Since nth-child can only take the nth child no matter what, not nth with some attribute, we're going to have to give up on this part of the CSS. There is also nth-of-type, which selects the nth child of the same element type, but that's all you can do.
#mytable tr:nth-child(odd) {
background-color: #000;
}
And there you have it.

Answer 2

If you are using JQuery to change the visibility of rows you can apply the following function to the table to add an .odd class where appropriate. Call it each time the rows visible is different.
function updateStriping(jquerySelector){
$(jquerySelector).each(function(index, row){
$(row).removeClass('odd');
if (index%2==1){ //odd row
$(row).addClass('odd');
}
});
}
And for the css simply do
table#tableid tr.visible.odd{
background-color: #EFF3FE;
}

Answer 3

While you can't Zebra stripe a table with hidden rows using CSS3 you can do it with JavaScript. Here is how:
var table = document.getElementById("mytable");
var k = 0;
for (var j = 0, row; row = table.rows[j]; j++) {
if (!(row.style.display === "none")) {
if (k % 2) {
row.style.backgroundColor = "rgba(242,252,244,0.4)";
} else {
row.style.backgroundColor = "rgba(0,0,0,0.0)";
}
k++;
}
}

Answer 4

For a jquery way, you could use this function which iterates through the rows in your table, checking the visbility of the row and (re)setting a class for visible odd rows.
function updateStriping(jquerySelector) {
var count = 0;
$(jquerySelector).each(function (index, row) {
$(row).removeClass('odd');
if ($(row).is(":visible")) {
if (count % 2 == 1) { //odd row
$(row).addClass('odd');
}
count++;
}
});
}
Use css to set a background for odd rows.
#mytable tr.odd { background: rgba(0,0,0,.1); }
Then you can call this zebra-striper whenever by using:
updateStriping("#mytable tr");

Answer 5

I came up with a sort of solution but it's reliant on the fact that the table can only ever have a maximum number of hidden rows and comes with the downside of requiring 2 additional CSS rules for each possible hidden row. The principle is that, after each hidden row, you switch the background-color of the odd and even rows around.
Here's a quick example with just 3 hidden rows and the necessary CSS for up to 4 of them. You can already see how unwieldy the CSS can become but, still, someone may find some use for it:
table{
background:#fff;
border:1px solid #000;
border-spacing:1px;
width:100%;
}
td{
padding:20px;
}
tr:nth-child(odd)>td{
background:#999;
}
tr:nth-child(even)>td{
background:#000;
}
tr[data-hidden=true]{
display:none;
}
tr[data-hidden=true]~tr:nth-child(odd)>td{
background:#000;
}
tr[data-hidden=true]~tr:nth-child(even)>td{
background:#999;
}
tr[data-hidden=true]~tr[data-hidden=true]~tr:nth-child(odd)>td{
background:#999;
}
tr[data-hidden=true]~tr[data-hidden=true]~tr:nth-child(even)>td{
background:#000;
}
tr[data-hidden=true]~tr[data-hidden=true]~tr[data-hidden=true]~tr:nth-child(odd)>td{
background:#000;
}
tr[data-hidden=true]~tr[data-hidden=true]~tr[data-hidden=true]~tr:nth-child(even)>td{
background:#999;
}
tr[data-hidden=true]~tr[data-hidden=true]~tr[data-hidden=true]~tr[data-hidden=true]~tr:nth-child(odd)>td{
background:#999;
}
tr[data-hidden=true]~tr[data-hidden=true]~tr[data-hidden=true]~tr[data-hidden=true]~tr:nth-child(even)>td{
background:#000;
}
<table>
<tbody>
<tr><td></td><td></td></tr>
<tr><td></td><td></td></tr>
<tr data-hidden="true"><td></td><td></td></tr>
<tr><td></td><td></td></tr>
<tr><td></td><td></td></tr>
<tr><td></td><td></td></tr>
<tr><td></td><td></td></tr>
<tr data-hidden="true"><td></td><td></td></tr>
<tr data-hidden="true"><td></td><td></td></tr>
<tr><td></td><td></td></tr>
<tr><td></td><td></td></tr>
<tr><td></td><td></td></tr>
<tr><td></td><td></td></tr>
<tr><td></td><td></td></tr>
<tr><td></td><td></td></tr>
<tr><td></td><td></td></tr>
</tbody>
</table>

Answer 6

in jquery ..
var odd = true;
$('table tr:visible').each(function() {
$(this).removeClass('odd even').addClass(odd?'odd':'even');
odd=!odd
});

Answer 7

You can easily fake the zebra stripes if you apply a vertically repeating gradient on the parent table, positioned exactly to match the rows' height (the rows would have to be transparent). That way the table won't care if anything's hidden, it will repeat no matter what.

Answer 8

If anyone tries to do something like me, where I have alternating hidden and visible rows, you can do this:
.table-striped tbody tr:nth-child(4n + 1) {
background-color: rgba(0,0,0,.05);
}
This will get every 4th element starting with the 1st one, and allows you to maintain striping with hidden rows between each visible row.

Answer 9

Here is a 2022 version of a javascript version
let cnt = 0;
document.querySelectorAll("#mytable tbody tr").forEach(tr => {
cnt += tr.hidden ? 0 : 1;
tr.classList.toggle("odd",cnt%2===0);
});
.odd { background-color: grey; }
<table id="mytable">
<thead><tr><th>Num</th></tr></thead>
<tbody>
<tr><td>1</td></tr>
<tr><td>2</td></tr>
<tr><td>3</td></tr>
<tr hidden><td></td></tr>
<tr><td>5</td></tr>
<tr><td>6</td></tr>
</tbody>
</table>

Answer 10

I add in css:
tr[style="display: table-row;"]:nth-child(even) {
background-color: #f3f6fa;
}
and on create tr add in tag
style="display: table-row;"

Answer 11

Jquery codes for zebra color in html table
$("#mytabletr:odd").addClass('oddRow');
$("#mytabletr:even").addClass('evenEven');
And CSS you can do
.oddRow{background:#E3E5E6;color:black}
.evenRow{background:#FFFFFF;color:black}