Amra Amra - 1 month ago 7
jQuery Question

Set width of td depending on No of td using jquery

I have several web pages with different amount of tables with different amount of columns.

I was looking on the net for a jquery spinet which gets the number of columns of the table and depending on the number of columns will define the width of each column.

Ex.

if (noOfTdOnTable == 2) {
tdWidth = "50%";
}
if (noOfTdOnTable == 3) {
td1Width = "40%";
td2Width = "40%";
td3Width = "20%";
}
if (noOfTdOnTable == 4) {
td1Width = "35%";
td2Width = "25%";
td3Width = "15%";
td4Width = "15%";
}


Update

Using the only answer I was given I have this at the moment but only works when there is one table on the page and I could not figure out how to apply when there are two columns.

var num = $("table > td").length;


if (num % 4 == 0) {
$("table > td:eq(0)").css("width", "50%");
$("table > td:eq(1)").css("width", "30%");
$("table > td:eq(2)").css("width", "10%");
$("table > td:eq(3)").css("width", "10%");
}
if (num % 3 == 0) {
$("table > td:eq(0)").css("width", "50%");
$("table > td:eq(1)").css("width", "40%");
$("table > td:eq(2)").css("width", "10%");
}


This is an example of the html, but the code will apply to lots of pages with different No of Tables but all tables will have either 2,3 or 4 columns.

<html>
<table>
<tr>
<td>text</td>
<td>text</td>
<td>text</td>
</tr>
</table>

<table>
<tr>
<td>text</td>
<td>text</td>
</tr>
</table>

<table>
<tr>
<td>text</td>
<td>text</td>
<td>text</td>
<td>text</td>
</tr>
</table>
</html>

Answer

To get the number of columns:

var num = $("#table > tr > td").length;

To specify the width:

$("#table > tr > td").width(w + "px");

I hope this was what you were looking for

Edit:

To specify the width to a specific column:

//if you've specified an id to each td

$("#td1").width(td1Width+"px");

//if you just use classes to identify them

$("td.td1", "#table1").width(td1Width+"px");

I would also recommend you to look into find() and end() to select columns in an efficient way. since making a $() call is a more expensive operation:

$("#table1").find("#td1").width(td1Width+"px").end().find("#td2")...

Edit 2

try this instead

$("table > tr > td:eq(0)").css("width", "50%");

or even better

$("table > tr > td").eq(0).css("width", "50%").end()
                    .eq(1).css(...etc;

"table" will select all tables on the page; use id or class to identify which table you require.

Edit 3 (final!!)

Ok, now i can see all the code i can give a better answer. try this:

var num;
var $tds;
$("table").each(function(i, t) {
   $tds = $("td", t);
   num = $tds.length;

   if (num % 4 == 0) {
        $tds.eq(0).css("width", "50%").end()
            .eq(1).css("width", "30%").end()
            .eq(2).css("width", "10%").end()
            .eq(3).css("width", "10%");
    }
    if (num % 3 == 0) {
        //etc
    }
});

I hope this is a better answer :)