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>
开发者_如何学Go <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>
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 :)
精彩评论