How to return the result without altering the source?
Question
My HTML and JavaScript code look like this:
<html>
<!--...
many code
...-->
<button onclick="alert($('#mytable').html().wrap('<html/>') );">Show HTML Table</button>
<table id="myTable">
<tr's and td's>
</table>
<!--...
many code
...-->
</html>
I want my javascript to return the table wrapped by the HTML tags but I do not want the table itself to be changed.
5 answers
solution1
4 2011-06-28 16:00:00
You could take a copy of the table first:
$('#mytable').clone()...
To get the actual HTML of the tag you'd need something like this plugin which I posted in another answer yesterday :
(function($) {
$.fn.outerhtml = function() {
return $('<div/>').append(this.clone()).html();
};
})(jQuery);
So you can then do:
alert('<html>' + $('#myTable').outerhtml() + '</html>');
See http://jsfiddle.net/alnitak/2y988/ for a working demo.
solution2
3 ACCPTED 2011-06-28 16:01:50
This does not work. .html() returns a string, not a jQuery object. So you cannot call wrap on it.
The other problem is that .html() only returns the inner HTML , it does not include the table tag.
You could .clone() the node, attach it to some dummy element and return the .html() :
var html = ['<html><body>',
$('<div/>').append($('#mytable').clone()).html(),
'</body></html>'].join('');
solution3
1 2011-06-28 16:01:00
Maybe this jQuery outerHTML plugin will help you. It will give you the code for the table, including the enclosing <table> tags. You can maybe do something like alert("<html>" + $("#myTable").outerHtml() + "</html>") .
solution4
0 2011-06-28 15:58:42
Why not just do the following?:
alert('<html>'+$('#mytable').html()+'</html>');
solution5
0 2011-06-28 15:59:33
$("#myTable").wrap("...");
That will wrap the table in the tag supplied to the wrap function, without altering the table itself.
For more information, see the jQuery API for the wrap function.
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