在Jquery .filter（）函数内更改变量值

Question

I have a table which has id 'myTable'. Now there is a textbox in which I write down the item name to find whether it exists in the myTable cell or not. To do this I have written the following code -

var retval=0;   
var search = $("#searchitem").val().trim();
$("table#myTable tr td").filter(function() {
    if($(this).text() == search){retval=1;alert('Same item found');}
else{retval=0;}
});

The problem is that, when it finds the same item in the table cell it shows the alertbox. But the value of my variable retval never changes. It always shows 0 . How can I solve it?

Answer 1

I think what you are trying to accomplish could be done by using :contains() selector, like this:

var search = $("#searchitem").val().trim();
var retval = $("table#myTable tr td:contains('"+search+"'") ? 1 : 0;

I haven't tested it but I'm almost sure it works. It's a much more cleaner approach and surely more readable.

jQuery :contains() docs: http://api.jquery.com/contains-selector/

Answer 2

please try return false like this

var retval=0;   
var search = $("#searchitem").val().trim();
$("table#myTable tr td").filter(function() {
    if($(this).text() == search){
        retval=1;
        alert('Same item found');
        return false;
    }else{
        retval=0;
    }
});

Answer 3

change the variable value only if found , because you assign the variable to 0 at the beginning.

var retval=0;   
var search = $("#searchitem").val().trim();
$("table#myTable tr td").filter(function() {
    if($(this).text() == search){
        retval=1;
        alert('Same item found');
    }
});

Answer 4

Try this code

var search = $("#searchitem").val().trim();
var filteredData = $("table#myTable tr td").filter(function(element, index) 
{
    return $(element).text() == search;
});

now this code return array of TD elements which satisfy the condition.