Finding difficult in looping table

Question

I want to get all <input>, select the name and all data-attributes which are inside the column in the table. I'm stuck, my table loop starts from 5th row instead of starting from 2nd;

var rowNo = jQuery(this).attr("data-row"); jQuery("#example1 table tbody").find('tr').each(function (key, value) { jQuery(this).find('td').each(function (key, val) { var tName = jQuery(this).find('input,select').attr('name'); console.log(tName); }); });

and

<div id="example1"> <table id="Invoice"> <tbody> <tr class="header_table"> <th></th> <th>InvoiceNo</th> <th>InvoiceDate</th> </tr> <tr class="even" name="tr_2"> <td><i class="fa fa-trash-o icon-class" data-row="2"></i></td> <td><input data-cols="1" data-row="2" name="input_[2][InvoiceNo]" type="text"></td> <td><input data-cols="2" data-row="2" name="input_[2][InvoiceDate]" type="text"></td> </tr> ..... ... </tbody> </table> </div>

output

"select_[5][Supplier]" "input_[5][Description]" "input_[5][Serial]" "select_[5][AssetType]" "select_[5][PurchasePrice]" "input_[5][PONumber]"

tested my the html table

 jQuery("table").find('tr').each(function(key,value){ jQuery(this).find('th,td').each(function(key1,val){ console.log(key1); }); });

I get output as:

3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 . ... ... ..

it just typo in question i even cannot loop from column 1 the table looping start from 3 — MayuriS
– MayuriS, Commented May 24, 2015 at 11:34
@MayuriS, see here (jsfiddle.net/2296gdru). after fix your html it works fine. — Mahedi Sabuj
– Mahedi Sabuj, Commented May 24, 2015 at 11:47

Tomalak · Accepted Answer · 2015-05-24 12:05:49Z

From the comments

i want retrieve all td data from table

I translate that as "I want to have a collection of all data in all input fields in my table".

This could look as simple as:

var data = {}; jQuery("#example1 table input").each(function () { data[this.name] = this.value; });

This produces an object with keys and values for further processing.

If the data is directly for transmission to a server, it becomes as simple as

var data = jQuery("#example1 table").serialize(); $.post("url", data);

@MayuriS Notice the difference between what you want to do and what you want to achieve? Next time you ask a question here, try to concentrate more on the desired result you want to have in the end (and less on the activities you think you need for that).
yes sure hey once again thanks for showing example of data transmission

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