DataTable in php, Mysql and Ajax example by WebTuts

[@Egyptiano200]

nice tut pro

[@juniorWeb247]

thanku brother , its working Properly

[@sakiljomadder8725]

u made it simple, thanks a lot

[@rotanapisey4953]

how to use database query with join in server side processing?

[@lovewithcloud]

Thanks, can we send a post request with a parameter to any API to get the data ?

[@samirpatel2791]

Sumanta Sharma yes you can, all you need to check the documentation of site from which you want to retrieves the data like url and parameters. once you get response in json format , follow above tutorial to display data.

[@mustaphasodruldeen1076]

doesnt work , error DataTables warning: table id=myTable - Requested unknown parameter '0' for row 0, column 0

[@webtuts7548]

Web Tuts
Could you run the debugger or link to the page please? That will hopefully give me enough information to understand why it isn't working.

[@vishwajeetmandal8243]

I tried this but it throws an invalid json response error. kindly help

[@webtuts7548]

you can troubleshoot through developer tool under chrome(see the request in network). hope it'll help you or get back

[@webtuts7548]

you can check your json response following site.
jsonlint.com/
and see you have valid json or not

[@lumierre0791]

What if my approach was PDO and OOP? And All of my methods are in one php file?

[@cloudplayro]

As long as you're displaying a json it doesn't matter where your functions are :)

[@rzq8896]

How to add number per row?

[@muhammadaqil5724]

error DataTables warning: table id=myTable - Requested unknown parameter '0' for row 0, column 0. what the problem??

[@webtuts7548]

Could you run the debugger or link to the page please? That will hopefully give me enough information to understand why it isn't working.

[@rizz9737]

thanks

[@Avocat-du-diable-83]

this is not good, you must use the serverSide option and processing, and using LIMIT in the query, not selecting all the data from table. DataTable send variables like $_REQUEST["start"] which is the offset start of selection and $_REQUEST["length"] which is the LIMIT number.

[@christiantampus9798]

how to display data having the same age? pls help thank you

[@webtuts7548]

if you are talking about mysql then you need to write this query
SELECT * FROM table_name WHERE age = ENTER_AGE;

[@christiantampus9798]

ive try it sir but error says datatable error invalid json response

[@webtuts7548]

you can check your json response following site.
jsonlint.com/
and see you have valid json or not

[@Mr.Albertos]

Not working.

[@abeabe4953]

10 Q !!!!!!!!!!!!!!!!!

[@WPANDSEOTUTORIALS]

Not working !