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MYSQL Events (cron job)

I have a chatting system working fine. Two persons are chatting, one is visiting, and the other is a representative. when a visitor tries to chat, a random representative is selected from the online representatives.

What you need to updated:

if a representative is already chatting with someone else, he should not be assigned to anyone else.

if a representative does not responds for 2 minutes, he should be made offline by force.

if a visitor does not responds for 2 minutes, he should be made offline by force.

and anything else to make the chatting system perfect.

if a bidder does not explains the problem in his bid and his suggestions, his bid will be ignored and will be marked as spam

Kĩ năng: PHP, MySQL, HTML, Thiết kế trang web, SQL

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Về Bên Thuê:
( 2 nhận xét ) Rawalpindi, Pakistan

ID dự án: #30949957

7 freelancer chào giá trung bình$24 cho công việc này

(176 Nhận xét)

Hi , This is Ajay ! I can do this functionality easily as per your given requirement ! it would be great if we can connect over chat ! Having much expertise on PHP, Laravel ,CI Frameworks Websites. If you need assi Thêm

$30 USD trong 1 ngày
(78 Nhận xét)

Hello, I’m an experienced freelancer on this platform willing to help you with your problem. Changing the online status of your representatives within 2 minutes as well as marking them as occupied will be no problem f Thêm

$30 USD trong 1 ngày
(23 Nhận xét)
(2 Nhận xét)
(4 Nhận xét)

Hi, I will complete the conditions of assigning the representative to visitor and mark it offline as you have mentioned if he did not start the chat in 2 minutes. I can start right now. I am a skilled full-stack PHP Thêm

$25 USD trong 2 ngày
(1 Nhận xét)

Hi there! This is Sujan. Web Developer from Nepal. I worked as a Website Developer/Designer, Logo Designer, WordPress Designer, JQuery, PHP, JavaScript, HTML, Graphic Designer and so on. I can do Google Adsense and Goo Thêm

$20 USD trong 7 ngày
(0 Nhận xét)