For everyone saying it doesn't work: 1. make sure you have set $user_id somehow 2. in the SQL statements put ' ' around the variables (variable = $...)
thanks a lot for this video. but plz do a voice over your work for future tutorials. i had a hard time understanding it for the first time. i've been watching it for the last 4 days
@@hazirahmusa2840 yes I can, see if the path in which you call the icon corresponds to the icon folder, if you don't resolve it call me here we will try to resolve it
@@hazirahmusa2840 the action function of like and unlike in the javascript file is the same as the video in minute 14:28? do you have github ?, could you give me this code for me to see?
uhm pls whenever i click on the like button and then on the dislike button the like button remains the same, i have checked every thing and it doesn't show any error, pls can you tell me where to check, thanks
Great Content Awa. thank God i found someone with a Nigerian assent like myself lol... Indians are difficult to understand sometimes... PLEASE I HAVE A REQUEST CAN YOU MAKE A PHP MYSQL ECOMMERCE APP WITH REMITTA OR PAYSTACK CHECKOUT | CHART | ???
. this code doesn't work for me . tried to debug this but the result is the same , i can't have two users on the same post id . but this one is still useful though , on this vid learned how to use data-id attr that will help me a lot
. yes but the whole code for me seems doesn't work . if your asking how did i done it just check your jquery syntax , the brackets, y spelling of the of statements . he's code is working until the loop
@Clash Guy in mine the situation was about users liking the same comment . for sure it's inside the loop . you can also try if your havind bad creating each function on each comment section to insert on database . that's what i did
Very much use full video, sirji...thankyou so much sir..for this clarity and for source code🙏
Amazing clarity. You probably maxed this out Bro.. It's so nice to see a new version of this demonstration in every 1-2 years from you. Respect, W
Thank you! A very nice tutorial with quite an elegant code. It's such a pity the video is mute.
Good job - it does take a short moment to apply the like after clicking on it - but it works good, thank you!
Your tutorials are always on point. Thanks a million!
For everyone saying it doesn't work:
1. make sure you have set $user_id somehow
2. in the SQL statements put ' ' around the variables (variable = $...)
still doesnt work for me
at last i found your video... the only video that help me a lot
It's the foreach. For sure it's the reason why I can only like once using the same user in all of the comments
JSON.parse() error shows unexpected character at line 1 column 1 of JSON data ??
thanks a lot for this video. but plz do a voice over your work for future tutorials. i had a hard time understanding it for the first time. i've been watching it for the last 4 days
Este agora e o que eu queria........ obrigado....
Like dislike buttons not working. How can I fix that?
thank you so much bro helped me a lot, good video and best
teaching, sorry my english hahaha
hi can you help me? i dont know why my code doesn't show the thumbs up & down icon
@@hazirahmusa2840 yes I can, see if the path in which you call the icon corresponds to the icon folder, if you don't resolve it call me here we will try to resolve it
@@kevenescovedo6585 so far i can see the icon, but it doesn't show any action when i clicked it..
@@hazirahmusa2840 sometimes javascript was not even called, to see this try to alert and appear and why it was called
@@hazirahmusa2840 the action function of like and unlike in the javascript file is the same as the video in minute 14:28? do you have github ?, could you give me this code for me to see?
If anyone has multiple comments and from comments, if it has numerous replies then the rating_info table has to include more columns.
uhm pls whenever i click on the like button and then on the dislike button the like button remains the same, i have checked every thing and it doesn't show any error, pls can you tell me where to check, thanks
THANKS for your tutorials. Next time can you make tutorials in PDO not mysqli or mysql :) I wasted few hours to read about PDO and to convert to PDO.
Awa, you have an awesome content you win a sub, go on with the great work.
Fatal error: Call to undefined function mysqli_fetch_all() in /home1/softskvd/public_html/like/server.php on line 119
But Why?
This code dosnt work
Great Content Awa. thank God i found someone with a Nigerian assent like myself lol... Indians are difficult to understand sometimes... PLEASE I HAVE A REQUEST CAN YOU MAKE A PHP MYSQL ECOMMERCE APP WITH REMITTA OR PAYSTACK CHECKOUT | CHART | ???
the data returned is not jason how could i fix that??
thank you very much very useful
mine have unexpected token error at JSON part. Does anyone face the similar problem before and already find a solution pls help me
Although I have the error but after refresh the page, the rating still can be shown in the webpage and store in the database
Use stringyfy instead of parse
How to do it on web?
I am not able to download source code.
veryyyyyy gooooood
wheres MySQL?
. this code doesn't work for me . tried to debug this but the result is the same , i can't have two users on the same post id . but this one is still useful though , on this vid learned how to use data-id attr that will help me a lot
Good job on this!!!!
Thank you!
good job man
your tutorials are very helpful :).. can you make a tutorial on how to make a Forum?
mybb
best tutorial. subscribe your channel thank you
If code give you "Unexpected token < in JSON at position 0" error change "JSON.parge" with "JSON.stringify"
same counter on all device
Well done
2x speed with some music is better
Please create FILTER.
Hey Rey Basilio, Thanks for watching! what kind of filter are you referring to? Can you please elaborate?
Using Option button. Example in school. When i choose IT, all the students of IT only will display in the form.
Ahh. I think it was a JQuery Filter? Using option button.
Doesn't work anymore!!!
y
ahaha. I'm not having any inserted data on my database whenever I clicked the like button
. yes but the whole code for me seems doesn't work . if your asking how did i done it just check your jquery syntax , the brackets, y spelling of the of statements . he's code is working until the loop
@Clash Guy in mine the situation was about users liking the same comment . for sure it's inside the loop . you can also try if your havind bad creating each function on each comment section to insert on database . that's what i did
Great script, thanks - but i can only like it once even if i like it from another device like from my mobilephone.