Modified , I have attached link to the program codes to this post - Google drive , nairaland didn't allow txt file containing SQL query. l

I have an issue that I can't seem to find solution to for 3 days now.

Please help!

Overview : This a product rating / review system for a test website.

In the first screenshot below, the query on line 54 runs well.

In the second screenshot, I tried to get a value from a form in the previous page as use it as a variable in the query, then the query will not work.

The third screenshot, I realized that the if I put the codes such as below anywhere in the code, then the query won't work(including line 54 that was working before, and it doesn't return any error.

$work_id = $_POST['work_id'];


Link to the program file

Index
https://drive.google.com/file/d/1il156QZR3aWrCTg8q6rVCChvNWTD7YwO/view?usp=drivesdk

Submit rating
https://drive.google.com/file/d/1ibuXPrDmPYaIm5iLCxjImZjtZNJN3JNx/view?usp=drivesdk

Comment out the other part of the code, First is to know if the form is getting submitted, try to echo something out if submit is successful e.g echo "form submitted"

Or you can just return "form submitted";

Without commenting out anything

If unsuccessful, fix the form submit or the action path maybe.

If successful

Try to know if you're getting anything for $_POST ['work_id'], echo it

Or just echo Json_encode all the post it's getting, see it work_id is posted

If it's not working go to your form and fix the name probably

If all the above work correctly, then it's your sql statement most likely

Use a try catch to catch the exception abi na error you dey call am

Submit your code if nothing works for you make I help you check am...

Shalom

Mrchusenvessel:
Comment out the other part of the code, First is to know if the form is getting submitted, try to echo something out if submit is successful e.g echo "form submitted"

Or you can just return "form submitted";

Without commenting out anything

If unsuccessful, fix the form submit or the action path maybe.

If successful

Try to know if you're getting anything for $_POST ['work_id'], echo it

Or just echo Json_encode all the post it's getting, see it work_id is posted

If it's not working go to your form and fix the name probably

If all the above work correctly, then it's your sql statement most likely

Use a try catch to catch the exception abi na error you dey call am

Submit your code if nothing works for you make I help you check am...

Shalom

Thanks for your response,

The form is getting submitted in the database actually.

I also echoed : $_POST['work_I'd] and I got a response, the issues comes when I try to use it as a variable in the query. Then it won't work. This is a big headache for me!

Link to the program file

Index
https://drive.google.com/file/d/1il156QZR3aWrCTg8q6rVCChvNWTD7YwO/view?usp=drivesdk

Submit rating
https://drive.google.com/file/d/1ibuXPrDmPYaIm5iLCxjImZjtZNJN3JNx/view?usp=drivesdk

AllyPolly:
I have an issue that I can't seem to find solution to for 3 days now.

Please help!

In the first screenshot below, the query on line 54 runs well.

In the second screenshot, I tried to get a value from a form in the previous page as use it as a variable in the query, then the query will not work.

The third screenshot, I realized that the if I put the codes such as below anywhere in the code, then the query won't work(including line 54 that was working before, and it doesn't return any error.

$work_id = $_POST['work_id'];

Why not use a var_dump($connect); die;

After you make the query
And see what happened to the connection

Also you could add an if clause just incase $result was false
You’re making some assumptions in your code
First
you’re thinking $result would always be an array

secondly you’re thinking
$connect->query() always would run successfully.

AllyPolly:


Thanks for your response,

The form is getting submitted in the database actually.

I also echoed : $_POST['work_I'd] and I got a response, the issues comes when I try to use it as a variable in the query. Then it won't work. This is a big headache for me!

Lol... fixed yet? submit your code, you probably didn't turn on error reporting

Mrchusenvessel:


Lol... fixed yet? submit your code, you probably didn't turn on error reporting

No, please how can I do that?

I sent u a pm

stanliwise:

Why not use a var_dump($connect); die;

After you make the query
And see what happened to the connection

Also you could add an if clause just incase $result was false
You’re making some assumptions in your code
First
you’re thinking $result would always be an array

secondly you’re thinking
$connect->query() always would run successfully.

Nothing seem to be working

AllyPolly:


Nothing seem to be working

When you dump $result what happened?
Was it an array?
You also didn’t show use the dump of $connect
After you pushed the query

"SELECT * FROM review_table WHERE work_id = 'work_id'"

Where is the connection file you didn't include any connection file

Has it been solved?
If NO, the problem actually is that MariaDB does not support that syntax(I mean concatenating) due to some problems

WhatsApp 08122647016 for better communication

Nurudeen06:
Has it been solved?
If NO, the problem actually is that MariaDB does not support that syntax(I mean concatenating) due to some problems

WhatsApp 08122647016 for better communication

Update
After 6 days of trying, my boss here took it upon himself to make it work. You are God sent sir, and may God bless you. Even when I wanted to give up on the project, you said 'no, it must work', and you made it work!

Thanks so much

Mrchusenvessel:


Lol... fixed yet? submit your code, you probably didn't turn on error reporting

Thanks to this boss also, you really tried on my case, no be small. I really appreciate, and may God reward you abundantly.