Idea

$review = $_POST['review'];

insertIntoDatabase($review); Does anybody know how to do this part as it does seem to be working it either says

That's not it. You need to pass the $review variable through the escape_tags(); function.

Don't forget to check your notepad!

Or it Says...

Having trouble with the syntax? Here's an example of passing a variable through a function

$myVariable = theFunction($myVariable);

Don't forget to check your notepad!

Which is sooooooooooo annoying

Reply 1

NYANCAT_2.0

OP

6

Plz help I’ve been trying to stick my head around idea for a long time and i still dont get much -

I also done most my badges on silver award just not this one + another but that’s first things first.
In case you can follow me and see other problems that MAAAAANY people have on idea.

Reply 2

frvbee

4

LMAOO i'm stuck on that one rn too, have you figured it out in the past 10months?

Reply 3

Rinsler

2

Me to do you might have a clue for this

Reply 4

Rinsler

2

I have also tried $myvariable=escape tags($myvariable)I but it didn't work

Reply 5

tom foz

1

right ive done ityou need to out in $myFormData = escape_tags($myFormdata);insertIntoDatabase($myFormData) $ but replace the $myFormData with $review so it looks like $review = escape_tags($review);insertIntoDatabase($review) $

doesn't work for me

it doesn't work for me

Reply 8

sealysealy1

1

help I've been stuck on this forever

Reply 9

HelenHayes1

1

Original post by tom foz

right ive done ityou need to out in $myFormData = escape_tags($myFormdata);insertIntoDatabase($myFormData) $ but replace the $myFormData with $review so it looks like $review = escape_tags($review);insertIntoDatabase($review) $

This worked for me! Thank you my daughter has been stuck on it for days!

Reply 10

sandd

1

i dont get it

Idea

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