PHP Parse error: syntax error, unexpected variable "$link_id", expecting ")" in file | error

Question

I'm new to PHP and I tried to make the code at the very bottom work but I can't make it work so I'm asking for help here. so please help me

php version i use: 8.1.2

errors i get while i try to load the page with the code:

[Thu Feb 17 12:51:36.272143 2022] [proxy_fcgi:error] [pid 473273] [client (ip)] AH01071: Got error 'PHP message: PHP Parse error: syntax error, unexpected variable "$link_id", expecting ")" in /var/www/legionbooter/public_html/lib/class_db.php on line 86'

[Thu Feb 17 12:52:21.093521 2022] [proxy_fcgi:error] [pid 473263] [client (ip)] AH01071: Got error 'PHP message: PHP Parse error: syntax error, unexpected variable "$database", expecting ")" in /var/www/legionbooter/public_html/lib/class_db.php on line 86'

the /var/www/legionbooter/public_html/lib/class_db.php file (the code that doesn't work): https://pastebin.com/Es33qTiw

the code you posted on line 84 there's a typo mistake...function select_db($link_id, $database,) it should be instead this function select_db($link_id, $database) — Rao DYC
– Rao DYC, Commented Feb 17, 2022 at 13:22
welcome to SO. you don't need to ask such questions directly here in SO, in this situation you its better to try the parser/compiler error and most probably you can solve it yourself. — Ebrahim
– Ebrahim, Commented Feb 18, 2022 at 11:15
Does this answer your question? PHP parse/syntax errors; and how to solve them — Perry
– Perry, Commented Feb 27, 2022 at 2:36

Isha · Accepted Answer · 2022-02-17 13:34:31Z

0

remove comma - function select_db($link_id, $database**,**)
first parameter should be connection and second should be database name, so correct this line also - if (!$this->select_db($this->database, $this->link_id))

answered Feb 17, 2022 at 13:34

Isha

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4 Comments

bro Over a year ago

i tried that and i think i did something wrong the new code: pastebin.com/Dcbu9AY6 errors: pastebin.com/D4x2ZbnN

Isha Over a year ago

previous code was correct, you just have to remove comma after $database in line 86(previous code) and instead of if (!$this->select_db($this->database, $this->link_id)), convert it if (!$this->select_db($this->link_id,$this->database)) in line 50

bro Over a year ago

i did everything as u said and now i have these errors: pastebin.com/LfR7nvkU

Isha Over a year ago

can you show your code of line 86

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