Two PHP problems, who will solve it ~~~

剑下亡魂

Two questions
Question one
@#What is it
require_once (@ # conn.php@#);
session_start ();
$ username = $ _ POST [@ # username @ #];
$ password = $ _ POST [@ # password @ #];
$ exec = "select * from admin where username = @ #". $ username. "@ #";
 
Question two
. Represents a connector
But what do you mean here
if ($ gradnum> 0) {
for ($ i = 0; $ i <$ gradnum; $ i ++) {
$ forums ['li_add']. = '<ul>';
}

jinjialin

1. I don't know
2. $ forums ['li_add']. = '<Ul>'; equivalent to $ forums ['li_add'] = $ forums ['li_add']. '<Ul>';

獠牙精英

1. It should be replaced ~~ Replace @ # with 'to restore!
2. Upstairs has already been said, equivalent to $ forums ['li_add'] = $ forums ['li_add']. '<Ul>';

爱迪生

1 question
   Using @ # as single quotes' should prevent the program's url from invading. . However, it is unclear what line is running for replacement.

2 questions
   The final result is that $ forums ['li_add'] contains a string separated by <ul>

   If it is a 1-dimensional array, it can be compressed into a string separated by <ul>

   $ forums ['li_add'] = imploce ('<ul>', $ someSingleArray);

爱迪生

It's implode, wrongly written ~~ sorry

qijun2005

1 I do n’t know, you did n’t give the full code

2 $ forums ['li_add'] = $ forums ['li_add']. '<Ul>';

gameyum

I compare dishes. . .
According to my observation, the role of @ is
No warning prompt is displayed when executing some potentially dangerous statements. . . .

I just cooked this comparison dish. . .

jas5454sdsa

It should not be as the upstairs said, @#成成 appears~ I am not clear

yezi4321

1 @ is an error control character, and control errors are not output. # is a single-line comment symbol for *nix, and it can also be used as a single-line comment in PHP. However, @# appears together and has not been seen.
Can you run the program containing this code?

2 +=,.=This is a matching operator, someone explained the usage upstairs. It is also often used in other languages.

vothailam

edit by:
$exec="select * from admin where username='$username'";