How do i check a variable for a character?

SatanCow

Basicaly what i need is to figure out how to write an IF statement that will check the variable $userid to see if the last character is the number 2. I have no idea where to even start to find this information.

If someone could provide me with sample syntax i can take a look at that will help i would be quite greatfull.

LordShryku

[man]substr[/man] when looking at a specific piece of the string.

if(substr($userid, -1) == "2") {
   // true
}

TheDefender

$userid = "string2";

IF((substr($userid, -1)) == 2):
	echo "Yep";
ELSE:
	echo "Nope";
ENDIF;

SatanCow

Thanks so much, that worked. I have another problem in the same script later on. Should i post it in here? or make a totally new thread?

Either way thank you so much for the help I am EXTREMELY greatful.

LordShryku

Go ahead post it here. You already got our attention :p

TheDefender

Sorry to post over you like that LordShryku, I was composing when you posted... 😉 I will bow out when SatanCow posts the next question to this thread. (maybe)

SatanCow

$querystring = "SELECT COUNT (userid) AS count FROM inviter WHERE userid = '$userid'"; 
$query = mysql_query($querystring); 

while($queryobject = mysql_fetch_object($query)){ //line 16
   if($queryobject->count = 3){ 
      print "You can't invite any more friends you loser."; 
   } else { 
      $invitesleft = 3-$queryobject->count; 
      print "You can invite ".$invitesleft." more people."; 
   }; 
};

What this code is susposta do is check a table in the database called inviter, it will look to see if under the userid (not the primary key in the table) the same userid shows up 3 times or more, thus not allowing anyone to login and submit an invite for more then 3 users.

A bit more clarification, a user can login 3 times and invite 3 people, every time they invite someone there is an entry put in that table that is simular to newmember, newmemberemail, userid, email. The newmember is the person being invited (added later) the newmemberemail is the email of the person being invited (also added later) and the last 2 are the userid and email of the person inviting.

The script causes this error: Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in D:\Program Files\Abyss Web Server\htdocs\phpinvite\login_success.php on line 16

I put a comment on line 16 so you guys would know what one like 16 is.

hope this isin't too confusing.

LordShryku

Usually means a bad query/connection/db selection. Use [man]mysql_error[/man] to get a more detailed error.

$query = mysql_query($querystring) or die(mysql_error());

And no problem Defender, it happens. You may have a better solution than me, so post away.

SatanCow

that's kewl guys, i like the fact that both of you replied, it gave me a good idea of what the code could look like from 2 different prespectives. I find with coding if you are always shown how do do something 1 way if you ever have to do it a slightly different way your kinda screwed. So it's all good with me.

TheDefender

LOL... thanks guys. LordShryku, I still have my tail tucked between my legs after this afternoon 😉

I hear what you are saying SatanCow... LordShryku and I do have a different coding method that I noticed...
The glaringly obvious ones:

He uses the curly brackets, and I prefer the alternate syntax methods:
Him:
IF(){
}else{
}

Me:
IF():
ELSE:
ENDIF;

Oh, there is the other thing... More often than not, he is right, and I am not... LOL

SatanCow

Great that helped with the previous error but i think another peice of my code is messed up.

I get this error: You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near '(userid) AS count FROM inviter WHERE userid='SatanCow'' at line... it gets cut off at "line"

the code it's having a hard time with is:

$querystring = "SELECT COUNT (userid) AS count FROM inviter WHERE userid='$userid'";

LordShryku

Try pulling the space from between the count and paren, like count(userid).

And Defender, I may have whipped you like a lil girl this afternoon 😃
but I'd never go so far to think I know what I'm doing with this stuff :p

TheDefender

This may be one of those times where LordShryku and I post together, and mine is wrong and his is right, but I am going to jump back in here:

I believe you need to remove the space in your query between COUNT and (userid). Also, I write my code to escape the quotes on my variables too...

$querystring = "SELECT COUNT(userid) AS count FROM inviter WHERE userid='".$userid."'";

TheDefender

Dangit! Always the bridesmaid, never the bride😉 <<<SIGH>>> Beaten like a mule again! LOL I am going to bed now. I ain't had a beaten like that since some put a banana in my pants and let a monkey loose ;-) (Vegas Vacation)

LordShryku

I am super-fast-typing-man! 😃

The variable concatination like he showed is a good thing to get in the habit of using, with any string. You'll find yourself having trouble with arrays if you don't. You can also use braces around the variables. See here for some examples and explanations.

TheDefender

When I lived in Phoenix, I never could type that fast, because I couldn't move quick enough without burning up... You must have a real good Air Conditioner....

LordShryku

Yes, and a real high electric bill to go along with it 🙁

TheDefender

I hear ya... Mine averaged around 300+ a month in April?!? What's up with that? Ah, but I would move back there in a heartbeat if my wife didn't hate it so much... She couldn't stand how everything was brown.. She loves the grass, and ocean, and the salt river just didn't do it for her :p

Sorry guys, didn't mean to go off topic. SatanCow, how'd the advice work for you?

SatanCow

Ok i see, that actually makes some sence to me. The previous problem has been fixed. Unfortunatly for some reason the code always returns the value as true. As in if the user is in the database 0 times it counts that there in there too much, and if there in there 3 times the same thing.

The code (with alot of help from you 2) is looking like this:

$querystring = "SELECT COUNT(userid) AS count FROM inviter WHERE userid='".$userid."'";
$query = mysql_query($querystring) or die(mysql_error());

while($queryobject = mysql_fetch_object($query)){ 
   if($queryobject->count = 3){ 
      print "You can't more than 3 members."; 
   } else { 
      $invitesleft = 3-$queryobject->count; 
      print "You can invite ".$invitesleft." more people."; 
   }; 
};

I'm kinda glad to know that it's not working, cuz i would like to find out why a piece of code like this can return true no matter what. If i can figure out how to start to spot that kind of problem i think it would make a big difference in my coding.

LordShryku

I'm like that. My job's the only real thing keeping me here right now. The economy isn't really strong enough, and I'd have to find something with a lot of perks to leave. But I've been in AZ for 15 years, and I miss the small town I grew up in. Only a matter of time before I go back to it.