$5 paypal to first fixer,simple; How to see if someone already added that friend

Alright, i have a seperate table in my DB called: friends

It has 3 rows

1. id ( incremented row id .. nothing special )
2. user ( the user who added the other user as their friend )
3. myfriend ( the friend of user that they added )

Right now in the add friend file, all i have for adding the friend ( works fine ) is:

mysql_query("INSERT INTO friends 
(user, myfriend) VALUES('$uid', '$mem' ) ") 
or die(mysql_error());

echo "<b><center><font size='4'><u>$mem</u> is now your friend!</font></u></center></b><br>";

echo "<br><br><center><b><font size='5'><a href='member.php?msg=added'>Back to Member Area</a></font></b></center><br><meta http-equiv='refresh' content='3;url=http://www.**********/member.php?msg=added'>";

Code (markup):

But i need it to check if user has already added myfriend , and if they have, don't add them again and echo a message saying " $mem is already your friend! "

So, it checks if " user " has added " myfriend " before in the DB, and if they have, it wont show/do the insert into friends , and instead echo the message saying " $mem is already your friend! " error.

$5 paypal to the first person who get's me a working code that i try and am happy with!

 

I'm not a MySQL expert, so my way may not be the best, but it should work.

$result = mysql_query("SELECT FROM friends WHERE user='$uid' AND myfriend='$mem') 
if (!$result)
echo "Query failed!" . mysql_error();
else

if ($row = mysql_fetch_array($result)) {
  echo("$mem is already your friend");
} else {
mysql_query("INSERT INTO friends 
(user, myfriend) VALUES('$uid', '$mem' ) ") 
or die(mysql_error());

echo "<b><center><font size='4'><u>$mem</u> is now your friend!</font></u></center></b><br>";
}

Code (markup):

I'll check and verify that this works, but go ahead and try it out.

If it does, my paypal is

 

Im getting:
Parse error: syntax error, unexpected T_STRING in /home/ffilehos/public_html/***/maddfriend.php on line 117

Which is:

116: if (!$result)
117: echo "Query failed!" . mysql_error();
118: else

 

Just remove those 3 lines and see how it goes. They aren't awfully important, just a test to make sure that the query works first. Let me know how it works after that.

 

Parse error: syntax error, unexpected T_VARIABLE in /home/ffilehos/public_html/*****/maddfriend.php on line 118

 

117: if ($row = mysql_fetch_array($result)) {
118: echo("$mem is already your friend");
119: } else {

 

Check and make sure that there is a ";" at the end of each line... I think I missed one.

 

Which lines ? Im using this below, removed the 3 lines you said:

$result = mysql_query("SELECT FROM friends WHERE user='$uid' AND myfriend='$mem')

if ($row = mysql_fetch_array($result)) {
echo("$mem is already your friend");
} else {
mysql_query("INSERT INTO friends
(user, myfriend) VALUES('$uid', '$mem' ) ")
or die(mysql_error());

echo "<b><center><font size='4'><u>$mem</u> is now your friend!</font></u></center></b><br>";

 

Just that first line was missing it.

 

Wait, also need to add something else.

$result = mysql_query("SELECT FROM friends WHERE user='$uid' AND myfriend='$mem'");

I forgot the double quote at the end.

 

Still:

Parse error: syntax error, unexpected T_VARIABLE in /home/ffilehos/public_html/insanetext/maddfriend.php on line 118

w/o the 3 lines

 

What does it look like now? We might need to pull the variables out of the string...

 

Exactly:

$result = mysql_query("SELECT * FROM friends WHERE user='$uid' AND myfriend='$mem');

if ($row = mysql_fetch_array($result)) {
echo "$mem is already your friend";
} else {
mysql_query("INSERT INTO friends (user, myfriend) VALUES('$uid', '$mem')")
or die(mysql_error());

echo "<b><center><font size='4'><u>$mem</u> is now your friend!</font></u></center></b><br>";
}

Code (markup):

 

ok if you need help pm me i cant know where you are at now so simply pm me if you still need help.

 

There was something that I had missed earlier, that might be causing it. Try this first:

$result = mysql_query("SELECT * FROM friends WHERE user='$uid' AND myfriend='$mem'");

Add the double quote to the end of the line just before the closing parenthesis.

 

I think its working! No error codes, and i added someone who is already my friend and it didnt do so, and showed the error let me test it for like two more minutes, i already have your paypal but ill let you know anyway!

 

Good deal. Sorry for missing that earlier, I guess I was just typing too quick so I could be the first to reply.

 

$query = "INSERT INTO friends(id, user, myfriend) VALUES('', '$uid', '$mem' ) ";
if(!$query)
{
echo "error";
}
else
{
echo "<b><center><font size='4'><u>$mem</u> is now your friend!</font></u></center></b><br>";

echo "<br><br><center><b><font size='5'><a href='member.php?msg=added'>Back to Member Area</a></font></b></center><br><meta http-equiv='refresh' content='3;url=http://www.**********/member.php?msg=added'>";
}

 

Sending payment now, thread "closed."