Checking a username

[yogibear]

Hi I have been trying to get a check on a username field so that the username is unique
this is what i have but im getting errors
[code]<?php
$con = mysql_connect('localhost','web22-james','***');
if (!$con)

$username = $_POST['username'];
$email = $_POST['email'];
$pword1 = $_POST['pword1'];
$pword2 = $_POST['pword2'];

$query = "SELECT * FROM usersinformation WHERE username='$username' LIMIT 1";    // check if username already 
    $result = mysql_query($query);
if(0 != mysql_num_rows($result))
        echo "Our database shows this username has already been registered. Please choose another username.";
    } else {
            echo "WORKING!";
    }     
    mysql_close();
?> [/code]

any ideas what is the problem

many thanks

yogi

[c4onastick]

You need to bracket that first 'if', for starters:
[code]<?php
$con = mysql_connect('localhost','web22-james','stash');
if (!$con) {
  die("Connection failed.");
}else{

$username = $_POST['username'];
$email = $_POST['email'];
$pword1 = $_POST['pword1'];
$pword2 = $_POST['pword2'];

$query = "SELECT * FROM usersinformation WHERE username='$username' LIMIT 1";    // check if username already 
    $result = mysql_query($query);
if(0 != mysql_num_rows($result))
        echo "Our database shows this username has already been registered. Please choose another username.";
    } else {
            echo "WORKING!";
    }   
    mysql_close();
}
?> [/code]

What kind of errors are you getting? Post them, we'll help you sort them out.

[yogibear]

Hi this is the error

Parse error: syntax error, unexpected T_ELSE in /home/sites/***/public_html/registerconfirm.php on line 16

many thanks

yogibear

[simcoweb]

You're missing a { here:

[code]if(0 != mysql_num_rows($result)) [b]{[/b][/code]

And that's assuming you adapted c4onastick's changes also as that open 'if' statement will produce another error.

If statements always have code to run and that code is between curly's:

if ($this_does_not_work == true)
{
  echo "This does not work";
}

[yogibear]

hi im getting a error on line 14 now
[code]
<?php
$con = mysql_connect('localhost','***','***');
if (!$con) {
  die("Connection failed.");
}else{

$username = $_POST['username'];
$email = $_POST['email'];
$pword1 = $_POST['pword1'];
$pword2 = $_POST['pword2'];

$query = "SELECT * FROM usersinformation WHERE username='$username' LIMIT 1";    // check if username already 
    $result = mysql_query($query);
if(0 != mysql_num_rows($result)) [b]{[/b]
        echo "Our database shows this username has already been registered. Please choose another username.";
    } else {
            echo "WORKING!";
    }   
    mysql_close();
}
?>
[/code]

[Jessica]

This is why you don't just copy and paste code without understanding it. [b] isn't php, that's a bold tag.
Take that stuff out, and in the future POST the error, not just that you got one.

[yogibear]

[quote author=jesirose link=topic=123684.msg511547#msg511547 date=1169576103]
This is why you don't just copy and paste code without understanding it. [b] isn't php, that's a bold tag.
Take that stuff out, and in the future POST the error, not just that you got one.
[/quote]

hi i did post the error it was in my second post and i also tried it with and without the bold tag i didnt just Copy and Paste but that is why people get help because they dont understand

thanks for help

[marcus]

[quote author=yogibear link=topic=123684.msg511554#msg511554 date=1169576651]
[quote author=jesirose link=topic=123684.msg511547#msg511547 date=1169576103]
This is why you don't just copy and paste code without understanding it. [b] isn't php, that's a bold tag.
Take that stuff out, and in the future POST the error, not just that you got one.
[/quote]

hi i did post the error it was in my second post and i also tried it with and without the bold tag i didnt just Copy and Paste but that is why people get help because they dont understand

thanks for help
[/quote]

Posting the error would be like the error you posted before ERRORNAME on LINE ### on FILENAME

[yogibear]

This is the code
[code]<?php
$con = mysql_connect('localhost','***','***');
if (!$con) {
   die("Connection failed.");
}else{

$username = $_POST['username'];
$email = $_POST['email'];
$pword1 = $_POST['pword1'];
$pword2 = $_POST['pword2'];

$query = "SELECT * FROM usersinformation WHERE username='$username' LIMIT 1";    // check if username already 
    $result = mysql_query($query);
if(0 != mysql_num_rows($result)) {
        echo "Our database shows this username has already been registered. Please choose another username.";
    } else {
            echo "WORKING!";
    }     
    mysql_close();
}
?> [/code]

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/sites/***/public_html/registerconfirm.php on line 14

Many thanks

yogi

[simcoweb]

I was gonna say, personally i've not seen that 'if' statement written in that format. Here's what I use;

$num_rows = mysql_num_rows($result);

if ($num_rows => 0) {
    echo "Our database shows this username has already been registered. Please choose another username.";
} else {
    echo "WORKING!";
}

Also, your mysql connection is missing one vital ingredient. The database name:

mysql_select_db("database_name");

[Jessica]

Indeed, you need to select a db - do
$result = mysql_query($query) OR die(mysql_error());
This will show you when you have an SQL error. It would tell you something about no database.

[kenrbnsn]

The error you are receiving indicates that the previous mysql_query did not work correctly, change this line:
[code]<?php
$result = mysql_query($query);
?>[/code]
to
[code]<?php
$result = mysql_query($query) or die("Problem with the query:<pre>$query</pre><br>" . mysql_error());
?>[/code]
The output should give you a hint as to why the query didn't work correctly. If not post the output here.

Ken

[yogibear]

Hi this is my code i have no idea about it i have changed removed and added bits and now im lost with it. [code]<?php
$con = mysql_connect('localhost','***','***');
mysql_select_db("web22-james");

if (!$con) {
  die("Connection failed.");
}else{

$username = $_POST['username'];
$email = $_POST['email'];
$pword1 = $_POST['pword1'];
$pword2 = $_POST['pword2'];

$query = "SELECT * FROM usersinformation WHERE username='$username' LIMIT 1";

$result = mysql_query($query) or die("Problem with the query:<pre>$query</pre><br>" . mysql_error());

$num_rows = mysql_num_rows($result);
if ($num_rows => 0) {
    echo "Our database shows this username has already been registered. Please choose another username.";
} else {
    echo "$results";
}
    mysql_close();
?>[/code]

this is the html form
[code]<form action="registerconfirm.php" method="POST">
Username: <input type="text" name="username" maxlength="16" /><br>
E-mail: <input type="text" name="email"  /><br>
Password: <input type="text" name="pword1"><br>
Password (again): <input type="text" name="pword2"><br>
<input type="submit" value="Register">
</form>[/code]

when i click submit it says Parse error: syntax error, unexpected $end in /home/sites/***/public_html/registerconfirm.php on line 25

[Jessica]

if ($num_rows => 0) {

=> is syntax for assigning array values. You want >=

[yogibear]

[quote author=jesirose link=topic=123684.msg511609#msg511609 date=1169579762]
if ($num_rows => 0) {

=> is syntax for assigning array values. You want >=
[/quote]

I'm still getting the same error

thanks

yogi

[simcoweb]

You're missing a closing } to your first 'else' statement. Put it just before your closing ?> php tag.

[simcoweb]

Thanks, jesirose.. my fingers got ahead of me :)

[kenrbnsn]

You're missing the terminating "}" on the else statement:
[code]<?php
if (!$con) {
  die("Connection failed.");
}else{
?>[/code]
Put the "}" just before the "?>" at the end of your script.

Ken

[Jessica]

[quote author=yogibear link=topic=123684.msg511611#msg511611 date=1169579981]
[quote author=jesirose link=topic=123684.msg511609#msg511609 date=1169579762]
if ($num_rows => 0) {

=> is syntax for assigning array values. You want >=
[/quote]

I'm still getting the same error

thanks

yogi
[/quote]

Actually, I think you'd just want > - you don't want it to go if it's 0, just if it's > 0

[simcoweb]

he could also use

if ($num_rows == 1) {

[yogibear]

[quote author=jesirose link=topic=123684.msg511617#msg511617 date=1169580331]
[quote author=yogibear link=topic=123684.msg511611#msg511611 date=1169579981]
[quote author=jesirose link=topic=123684.msg511609#msg511609 date=1169579762]
if ($num_rows => 0) {

=> is syntax for assigning array values. You want >=
[/quote]

I'm still getting the same error

thanks

yogi
[/quote]

Actually, I think you'd just want > - you don't want it to go if it's 0, just if it's > 0
[/quote]

IT WORKS THANK YOU ;D simcoweb got it going but it always said it was in the database. jesirose fixed the last problem many thanks everyone