Hello. - Printable Version

Hello. - Printable Version

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Hello. - iGetty - 06.06.2012

Edit:

I have a new problem.

Well; two.

1) How do I get multiple results from a sql database and then store them into a string?

I have this:

PHP код:



$masql = mysql_query("SELECT * FROM Accounts WHERE MasterAccount='$user'");

$marow = mysql_fetch_row($masql);

And

PHP код:



<div align="center"><h5>Current Characters</h5>

                <p><b><?php echo $marow[1]; ?></b></p></div>

IT should show up to 4 characters, not just one.

Also;

PHP код:



$sql = mysql_query("INSERT INTO Accounts (Username,Password,IP,MasterAccount) VALUES('$charname','$password', '$ip', '$user')") or die (mysql_error());

That doesn't store the result of $user into the MasterAccount section on the Table. It just leaves it blank.

Thanks!

Re: Hello. - DaRealShazz - 06.06.2012

I am new to PHP, but wouldn't you wanna put your variable outside of the quotation marks?

Re: Hello. - Stereotype - 06.06.2012

if u make something in php for email it should be like this .. $email

i mean

PHP код:



<?php echo "\$" . $row['$email']; ?>

That what u maked means like you searching all from the base

See here : http://www.learnphp-tutorial.com/Sessions.cfm

or try like this

PHP код:



echo "<div align='center'>$email</div>

Re: Hello. - Cell_ - 06.06.2012

Hello.

Re: Hello. - kikito - 06.06.2012

PHP код:



$newsql = mysql_query("SELECT * FROM Member WHERE id = '$userid'");

$row = mysql_fetch_array($newsql);  

echo '<p><center><b>E-mail Address: </b></center><div align="center"><b>'. $row['email'].'</b></div></p>';

Just to know, did you escaped the $userid?

Re: Hello. - iGetty - 06.06.2012

It's fine, I got it all working. Also Kikito, yes I did

Re: Hello. - iLinx - 06.06.2012

Just saying if you wanted to put a variable inside a double-quote string (this is possible), this is the correct format:

PHP код:



echo "<div align='center'>{$email}</div> ...

Not including the braces can confuse the compiler in some cases.

Re: Hello. - Mauzen - 06.06.2012

The query is the problem I guess. Variables in strings arent parsed afaik, use . to concatenate strings:

PHP код:



$newsql = mysql_query("SELECT `email` FROM Member WHERE id='".$userid."'");

$row = mysql_fetch_assoc($newsql);

NVM just read you already solved it

Re: Hello. - iGetty - 06.06.2012

Thanks guys, I've edited the first post.

Re: Hello. - donB - 07.06.2012

Quote:

Originally Posted by iGetty

Edit:

I have a new problem.

Well; two.

1) How do I get multiple results from a sql database and then store them into a string?

I have this:

PHP код:



$masql = mysql_query("SELECT * FROM Accounts WHERE MasterAccount='$user'");

$marow = mysql_fetch_row($masql);

And

PHP код:



<div align="center"><h5>Current Characters</h5>

                <p><b><?php echo $marow[1]; ?></b></p></div>

IT should show up to 4 characters, not just one.

I understand that you're trying to retrieve all the field values for the 'MasterAccount' $user, and displaying them in php.
If that's the case, try this:

PHP код:



$masql = mysql_query("SELECT * FROM Accounts WHERE MasterAccount='$user'");

if($masql)

{

while($marow = mysql_fetch_array($masql, MYSQL_NUM))

{

$field1 = $marow[0];

$field2 = $marow[1];

// and so on.. to whatever values you need in the table 'Accounts'


}}

Now, displaying them in php,

PHP код:



<div align="center"><h5>Current Characters</h5>

                <p><b><?php echo "$field1 <br /> $field2 </br> And so on.."; ?></b></p></div>

Quote:

Originally Posted by iGetty

Also;

PHP код:



$sql = mysql_query("INSERT INTO Accounts (Username,Password,IP,MasterAccount) VALUES('$charname','$password', '$ip', '$user')") or die (mysql_error());

That doesn't store the result of $user into the MasterAccount section on the Table. It just leaves it blank.

Thanks!

Do the remaining values store the result as expected?