Echo variable

[Phpfr3ak]

I'm unsure as to why thsi doesnt work.. any clues?

$scoretype = mysql_real_escape_string($users['scoretype']); 
<?php echo $scoretype ?>

 

 

[PFMaBiSmAd]

Since you forgot to tell us what it does do vs what output you expected, no one here can directly help you. It also appears that those two lines of code are taken out of their actual context, so there could easily be a dozen different problems in the code you did not post that could cause the two lines you did post to not work.

 

Post the actual code that demonstrates the problem and tell us what the actual symptom is.

[Phpfr3ak]

Sorry, the code is to echo a certain aspect from the database that is selected i.e /playerranks.php?ReportType=score would echo the score colum, /playerranks.php?ReportType=cash_Stolen would echo the cash_stolen colum and so on, Sorry for not being so clear.

[arbitter]

Because there is no ';' ?

[PFMaBiSmAd]

To reference the value from the url for ?ReportType= xxxxx you would need to use $_GET['ReportType'] in your php code.

[Phpfr3ak]

Hmmm odd i tried:

Notice: Undefined index: $ReportType in C:\Program Files\EasyPHP-5.3.3\www\htdocs\playerranks.php on line 60

if i echo $reporttype the actual colum is echo'd, but i cant echo the specific reportype based for the users, Rather odd.

<?php $ReportType = mysql_real_escape_string($_GET['ReportType']); ?>
<?php echo $users['$ReportType'] ?>

[PFMaBiSmAd]

Php variables are not replaced with their value when they are enclosed by single-quotes.

 

Unless you are building a string that also contains literal characters in addition to php variables, there's no point in putting any quotes around a php variable.

[Phpfr3ak]

That was it, Thanks for that! Taught me something i didn't know :]