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syntax error, unexpected T_VARIABLE....

 

syntax error, unexpected T_VARIABLE....
Decypher

2 Jul, 2009 - 05:03 AM
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CODE
        $time = rand(30,90);
        
        $jailtime="INSERT INTO jail (username, time, length) VALUES ('".$username."', NOW(), '"$time"')";
        $jailresult = mysql_query($jailtime);


error:
Parse error: syntax error, unexpected T_VARIABLE in /home/a6490674/public_html/Mafia/crime.php on line 60

I know it means usually a semi-colon is missin but I can't see one missin here...any ideas?

EDIT: Title edited to be more descriptive

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Decypher

RE: Syntax Error, Unexpected T_VARIABLE....

2 Jul, 2009 - 05:34 AM
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[edit] resolved the problem
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dimjaxor

RE: Syntax Error, Unexpected T_VARIABLE....

2 Jul, 2009 - 05:35 AM
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QUOTE(Decypher @ 2 Jul, 2009 - 05:03 AM) *

CODE
        $time = rand(30,90);
        
        $jailtime="INSERT INTO jail (username, time, length) VALUES ('".$username."', NOW(), '"$time"')";
        $jailresult = mysql_query($jailtime);


error:
Parse error: syntax error, unexpected T_VARIABLE in /home/a6490674/public_html/Mafia/crime.php on line 60

I know it means usually a semi-colon is missin but I can't see one missin here...any ideas?


unexpected T_VARIABLE could mean a few things, missing ; or you didn't close brackets, but I am going to go with you left out the periods for the $time variable

CODE
        $time = rand(30,90);
        
        $jailtime="INSERT INTO jail (username, time, length) VALUES ('".$username."', NOW(), '"$time"')";  //RIGHT HERE -- >   '".$time."'
        $jailresult = mysql_query($jailtime);

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Decypher

RE: Syntax Error, Unexpected T_VARIABLE....

2 Jul, 2009 - 07:54 AM
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nah it was the fact I forgot to add the $ sign on the username part when I first dragged from the database... thanks anyways...



This post has been edited by Decypher: 2 Jul, 2009 - 08:10 AM
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