Maq

You need to put it in the while loop: while($sql_assoc = mysqli_fetch_assoc($sql_run)) { echo $sql_assoc['category']; }

GRooVeZ, in the future, place OR tags around your code.

Post the code you tried and the error messages you received.

Yes we need to see the actual code. It looks like your query starts out: INSERT into zen_blog (id, content, active) VALUES ('',content, Assuming "content" is a variable then you're missing the '$', if not, then you need quotes around it.

Repeating what everyone else said, users have a sufficient amount of warnings and chances to check their posts. Anyway, if it's reported it usually gets looked at within minutes.

Check out Github, it's a very popular and well designed form of version control.

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A lot of IDEs allow you to import custom formatting (usually in XML) and you can hotkey to quickly indent and format code.

sleepnot, In the future, place OR tags around your code.

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vjmehra, in the future, place OR tags around your code.

Have you read this sticky? http://www.phpfreaks.com/forums/index.php?topic=37442.0

This error is usually caused because your SQL is invalid. Try changing this line to: $result = mysql_query("Select words from CurrentWords") or die(mysql_error());

Yes, that's how you would use that function and yes you can use mysql_fetch_assoc in a function.

Inserting equal signs shouldn't be a problem. Can you post your code and what type of column you're inserting into?

If they didn't get any details then probably nothing... They could notify the police or local authorities with a description of your car but probably not.

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Maybe it's the way you CnP your code but you should properly format and indent your code to detect such errors.

You should be joining these tables: http://dev.mysql.com/doc/refman/5.0/en/join.html

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