Help! Can't find error in query string

crystalsilver

I hope somebody's out there surfing these forums who's bored to death because I have a short query string that keeps giving me errors, and I don't know if it's just that I've been staring at it too long and my eyes are glazing over the obvious faux-pas or that I'm just not paying close enough attention, but it's about to make me pull my hair out! Hope someone sees the mistake right away!

🙂

$sections = mysql_query("CREATE TABLE sections (
unique_id mediumint(10) unsigned PRIMARY KEY AUTO_INCREMENT,referrer mediumint(10) unsigned,priority mediumint(10) unsigned,active enum('YES','NO'),section varchar(75),category varchar(50),class varchar(50),topic varchar(25),subtopic varchar(25),date date,title varchar(100),keywords varchar(255),description varchar(255),stylesheet varchar(25),tagline varchar(255),caption varchar(100),intro varchar(255),body text,url varchar(100),resource varchar(255),publication varchar(50),editor varchar(25),amount decimal(8,2),fname varchar(25),lname varchar(25),position varchar(100),company varchar(100),address1 varchar(50),address2 varchar(50),postal mediumint(10) unsigned,country varchar(50),phone mediumint(25) unsigned,email varchar(100),list varchar(100),$link");

$schema = mysql_query($sections,$link);
if ($schema)
{
echo "Database Schema Created";
}
elseif (!$schema)
{
echo mysql_errno().": ". mysql_error ()."";
return 0;
}

I'm getting the error at the first column name of the table creation (unique_id). It doesn't seem to like something I've used in the part of the query string that creates the table's columns.

I've input this create table query in the command line terminal, and it works fine. It must be an error where I'm forgetting to escape something that PHP doesn't understand...because MySQL understands the query perfectly.

I've also tried other variations of ' and " and such, and I also tried ",$link); in place of ,$link");

I also tried separating out the create table query string into a separate variable that I could plug in its place immediately before ,$link because I thought it would make finding quotation errors easier for me - that didn't work either.

Finally, after making my way through a bunch of different errors I got each time I changed something, here's what it finally gave me after the portion of my script immediately preceding this part successfully does everything clear up to selecting the database in which to perform this query:

1065: Query was empty

I know, it's probably a really obvious oversight...but, that's what everyone's here for, right? 🙂

Maybe I should do the following instead?

$schema = mysql_query($sections,$link);

...and place the create table string all by itself in $sections ???

I could have sworn that's what I did in the first place, but maybe I missed a quote or something....

Anyway, any help would be greatly appreciated.

Thanks,

Crystal

crystalsilver

Instead of what was in my first post, I did the following:

$sections = "CREATE TABLE sections (
unique_id mediumint(10) unsigned PRIMARY KEY AUTO_INCREMENT,referrer mediumint(10) unsigned,priority mediumint(10) unsigned,active enum('YES','NO'),section varchar(75),category varchar(50),class varchar(50),topic varchar(25),subtopic varchar(25),date date,title varchar(100),keywords varchar(255),description varchar(255),stylesheet varchar(25),tagline varchar(255),caption varchar(100),intro varchar(255),body text,url varchar(100),resource varchar(255),publication varchar(50),editor varchar(25),amount decimal(8,2),fname varchar(25),lname varchar(25),position varchar(100),company varchar(100),address1 varchar(50),address2 varchar(50),postal mediumint(10) unsigned,country varchar(50),phone mediumint(25) unsigned,email varchar(100),list varchar(100)
);";
$schema = mysql_query($sections,$link);
$res = mysql_query($sections,$link);
if (!$res)
{
echo "Database Schema Created<BR>";
}
if ($res)
{
echo mysql_errno().": ". mysql_error ()."";
return 0;
}

...and it worked. I sometimes get confused about which is which when using "if" statements with "!" in front of variable.

The only problem is that now i have to paste in this same code for 4 other table names (different columns, of course) - at first, I tried doing this all in one shot with a single multi-dimensional array, but I kept getting errors I suspect were because I kept missing an incorrect single or double quote somewhere - but I got frustrated with myself and separated the arrays.

then I want to fetch the entire database to display everything that I've just added to it - as a cross-check. This is a portion of an install script that will eventually be have several queries in it, so I want to be able to display results of each in an organized table-like format so I don't have to go into shell or phpmyadmin every time I want to see if what I did worked.

Now, this is the part of my script that I'm having errors pop up in...any quick fixes will be greatly appreciated...otherwise, I'm off to try to figure out where I'm going wrong in the next query.

🙂

tha_mink

I don't get it. Are all these tables getting constantly created and droped or is it a one time deal?

If these aren't dynamic tables just use the mysql monitor from the command line. Make your table declarations in a text file and run them via the monitor with

. /path/to/textfile.txt

then
select * from table;

Why does this all have to happen through a php script?

crystalsilver

For my own purposes, it doesn't. I can run it in command line just fine...but this script is also going to be use by some people who have web sites who wouldn't understand that if I tried to explain it to them.

For these people, I'm creating an install script - they'll only have to edit config.php to type in username and password and database name.

The rest of the script that you haven't seen shows why this is important.

The tables themselves are going to be static, I hope, but if someone DOES need to change them, I'll be able to make a small modification to give the install script an input form or something - but at least I won't have to start from scratch.

The categories, classes, topics, and subtopics are going to need to be chosen and input by the admin. My goal is to make this as much GUI as possible for those who haven't a clue how to read any kind of code - and for those using virtual hosts who don't have shell access but only have ftp and web access.

As for SELECT * from TABLE, that is the query I'm using right now. I just haven't finished with it yet...

Besides, why does it matter WHY I want to perform a particular query? Shouldn't it only matter that I DO want to and am asking for help as to how to fix an error? Aren't my reasons for doing something irrelevant as long as they succeed at accomplishing what I've set out to do?

If you know of a way to make this dynamic WITHOUT all the steps I've used above in my previous posts, by all means, let me know.

But, I can't put it in a text file because then I'll have to constantly update the web sites of everyone I know - as nobody else will be able to figure out how to do it right and will probably screw it up...some people just don't have the knack for this sort of thing - no matter how well you explain it.

crystalsilver

No further need to wonder why I'm writing an install script...I fixed my problem on my own...usually how it goes...once you take the time to ask, the answer comes to you quicker than anyone has a chance to give you one of their own.

🙂

Thanks again.