PHP/MYSQL: What's Wrong With My Query/ PHP variable in query

Question

I've been coding with PHP and MySQL for about a year now and have gotten the hang of it pretty well; constructing really complicated queries with joins and calculated fields and all the other joys of MySQL hasn't been a problem for me in months.

BUT there's something syntactically screwy with the following chunk of code that I can't figure out - even though it's impossibly simple, and, even more infuriating, is closely related to other parts of the project that I'm working on (and which works flawlessly).

Here's the problem code I'm trying to run, followed by the bugchecking I've already done to try to isolate the problem.

If anyone has any suggestions, I'd be totally grateful because I'm beginning to lose my mind.

Problem:

I'm really losing my mind over this, so please don't laugh when you see the code:

$query="SELECT count(somefield) FROM db_name WHERE otherfield='".$myvariable."'";

My query finds no results when using a certain variable as part of a field search - even though I know that there are over 900 records in the database that should match.

Bugchecking:

• Because I know the value of the variable I'm passing to the query, I've tried hardcoding it into the query and it works fine.
• I've run the query in the MySQL console (again, of course, hardcoded instead of with the variable) and it works fine.
  • To my mind, these two facts eliminate the possibility that there's something syntactically incorrect with the PHP version of the query.
• In order to eliminate all possible database connection issues and to make sure the problem isn't related to iterating through the results returned, instead of trying to get the actual results, I've altered my original query to return only the count of the results and have incorporated the standard or die(mysql_error()) statements during the connection sequence. The query is executing, but is finding 0 results, so that eliminates the possibility that it's a connection issue.
• I've verified that the field I'm checking is the correct field for the information I'm looking for (like I said, it runs fine if I hardcode the variable into the query... which, of course, will not be an option in the finished code).
• I've checked the type of the variable before trying to pass it into the query (figuring that maybe, because it's pulled from a returned xml earlier in the script, that maybe it was showing up as an array or something). It typed as string.
• I've verified that the variable is formatted in the way that I expect it to be found in the database; strtoupper, etc.
• I've tried using LIKE '%".$myvariable."'"; still no dice.

Anyone have any suggestions for what I can do to figure out what the hell is going wrong? Thanks so much!

Answer 1

It's not a PHP issue so quotes have nothing to do with it. The query has no error, so you'll need to debug it step by step.

I. SELECT count(*) FROM table_name

II. SELECT count(*) FROM table_name WHERE field='$myvariable'

Where you're dealing with PHP variables in query, echo the query and run it directly in database to omit PHP's side of error.

III. SELECT count(somefield) FROM table_name WHERE field='$myvariable'

Answer 2

Is the $myVariable escaped? If not, escape it using

$escapedVariable=mysql_real_escape_string($myVariable);

and then run

$query="SELECT count(somefield) FROM db_name WHERE otherfield='$escapedVariable'";

Answer 3

Thanks so much to everyone who tried to help, but I figured it out several hours after posting: The first problem was that I forgot to use mysql_real_escape_string($myvariable)... the reasons for how and why I forgot are manifold, but there it is.

So, after plugging that guy back in (which I had ASSUMED had been in this particular module of my code in the first place, but that's where "assuming" gets you, lol), I thought I had the whole thing licked. Three hours later, and still nothing. THEN I realized that it had to be related to XML that was being parsed into the $myvariable... so around and around we went with that one for a few more hours.

FINALLY, I realized that the real culprit was my eyes (which aren't so great).. what looked like a perfectly legit quoted string while reading a debug echo of the query before running turned out to have leading and trailing white space (which, of course, I instantly removed with $myvariable=trim($myvariable, " "), and, that, of course, solved the entire problem...:< ... Yes, I am an idiot, and I'm sorry, but, after working over this UTTERLY INFURIATINGLY stupid line of code for over 48 hours (I'm used to writing things like:

$query="UPDATE db_one.table_one SET item1='".(string)$result_array[$i][1]."', item2='".(string)$result_array[$i][2]."' WHERE thing3=".(string)$result_array[$i][19];

... and other assorted fun nonsense), I had to resort to asking (because I - no pun intended - couldn't "see" the problem... ugh)... SO... I am an idiot, and I'm sorry (but encouraged by the efforts of all of you nice people who tried to help) and am sorry for wasting everyone's time. I need to learn how to handle XML much *more*.

Sorry and thanks again!

Answer 4

I always wrote it like this

$query="SELECT count(somefield) FROM db_name WHERE otherfield='$myvariable' ";

Try removing the double quotes and dot at the variable name

Answer 5

You shouldn't need the parenthesis around the var. Plus look at other changes.

$query="SELECT count(*) FROM table_name WHERE field='$myvariable'";