[RESOLVED] SQL Query with - 's in it

Wink-

Here's an easy one for you gurus.

I have a table with nicknames and fullnames. This query works perfectly fine for me except if there is a hyphen " - " in someones nickname. Such as T-bone. Works fine for Bob, Sal, and Winger.

How can I get my query to correctly search nicknames with hypens in them?

function nickname2fullname($nickname)
{
   $sql = "SELECT `FULLNAME` FROM `Peoples` WHERE `NICKNAME` LIKE    '$nickname'";
   $result = mysql_query($sql) or die('Error, query failed');
        if(mysql_num_rows($result) == 0)
        {
            echo "Database empty<br>";
        }
		$row = mysql_fetch_array($result);
   $fullname = $row['FULLNAME'];
   return $fulllname;
}

Weedpacket

Use = instead of LIKE; you're searching for something in particular, not just something that's like what you're searching for.

Wink-

Thanks for the tip! I've switched it over to = instead of LIKE and still have the same problem. You are correct in usage, the = will be better.

Wink-

Figured out the problem. If I add double quotation marks around my parameter, it works. Can anyone explain to me why this happens?

I just need to make the call like this:

nickname2fullname("T-Bone");
instead of:
nickname2fullname(T-Bone);

function nickname2fullname($nickname)
{
   $sql = "SELECT `FULLNAME` FROM `Peoples` WHERE `NICKNAME` LIKE    '$nickname'";
   $result = mysql_query($sql) or die('Error, query failed');
        if(mysql_num_rows($result) == 0)
        {
            echo "Database empty<br>";
        }
		$row = mysql_fetch_array($result);
   $fullname = $row['FULLNAME'];
   return $fulllname;
}

laserlight

Figured out the problem. If I add double quotation marks around my parameter, it works. Can anyone explain to me why this happens?

Oh, now I understand. The problem is that "T-Bone" is a string in PHP, but T-Bone is... well, I am not sure what it is, but the PHP interpreter probably (reluctantly) interpreted T as a string and Bone as another string, and the result of 'T'-'Bone' is 0. Effectively, you were calling nickname2fullname(0) instead of nickname2fullname("T-Bone").

I note that the PHP interpreter does this "reluctantly" since it will emit a notice, but you probably disabled notices in error_reporting, or worse still, you are programming with display_errors off. You should develop scripts with error_reporting set to at least E_ALL and display_errors on, and then switch display_errors to off for a production website and use logging instead.

Wink-

Thank you for the explanation laserlight. I really appreciate it.

That brings up a new question. How in the world can I take the raw form T-Bone and turn that into "T-Bone". I've tried every different version of concatenating on quotation marks I could think of.

Basically

nickname2fullname("T-Bone");

works fine, however

$foo = T-Bone;
$bar = " . $foo . ";

does not give $bar the value "T-Bone" which is the string I have to query.

laserlight

How in the world can I take the raw form T-Bone and turn that into "T-Bone".

The "raw form" is "T-Bone" (or 'T-Bone', which amounts to the same thing here).

If you leave out the quotes, you get a named constant instead of the string literal that you want. However, this named constant does not exist, so the PHP interpreter guesses that you made a mistake and decides to treat it as a string. You got away with this when you tested with Bob, but with T-Bone you end up with 'T'-'Bone'. The string 'T' when converted to an integer is 0. The string 'Bone' when converted to an integer is also 0. Since 0-0=0, the end result is 0 instead of your expected 'T-Bone'.

Wink-

laserlight, I'm really learning a lot from this. Thank you.

I read through the Type Juggling section in the PHP Manual hoping to figure out the answer but still no avail.

How can I assign a variable the value T-Bone and have it remain T-Bone instead of being interpreted as 0?

I have tried declaring the variable

CODE $raw = T-Bone;
echo $raw;[/CODE]

gives the output of 0 still.

laserlight

How can I assign a variable the value T-Bone and have it remain T-Bone instead of being interpreted as 0?

Write it like this:

$raw = 'T-Bone';
echo $raw;

I read through the Type Juggling section in the PHP Manual hoping to figure out the answer but still no avail.

Read the section on strings.

Wink-

Write it like this:

$raw = 'T-Bone';
echo $raw;

While yes that does work, I am going to be receiving my input for the function from a variable in an array which is being pulled from a mysql database.

$row['NICKNAME'] will have a value such as T-Bone. I need to put that value (T-Bone) in a sql query.

so essentially;

echo nickname2fullname($row['NICKNAME']);

has to interpret $row['NICKNAME'] as the string literal "T-Bone" instead of the value 0 so the query will query what I want it to.

laserlight

While yes that does work, I am going to be receiving my input for the function from a variable in an array which is being pulled from a mysql database.

$row['NICKNAME'] will have a value such as T-Bone. I need to put that value (T-Bone) in a sql query.

Yes, that is normal. $row['NICKNAME'] will have a value such as the string 'T-Bone', not a named constant T-Bone.

has to interpret $row['NICKNAME'] as the string literal "T-Bone" instead of the value 0 so the query will query what I want it to.

Therefore, you have nothing to worry about since the value of 0 will not happen unless the database field itself contains just a '0'. This is all about faulty test input, not a problem with SQL or your actual PHP code.

Wink-

Thank you for your help laser. You're right about everything, not that I doubted you!