Newbee question

Anon

I just began with learning php and i already have my first problem which i really do not understand.

I have the following code :

$result = mysql_query("SELECT name FROM user WHERE id=$id",$db) or die ("query failed");

echo "<center>";

$myrow = mysql_fetch_row($result);

echo $myrow[0];

echo "</center>";

With a form i enter a id (an int) and then it searches the db user for the name (a varchar) corresponding with the submitted id. Works perfectly.
However, if i switch id and name, and submit a name using the form and try to get the corresponding id, it always fails the query.
What am i doing wrong?
Why does it work with int's and not with strings?

Anon

Hopefully you are using seperate query structures for the two seperate queries!

To snag a name based on id, use the query you mentioned:

SELECT name FROM user WHERE id=$id

To snag an id base on name, change it around:

SELECT id FROM user WHERE name=$name

Anon

Well.. Brett is somewhat right.. But you forget about the '...

do it like this:

SELECT id FROM user WHERE name = '$name'

or else it wont work... you always got to have those ' when your working with strings...

Andreas Bernhardsen

Anon

After some trying i think it is just a problem with the syntax. I have the following code :

<body>

<?php

if ($submit)

{
$db = mysql_connect("localhost", "USERNAME", "********");

mysql_select_db("DBNAME", $db);

$result = mysql_query("SELECT name FROM user WHERE id=$id", $db) or die("fail");

printf(mysql_result($result,0,0));

}

else

{

?>
<form method="post" action="<?php echo $PHP_SELF?>">
<input type="text" name="id">
<input type="submit" name="submit" value="submit">
</form>
<?php

}

</body>

The table has two colums id (tinyint) and name(varchar).
When i use the above script and enter an int it works perfectly and returns the corresponding name.
Now i want to submit a name and have the id returned. I use the following code :

<body>

<?php

if ($submit)

{
$db = mysql_connect("localhost", "USERNAME", "********");

mysql_select_db("DBNAME", $db);

$result = mysql_query("SELECT id FROM user WHERE name=$name", $db) or die("fail");

printf(mysql_result($result,0,0));

}

else

{

?>
<form method="post" action="<?php echo $PHP_SELF?>">
<input type="text" name="name">
<input type="submit" name="submit" value="submit">
</form>
<?php

}

</body>

If i enter a name it ALWAYS fails the query.
However if i replace name=$name with name='ANYNAMEINDB' it gives the right id.

Why doesn't it work if i use a variable?
I think it has something to do with the syntaxis but what?

Anon

Just posted before i saw the new reply...

I tried the " ' " with the variable but then it didn't work, must be the spaces or something.

It works perfectly now!

Thanks all