cant find a sql/php query???

norok · #1 26-Jun-2003, 09:12

Hi ,
Tables like this:
table title
blabla
titelid,

table rel_user_titel
titelid
userid
usernr

location table
userid
prename
name

And need following:
to show the location, given the titelid i need a myasql
command in for php.
...select from
...
$result or somethin shoult contain
prename, name and userid,

Would be great to get something.

Matthias

ThX

JdS · #2 26-Jun-2003, 09:39

Try this:

PHP Code:


<?php

$sql = 'SELECT L.prename, L.name, L.userid
       FROM rel_user_titel AS R, `location` AS L
       WHERE R.titelid = '.$_GET['id'].'
       AND L.userid = R.userid';
?>

norok · #3 27-Jun-2003, 02:56

What do you mean with the letters?
t=title,l=location and what is "R"?
How does this command with "as" works?

Thanx

Matthias

norok · #4 27-Jun-2003, 03:02

I got it, I will test it
and then post again.

matthias

Garth Farley · #5 27-Jun-2003, 03:09

He's created a query that joins the 2 tables together. Using the AS command, he gives each table a temporary name (R and L), to make it easy to refer to it's contents.

Try to sketch out what happens when the query is run.

RETURN "prename", "name" and "userid" from the 'location' table (L).

WHERE "title_id" of 'rel_user_titel' is a certain value

AND "userid" of 'location' table = "userid" of 'rel_user_titel' table

So it returns the rows of the 2 tables which have the same userid, and has a certain value in "title_id" of 'rel_user_titel'

Hope this helps
GF

norok · #6 27-Jun-2003, 03:25

I changed the code to this:

PHP Code:


//query for Standort(location)
         
         $sql = "SELECT Standort.vorname, Standort.name, Standort.userid"
   line 244: "FROM rel_titel_user AS title, 'Standort' AS Standort"
               "WHERE title.titelid = '.$_GET['id']."
               "AND Standort.userid = title.userid";
          $result3 = mysql_query($sql);

Error

**** error: parse error, unexpected T_CONSTANT_ENCAPSED_STRING in d:\programme\apache group\apache\htdocs\bibli\find.php on line 244

Matthias

Garth Farley · #7 27-Jun-2003, 07:02

The entire SQL statement should be enclosed by a single pair of double quotes.

Read the error message. There's a string in an unexpected place. It's standing there on it's own, PHP doesn't realise it's part of the SQL.

Change the line to exactly how Jay posted it. I see you played about with it, I'm not sure it's safe to call Standort: Standort. mySql may get confused since you're renaming the table with the same name.

GF

norok · #8 27-Jun-2003, 08:26

No I need a short while or something construct to
"echo" the prename and the name.

Matthias

norok · #9 27-Jun-2003, 08:30

always only one name occurs, but i need a echo
like Location

rename name;

Thanx

Matthias

JdS · #10 27-Jun-2003, 18:29

try this:

PHP Code:


<?php
$title_id = intval( $_GET['id'] ); /*  Forgot to mention this before...
                                       make sure type is (int); IMPORTANT!  */
$sql = 'SELECT SO.vorname, SO.name, SO.userid
       FROM rel_titel_user AS RTU, Standort AS SO
       WHERE RTU.titelid = '.$title_id.'
       AND S.userid = RTU.userid';
$result3 = mysql_query( $sql );
// I suppose this could return more than one row, huh?
echo "<ol>\\n";
// to extend: use mysql_num_rows() to check if empty or n/a...
while( $row = mysql_fetch_assoc($result3) )
{
  echo     '  <li>Location: '.$row['vorname'].', '.$row['name']."</li>\\n";
}
echo "</ol>\\n";
?>

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cant find a sql/php query???

Thank you, but a question remains

ok

I tried it but errors occured

OK, I#ve fixed it

Ive tried some thing, but..

PHP MySQL's while() loop example