RE: [PHP-DB] Select Statement with output in different colors.
Thank you for the heads up. I did notice that is must have a ; after
every variable case entry, e.g. $bg = 'blue'; in order to work right.
That was at least how I had to get it working.
> -Original Message-
> From: Becoming Digital [mailto:[EMAIL PROTECTED]
> Sent: Tuesday, June 17, 2003 9:51 PM
> To: [EMAIL PROTECTED]
> Subject: Re: [PHP-DB] Select Statement with output in different
colors.
>
> Here's a slight alteration to Peter's code in "case" someone isn't
> familiar with
> switch statements. Ooh, such a bad pun, as if there's any other kind.
>
> $query = 'SELECT * FROM table
> $mysql_result = mysql_query($query, $link);
>
> while($row = mysql_fetch_array($mysql_result))
> {
> switch ($row["event"] )
> {
> case event_one:
> $bg = 'blue'
> break;
> case event_two:
> $bg = 'red'
> break;
> case event_three:
> $bg = 'green'
> break;
> default:
> $bg = 'white'
> }
> print '
> da da da
> ';
> }// end while
> ?>
>
>
> Edward Dudlik
> Becoming Digital
> www.becomingdigital.com
>
>
> - Original Message -
> From: "Peter Lovatt" <[EMAIL PROTECTED]>
> To: "Christopher Lyon" <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
> Sent: Tuesday, 17 June, 2003 17:42
> Subject: RE: [PHP-DB] Select Statement with output in different
colors.
>
>
> Try this
>
> $query = 'SELECT * FROM table
> $mysql_result = mysql_query($query, $link);
>
> while($row = mysql_fetch_array($mysql_result))
> {
>
>
> switch ($row["event"] )
> {
> case event_one:
> $bg = 'blue'
> break;
> case event_one:
> $bg = 'red'
> break;
> case event_one:
> $bg = 'green'
> break;
> default:
> $bg = 'white'
>
> }
>
>
> print '
> da da da
> ';
>
>
> }// end while
>
>
>
>
> HTH
>
> Peter
>
>
>
> -Original Message-
> From: Christopher Lyon [mailto:[EMAIL PROTECTED]
> Sent: 17 June 2003 22:35
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] Select Statement with output in different colors.
>
>
> I have a php script that does some select statements, on mysql, and
> outputs that to a table. I would like to change the color of the table
> row depending upon one of the fields. The select statements are from a
> syslog database that I have and I would like to highlight key events
by
> changing the colors for that row. Example would be MAJOR, yellow and
> INFO, green. Does anybody have any examples of how to do this?
>
>
>
>
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Re: [PHP-DB] Select Statement with output in different colors.
Here's a slight alteration to Peter's code in "case" someone isn't familiar with
switch statements. Ooh, such a bad pun, as if there's any other kind.
da da da
';
}// end while
?>
Edward Dudlik
Becoming Digital
www.becomingdigital.com
- Original Message -
From: "Peter Lovatt" <[EMAIL PROTECTED]>
To: "Christopher Lyon" <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Tuesday, 17 June, 2003 17:42
Subject: RE: [PHP-DB] Select Statement with output in different colors.
Try this
$query = 'SELECT * FROM table
$mysql_result = mysql_query($query, $link);
while($row = mysql_fetch_array($mysql_result))
{
switch ($row["event"] )
{
case event_one:
$bg = 'blue'
break;
case event_one:
$bg = 'red'
break;
case event_one:
$bg = 'green'
break;
default:
$bg = 'white'
}
print '
da da da
';
}// end while
HTH
Peter
-Original Message-
From: Christopher Lyon [mailto:[EMAIL PROTECTED]
Sent: 17 June 2003 22:35
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Select Statement with output in different colors.
I have a php script that does some select statements, on mysql, and
outputs that to a table. I would like to change the color of the table
row depending upon one of the fields. The select statements are from a
syslog database that I have and I would like to highlight key events by
changing the colors for that row. Example would be MAJOR, yellow and
INFO, green. Does anybody have any examples of how to do this?
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RE: [PHP-DB] Select Statement with output in different colors.
Try this
$query = 'SELECT * FROM table
$mysql_result = mysql_query($query, $link);
while($row = mysql_fetch_array($mysql_result))
{
switch ($row["event"] )
{
case event_one:
$bg = 'blue'
break;
case event_one:
$bg = 'red'
break;
case event_one:
$bg = 'green'
break;
default:
$bg = 'white'
}
print '
da da da
';
}// end while
HTH
Peter
-Original Message-
From: Christopher Lyon [mailto:[EMAIL PROTECTED]
Sent: 17 June 2003 22:35
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Select Statement with output in different colors.
I have a php script that does some select statements, on mysql, and
outputs that to a table. I would like to change the color of the table
row depending upon one of the fields. The select statements are from a
syslog database that I have and I would like to highlight key events by
changing the colors for that row. Example would be MAJOR, yellow and
INFO, green. Does anybody have any examples of how to do this?
--
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[PHP-DB] Select Statement with output in different colors.
I have a php script that does some select statements, on mysql, and outputs that to a table. I would like to change the color of the table row depending upon one of the fields. The select statements are from a syslog database that I have and I would like to highlight key events by changing the colors for that row. Example would be MAJOR, yellow and INFO, green. Does anybody have any examples of how to do this? -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Select statement drops 10th digit during table lookup?
Why not use the built-in conversion functions in mysql?
>From the manual:
INET_NTOA(expr)
Given a numeric network address (4 or 8 byte), returns the dotted-quad representation
of the address as a string:
mysql> SELECT INET_NTOA(3520061480);
-> "209.207.224.40"
INET_ATON(expr)
Given the dotted-quad representation of a network address as a string, returns an
integer that represents the numeric value of the address. Addresses may be 4 or 8 byte
addresses:
mysql> SELECT INET_ATON("209.207.224.40");
-> 3520061480
Doug
On Fri, 13 Jun 2003 10:19:32 +0100, Ronan Chilvers wrote:
>Here's how I did it when playing with this...
>
>Table definition:-
>
>CREATE TABLE ip_list (
> ip_from double default NULL,
> ip_to double default NULL,
> country_code char(2) default NULL,
> country_name varchar(100) default NULL,
> KEY ip_from (ip_from,ip_to,country_code,country_name)
>) TYPE=MyISAM;
>
>Basic code to query it (takes a parameter 'ip' as GET paramter):-
> /*
> IP Lookup script
> * prints a country code and name for an ip in the $_GET array
> */
>
> //Vars & Constants
> define("HOST","myhost");
> define("USER","myuser");
> define("PASS","mypassword");
> define("DB","ips");
> define("IP",$_GET["ip"]);
>
> // Functions
> function dbconnect() {
> if (($conn=mysql_pconnect(HOST,USER,PASS))===false) {
> die("Couldn't connect to server");
> }
> if (!mysql_select_db(DB,$conn)) {
> die("Couldn't select database");
> }
> return $conn;
> }
> function dbclose($conn) {
> if (!mysql_close($conn)) {
> die("Couldn't close database connection");
> }
> return true;
> }
> function dosql($SQL) {
> if (($result=mysql_query($SQL))===false) {
> die("Couldn't do sql : $SQL");
> }
> return $result;
> }
>
> //Core
> $SQL = "SELECT COUNTRY_NAME AS cn, COUNTRY_CODE AS cc FROM ips.ip_list WHERE
> IP_NUM BETWEEN ip_from AND ip_to";
>
> $conn = dbconnect();
>
> $ip = "127.0.0.1";
>
> $QSQL = str_replace("IP_NUM",sprintf("%u",ip2long(IP)),$SQL);
>
> $result = dosql($QSQL);
> $data = mysql_fetch_array($result);
>
> echo "ip:".IP."";
> echo "cn:".$data["cn"]."";
> echo "cc:".$data["cc"];
>
> dbclose($conn);
>?>
>
>HTH !!
>
>Ronan
>e: [EMAIL PROTECTED]
>t: 01903 739 997
>w: www.thelittledot.com
>
>The Little Dot is a partnership of
>Ronan Chilvers and Giles Webberley
>
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Re: [PHP-DB] Select statement drops 10th digit during table lookup?
Here's how I did it when playing with this... Table definition:- CREATE TABLE ip_list ( ip_from double default NULL, ip_to double default NULL, country_code char(2) default NULL, country_name varchar(100) default NULL, KEY ip_from (ip_from,ip_to,country_code,country_name) ) TYPE=MyISAM; Basic code to query it (takes a parameter 'ip' as GET paramter):- "; echo "cn:".$data["cn"].""; echo "cc:".$data["cc"]; dbclose($conn); ?> HTH !! Ronan e: [EMAIL PROTECTED] t: 01903 739 997 w: www.thelittledot.com The Little Dot is a partnership of Ronan Chilvers and Giles Webberley -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Select statement drops 10th digit during table lookup?
On Friday 13 June 2003 15:48, G & E Holt wrote: > I hope this is something easy I am overlooking but here goes: > > I have a table called: country which has the following fields: > > FieldType > ip_from double(11,0) > ip_todouble(11,0) > country_code char(2) > country_name varchar(50) Shouldn't you be using some INTEGER type instead of DOUBLE? An unsigned INT will just about handle it (2^32). -- Jason Wong -> Gremlins Associates -> www.gremlins.biz Open Source Software Systems Integrators * Web Design & Hosting * Internet & Intranet Applications Development * -- Search the list archives before you post http://marc.theaimsgroup.com/?l=php-db -- /* It seems like the less a statesman amounts to, the more he loves the flag. */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Select statement drops 10th digit during table lookup?
On 13 Jun,2003 at 0:48 G & E Holt wrote:
> $resolved=$DB_site->query_first("SELECT * FROM country WHERE
> '$resolvingip' BETWEEN ip_from AND ip_to");
Don't think you should have the '' around the ip number. Does that help?
Ronan
e: [EMAIL PROTECTED]
t: 01903 739 997
w: www.thelittledot.com
The Little Dot is a partnership of
Ronan Chilvers and Giles Webberley
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[PHP-DB] Select statement drops 10th digit during table lookup?
I hope this is something easy I am overlooking but here goes:
I have a table called: country which has the following fields:
FieldType
ip_from double(11,0)
ip_todouble(11,0)
country_code char(2)
country_name varchar(50)
IP_FROM NUMERICAL (DOUBLE)Beginning of IP address range.
IP_TONUMERICAL (DOUBLE)Ending of IP address range.
COUNTRY_CODE CHAR(2) Two-character country code based on
ISO 3166.
COUNTRY_NAME VARCHAR(50) Country name based on ISO 31
Here are a few lines from the country table which starts at 33996344 and
ends at:
ip_from ip_to country_code country_name
33996344 33996351 GB UNITED KINGDOM
50331648 83886079 US UNITED STATES
94585424 94585439 SE SWEDEN
3714175488 3714175743 DK DENMARK
3715104768 3715629055 JP JAPAN
3717201920 3718840319 KR KOREA, REPUBLIC OF
The IP_FROM and IP_TO fields of the database are numeric representations of
the dotted IP address.
The formula to convert an IP Address of the form A.B.C.D to an IP Number is:
IP Number = A x (256*256*256) + B x (256*256) + C x 256 + D
So here is what I do in PHP:
$resolvingip=sprintf("%u",ip2long($ip));
$resolved=$DB_site->query_first("SELECT * FROM country WHERE '$resolvingip'
BETWEEN ip_from AND ip_to");
$resolvecountry = $resolved[country_code];
so I type in: 213.21.158.96 and $resolvingip shows as: 3574963808
when I check $resolved[ip_from] I find out the SELECT statement looks up
this line:
335544320 369098751 US UNITED STATES
instead of looking up this line:
3574956032 3574972415 IT ITALY
which really makes me think the SELECT statement is dropping the last
digit...
I tried ip: 10.25.215.30 and $resolvingip shows as 169465630 which looks up
correctly since it is only 9 digits.
When I try ip: 68.109.155.135 $resolvingip shows as 1148033927 but the
SELECT statement drops the last digit and goes to a 9 digit again:
114803392
Any help is greatly appreciated..
Thanks!
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RE: [PHP-DB] select statement
What's the error?
-Original Message-
From: Steve Dodkins [mailto:Steve.Dodkins@;ebm-ziehl.co.uk]
Sent: 25 October 2002 16:46
To: Php-Db (E-mail)
Subject: [PHP-DB] select statement
Hi
I get an error message with the following statement, if I remove the 'AND
works_order_part_no=parts_parts_no' I don't get an error message but also
get the wrong data ??
$link = mysql_connect($host, $user, $pass);
mysql_select_db("DB_EDMsystem", $link);
$result = mysql_query("SELECT * FROM labor, works_orders, parts where
labor_ord_no = works_orders_ord_no AND works_order_part_no =
parts_parts_no",$link);
echo "";
echo"Works OrderPart NumberOrder
QtyMade QtyOptions";
while($myrow = mysql_fetch_array($result))
Regards
Steve Dodkins
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[PHP-DB] select statement
Hi
I get an error message with the following statement, if I remove the 'AND
works_order_part_no=parts_parts_no' I don't get an error message but also
get the wrong data ??
$link = mysql_connect($host, $user, $pass);
mysql_select_db("DB_EDMsystem", $link);
$result = mysql_query("SELECT * FROM labor, works_orders, parts where
labor_ord_no = works_orders_ord_no AND works_order_part_no =
parts_parts_no",$link);
echo "";
echo"Works OrderPart NumberOrder
QtyMade QtyOptions";
while($myrow = mysql_fetch_array($result))
Regards
Steve Dodkins
IMPORTANT NOTICE The information in this e-mail is confidential and should
only be read by those persons to whom it is addressed and is not intended to
be relied upon by any person without subsequent written confirmation of its
contents. ebm-ZIEHL (UK) Ltd. cannot accept any responsibility for the
accuracy or completeness of this message as it has been transmitted over a
public network. Furthermore, the content of this e-mail is the personal
view of the sender and does not represent the advice, views or opinion of
our company. Accordingly, our company disclaim all responsibility and accept
no liability (including in negligence) for the consequences of any person
acting, or refraining from acting, on such information prior to the receipt
by those persons of subsequent written confirmation. In particular (but not
by way of limitation) our company disclaims all responsibility and accepts
no liability for any e-mails which are defamatory, offensive, racist or in
any other way are in breach of any third party's rights, including breach of
confidence, privacy or other rights. If you have received this e-mail
message in error, please notify me immediately by telephone. Please also
destroy and delete the message from your computer. Any form of reproduction,
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publication of this e-mail message is strictly prohibited. If you have
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RE: [PHP-DB] SELECT statement problem
Assign the SELECT statement to a variable and then echo the variable. Does
it contain values you expect? If the database is MySQL it can be very
silent about errors in the SELECT and simply returns no results.
Cheers - Miles
At 02:58 PM 8/15/2002 -0400, Evan S. Weiner wrote:
>It only outputs the first tech. It worked fine until I tried to add the
>end date ino it.
>
>Thanks for all teh help!
>
>// Get technician info
>$TECH_SQL = "SELECT * FROM employees";
>$TECH = MYSQL_QUERY( $TECH_SQL, $CONNECTION) OR DIE ( mysql_error() );
>
>while ( $row = mysql_fetch_array( $TECH ) )
>{
> $tech_id = $row["id"];
> $tech_name_last = $row["name_last"];
> $tech_name_first = $row["name_first"];
>
> $tech_initials_first = substr( $tech_name_first, 0, 1 );
> $tech_initials_last = substr( $tech_name_last, 0, 1);
> $tech_initials = "$tech_initials_first$tech_initials_last";
>
> $tech_name = "$tech_name_first $tech_name_last";
>
> echo " \n";
> echo " class=\"big\">$tech_name\n";
>
> // Reset counters for this tech
> $tech_hour = 0;
> $tech_cost = 0;
>
>
> // Start pulling ticket work
> $TICKET_WORK_SQL = "SELECT * FROM tickets_work
> WHERE employee_id = '$tech_id'
> AND start > '$start'
> AND end < '$end'
> ORDER BY start";
> $TICKET_WORK_INFO = MYSQL_QUERY( $TICKET_WORK_SQL, $CONNECTION )
> OR DIE ( mysql_error() );
>
> while ( $row = mysql_fetch_array( $TICKET_WORK_INFO ) )
> {
> $ticket_id = $row["ticket_id"];
> $start = $row["start"];
> $end = $row["end"];
> $total_hours = $row["total_hours"];
> $total_billable = $row["total_billable"];
>
>-Original Message-
>From: Miles Thompson [mailto:[EMAIL PROTECTED]]
>Sent: Thursday, August 15, 2002 2:26 PM
>To: [EMAIL PROTECTED]
>Subject: Re: [PHP-DB] SELECT statement problem
>
>
>SELECT * FROM tickets_work
>WHERE employee_id = '$tech_id' AND
> start > '$start' AND
> end < '$end'
>ORDER BY start
>
>Which will select everything from tickets_work for a given employee between
>start and end. What is not working?
>
>Miles Thompson
>
>At 02:15 PM 8/15/2002 -0400, Evan S. Weiner wrote:
> >Hello all,
> >
> >I have this SQL statement I am trying to use and no matter what
> >combination I use, I cannot get it to work. The start and end if clauses
> >need to be grouped together.
> >
> >SELECT * FROM tickets_work WHERE employee_id = '$tech_id' && ( start >
> >'$start' && end < '$end' ) ORDER BY start
> >
> >Anyone able to help me?
> >
> >Thanks!
> >
> >Evan
> >
> >
> >--
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>
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RE: [PHP-DB] SELECT statement problem
It only outputs the first tech. It worked fine until I tried to add the end date ino
it.
Thanks for all teh help!
// Get technician info
$TECH_SQL = "SELECT * FROM employees";
$TECH = MYSQL_QUERY( $TECH_SQL, $CONNECTION) OR DIE ( mysql_error() );
while ( $row = mysql_fetch_array( $TECH ) )
{
$tech_id = $row["id"];
$tech_name_last = $row["name_last"];
$tech_name_first = $row["name_first"];
$tech_initials_first = substr( $tech_name_first, 0, 1 );
$tech_initials_last = substr( $tech_name_last, 0, 1);
$tech_initials = "$tech_initials_first$tech_initials_last";
$tech_name = "$tech_name_first $tech_name_last";
echo " \n";
echo " $tech_name\n";
// Reset counters for this tech
$tech_hour = 0;
$tech_cost = 0;
// Start pulling ticket work
$TICKET_WORK_SQL = "SELECT * FROM tickets_work
WHERE employee_id = '$tech_id'
AND start > '$start'
AND end < '$end'
ORDER BY start";
$TICKET_WORK_INFO = MYSQL_QUERY( $TICKET_WORK_SQL, $CONNECTION ) OR DIE (
mysql_error() );
while ( $row = mysql_fetch_array( $TICKET_WORK_INFO ) )
{
$ticket_id = $row["ticket_id"];
$start = $row["start"];
$end = $row["end"];
$total_hours = $row["total_hours"];
$total_billable = $row["total_billable"];
-Original Message-
From: Miles Thompson [mailto:[EMAIL PROTECTED]]
Sent: Thursday, August 15, 2002 2:26 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] SELECT statement problem
SELECT * FROM tickets_work
WHERE employee_id = '$tech_id' AND
start > '$start' AND
end < '$end'
ORDER BY start
Which will select everything from tickets_work for a given employee between
start and end. What is not working?
Miles Thompson
At 02:15 PM 8/15/2002 -0400, Evan S. Weiner wrote:
>Hello all,
>
>I have this SQL statement I am trying to use and no matter what
>combination I use, I cannot get it to work. The start and end if clauses
>need to be grouped together.
>
>SELECT * FROM tickets_work WHERE employee_id = '$tech_id' && ( start >
>'$start' && end < '$end' ) ORDER BY start
>
>Anyone able to help me?
>
>Thanks!
>
>Evan
>
>
>--
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RE: [PHP-DB] SELECT statement problem
Can you echo out the sql going in and any errors you're getting when the query doesn't work then post that info to this list? It'll help both you and us diagnose what might be going wrong. In addition, what datatype are the columns named start and end in your database? (Assuming database is MySQL) Rich -Original Message- From: Evan S. Weiner [mailto:[EMAIL PROTECTED]] Sent: Thursday, August 15, 2002 2:15 PM To: [EMAIL PROTECTED] Subject: [PHP-DB] SELECT statement problem Hello all, I have this SQL statement I am trying to use and no matter what combination I use, I cannot get it to work. The start and end if clauses need to be grouped together. SELECT * FROM tickets_work WHERE employee_id = '$tech_id' && ( start > '$start' && end < '$end' ) ORDER BY start Anyone able to help me? Thanks! Evan -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] SELECT statement problem
SELECT * FROM tickets_work WHERE employee_id = '$tech_id' AND start > '$start' AND end < '$end' ORDER BY start Which will select everything from tickets_work for a given employee between start and end. What is not working? Miles Thompson At 02:15 PM 8/15/2002 -0400, Evan S. Weiner wrote: >Hello all, > >I have this SQL statement I am trying to use and no matter what >combination I use, I cannot get it to work. The start and end if clauses >need to be grouped together. > >SELECT * FROM tickets_work WHERE employee_id = '$tech_id' && ( start > >'$start' && end < '$end' ) ORDER BY start > >Anyone able to help me? > >Thanks! > >Evan > > >-- >PHP Database Mailing List (http://www.php.net/) >To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] SELECT statement problem
Hello all, I have this SQL statement I am trying to use and no matter what combination I use, I cannot get it to work. The start and end if clauses need to be grouped together. SELECT * FROM tickets_work WHERE employee_id = '$tech_id' && ( start > '$start' && end < '$end' ) ORDER BY start Anyone able to help me? Thanks! Evan -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] SELECT statement does not return rows
oops.forgot a semicolon on the first line of the multi-line code
segment
> -Original Message-
> $row = mysql_fetch_array($result)
> echo 'columnvalue';
> while(list($key, $val) = each($row)) {
> // $row has two key/value pairs per column -- one integer,
> one string
> if (is_string($key)) {
> echo "$key$val";
> }
> }
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RE: [PHP-DB] SELECT statement does not return rows
I am assuming that this code is NOT the file "do_mod_job.php".
What happens if you do this next line right after you execute your query?:
echo 'num rows fetched: '.mysql_num_rows($result).'';
also, the while loop shouldn't really be necessary as the query should only
return one record if I understand the naming of your fields correctly. Your
echo statements don't have to have the variable encapsulated with
double-quotes either--it is actually easier on PHP if they are not (if you
use single-quotes rather than double-quotes as well when appropriate).
If there is a row being returned from the query as discovered above, you
might try the following to make sure that what you are getting is what you
are expecting:
$row = mysql_fetch_array($result)
echo 'columnvalue';
while(list($key, $val) = each($row)) {
// $row has two key/value pairs per column -- one integer, one string
if (is_string($key)) {
echo "$key$val";
}
}
echo '';
exit;
basically, you need to verify that $row['id'] is not empty.
> -Original Message-
> Todd Williamsen <[EMAIL PROTECTED]> said:
>
> > Weird..
> >
> > I want to be able to edit records, which I have done in the
> past, and I
> > cannot see why it isn't working... I have tried single
> qoutes around the
> > $row = and that doesn't work either. here is the code:
> >
> > $db = @mysql_select_db($db_name, $connection) or die("Could
> not select
> > database");
> > $sql="SELECT * FROM Jobs WHERE id = '$id' ";
> > $result = mysql_query($sql,$connection) or die("Couldn't
> execute query");
> > while ($row = mysql_fetch_array($result))
> > {
> > $id = $row["id"];
> > $Industry = $row["Industry"];
> > $Other = $row["Other"];
> > $JobTitle = $row["JobTitle"];
> > $Description = $row["Description"];
> > $Location = $row["Location"];
> > $Date = $row["Date"];
> > }
> > ?>
> >
> >
> > Edit Job Posting
> >
> >
> > Edit Job Posting
> >
> > ">
> >
> >
> > Job Information and Location
> > Job Description
> >
> >
> >
> > Job Title:
> > " size=50
> > maxlength=75>
> > Industry:
> > " size=50
> > maxlength=75>
> > Other:
> > " size=50
> > maxlength=75>
> > Location:
> > " size=50
> > maxlength=75>
> > Date:
> > " size=30
> > maxlength=40>
> >
> >
> > Description:
> >> "$Description"; ?>
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
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Re: [PHP-DB] SELECT statement does not return rows
Is your id an integer or a char/varchar? If it is an integer, take the
quotes off $id in your select statement.
Todd Williamsen <[EMAIL PROTECTED]> said:
> Weird..
>
> I want to be able to edit records, which I have done in the past, and I
> cannot see why it isn't working... I have tried single qoutes around the
> $row = and that doesn't work either. here is the code:
>
> $db = @mysql_select_db($db_name, $connection) or die("Could not select
> database");
> $sql="SELECT * FROM Jobs WHERE id = '$id' ";
> $result = mysql_query($sql,$connection) or die("Couldn't execute query");
> while ($row = mysql_fetch_array($result))
> {
> $id = $row["id"];
> $Industry = $row["Industry"];
> $Other = $row["Other"];
> $JobTitle = $row["JobTitle"];
> $Description = $row["Description"];
> $Location = $row["Location"];
> $Date = $row["Date"];
> }
> ?>
>
>
> Edit Job Posting
>
>
> Edit Job Posting
>
> ">
>
>
> Job Information and Location
> Job Description
>
>
>
> Job Title:
> " size=50
> maxlength=75>
> Industry:
> " size=50
> maxlength=75>
> Other:
> " size=50
> maxlength=75>
> Location:
> " size=50
> maxlength=75>
> Date:
> " size=30
> maxlength=40>
>
>
> Description:
>"$Description"; ?>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
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> To unsubscribe, e-mail: [EMAIL PROTECTED]
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> To contact the list administrators, e-mail: [EMAIL PROTECTED]
>
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RE: [PHP-DB] SELECT statement does not return rows
Did you print out the value of $sql before executing it? Was it as you
expected?
If so, did you print out mysql_num_rows() to verify it's greater than 0?
-Original Message-
From: Todd Williamsen [mailto:[EMAIL PROTECTED]]
Sent: Friday, February 01, 2002 2:52 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] SELECT statement does not return rows
Weird..
I want to be able to edit records, which I have done in the past, and I
cannot see why it isn't working... I have tried single qoutes around the
$row = and that doesn't work either. here is the code:
$db = @mysql_select_db($db_name, $connection) or die("Could not select
database");
$sql="SELECT * FROM Jobs WHERE id = '$id' ";
$result = mysql_query($sql,$connection) or die("Couldn't execute query");
while ($row = mysql_fetch_array($result))
{
$id = $row["id"];
$Industry = $row["Industry"];
$Other = $row["Other"];
$JobTitle = $row["JobTitle"];
$Description = $row["Description"];
$Location = $row["Location"];
$Date = $row["Date"];
}
?>
Edit Job Posting
Edit Job Posting
">
Job Information and Location
Job Description
Job Title:
" size=50
maxlength=75>
Industry:
" size=50
maxlength=75>
Other:
" size=50
maxlength=75>
Location:
" size=50
maxlength=75>
Date:
" size=30
maxlength=40>
Description:
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[PHP-DB] SELECT statement does not return rows
Weird..
I want to be able to edit records, which I have done in the past, and I
cannot see why it isn't working... I have tried single qoutes around the
$row = and that doesn't work either. here is the code:
$db = @mysql_select_db($db_name, $connection) or die("Could not select
database");
$sql="SELECT * FROM Jobs WHERE id = '$id' ";
$result = mysql_query($sql,$connection) or die("Couldn't execute query");
while ($row = mysql_fetch_array($result))
{
$id = $row["id"];
$Industry = $row["Industry"];
$Other = $row["Other"];
$JobTitle = $row["JobTitle"];
$Description = $row["Description"];
$Location = $row["Location"];
$Date = $row["Date"];
}
?>
Edit Job Posting
Edit Job Posting
">
Job Information and Location
Job Description
Job Title:
" size=50
maxlength=75>
Industry:
" size=50
maxlength=75>
Other:
" size=50
maxlength=75>
Location:
" size=50
maxlength=75>
Date:
" size=30
maxlength=40>
Description:
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RE: [PHP-DB] Select statement only returns 1 record
It's worked for me.
You'll note that I added a new column "id". Did you put that in your
table??
What error did you get
-Original Message-
From: Todd Williamsen [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, January 30, 2002 2:18 PM
To: 'Rick Emery'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Select statement only returns 1 record
Nope, that doesn't work
-Original Message-
From: Rick Emery [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, January 30, 2002 1:01 PM
To: 'Todd Williamsen'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Select statement only returns 1 record
";
while( $row = mysql_fetch_array($result) )
{
$FirstName = $row['FirstName'];
$LastName = $row['LastName'];
$id=$row['id'];
print "$LastName, $FirstName";
}
?>
print "";
-Original Message-
From: Todd Williamsen [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, January 30, 2002 12:45 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Select statement only returns 1 record
I am trying to get data from two columns, FirstName and Last name and
displaying all the records LastName, FirstName in a drop down menu.
The weird thing is that it only displays one record. I thought the
table
was hosed, but its not. I tried it through another database and still
doesn't work.
Here is the code:
I would list the whole page but its long! There is another SQL
statement at
the top of the page:
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RE: [PHP-DB] Select statement only returns 1 record
Nope, that doesn't work
-Original Message-
From: Rick Emery [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, January 30, 2002 1:01 PM
To: 'Todd Williamsen'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Select statement only returns 1 record
";
while( $row = mysql_fetch_array($result) )
{
$FirstName = $row['FirstName'];
$LastName = $row['LastName'];
$id=$row['id'];
print "$LastName, $FirstName";
}
?>
print "";
-Original Message-
From: Todd Williamsen [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, January 30, 2002 12:45 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Select statement only returns 1 record
I am trying to get data from two columns, FirstName and Last name and
displaying all the records LastName, FirstName in a drop down menu.
The weird thing is that it only displays one record. I thought the
table
was hosed, but its not. I tried it through another database and still
doesn't work.
Here is the code:
I would list the whole page but its long! There is another SQL
statement at
the top of the page:
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Re: [PHP-DB] Select statement only returns 1 record
You will need to put the option tag in a loop to get all of the records in the table. (this can go anywhere but in while loop) "; $i++; } ?> Also, without the value in the option tag, you will not be passing anything to the form. HTH Maureen Biorn Todd Williamsen <[EMAIL PROTECTED]> said: > I am trying to get data from two columns, FirstName and Last name and > displaying all the records LastName, FirstName in a drop down menu. > > The weird thing is that it only displays one record. I thought the table > was hosed, but its not. I tried it through another database and still > doesn't work. > > Here is the code: > > $db = mysql_connect($dbserver, $dbuser, $dbpass); > mysql_select_db($dbname,$db); > $sortby = "name ASC"; > $sql="SELECT * FROM webl_players ORDER BY $sortby"; > $result=mysql_query($sql,$db); > $row = mysql_fetch_array($result); > $FirstName = $row["FirstName"]; > $LastName = $row["LastName"]; > ?> > > > > > > > I would list the whole page but its long! There is another SQL statement at > the top of the page: > > > $db = mysql_connect($dbserver, $dbuser, $$dbpass); > mysql_select_db($dbname, $db); > $sql = "SELECT * FROM Canidate SORT BY 'LastName'"; > $result = mysql_query($sql); > ?> > > > > -- > PHP Database Mailing List (http://www.php.net/) > To unsubscribe, e-mail: [EMAIL PROTECTED] > For additional commands, e-mail: [EMAIL PROTECTED] > To contact the list administrators, e-mail: [EMAIL PROTECTED] > -- -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
RE: [PHP-DB] Select statement only returns 1 record
";
while( $row = mysql_fetch_array($result) )
{
$FirstName = $row['FirstName'];
$LastName = $row['LastName'];
$id=$row['id'];
print "$LastName, $FirstName";
}
?>
print "";
-Original Message-
From: Todd Williamsen [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, January 30, 2002 12:45 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Select statement only returns 1 record
I am trying to get data from two columns, FirstName and Last name and
displaying all the records LastName, FirstName in a drop down menu.
The weird thing is that it only displays one record. I thought the table
was hosed, but its not. I tried it through another database and still
doesn't work.
Here is the code:
I would list the whole page but its long! There is another SQL statement at
the top of the page:
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Re: [PHP-DB] Select statement only returns 1 record
Check the manual for the mysql_fetch_array() function, it shows you how to extract data from $result. You will use the while loop (as shown in the manual's example) to fill your combo box / selct list /drop down menu, whatever we're calling that creature today. Miles At 12:45 PM 1/30/2002 -0600, Todd Williamsen wrote: >I am trying to get data from two columns, FirstName and Last name and >displaying all the records LastName, FirstName in a drop down menu. > >The weird thing is that it only displays one record. I thought the table >was hosed, but its not. I tried it through another database and still >doesn't work. > >Here is the code: > >$db = mysql_connect($dbserver, $dbuser, $dbpass); >mysql_select_db($dbname,$db); >$sortby = "name ASC"; >$sql="SELECT * FROM webl_players ORDER BY $sortby"; >$result=mysql_query($sql,$db); >$row = mysql_fetch_array($result); >$FirstName = $row["FirstName"]; >$LastName = $row["LastName"]; >?> > > > > > > >I would list the whole page but its long! There is another SQL statement at >the top of the page: > > >$db = mysql_connect($dbserver, $dbuser, $$dbpass); >mysql_select_db($dbname, $db); >$sql = "SELECT * FROM Canidate SORT BY 'LastName'"; >$result = mysql_query($sql); >?> > > > >-- >PHP Database Mailing List (http://www.php.net/) >To unsubscribe, e-mail: [EMAIL PROTECTED] >For additional commands, e-mail: [EMAIL PROTECTED] >To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP-DB] Select statement only returns 1 record
I am trying to get data from two columns, FirstName and Last name and displaying all the records LastName, FirstName in a drop down menu. The weird thing is that it only displays one record. I thought the table was hosed, but its not. I tried it through another database and still doesn't work. Here is the code: I would list the whole page but its long! There is another SQL statement at the top of the page: -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] SELECT statement
* Julio Cuz, Jr. ([EMAIL PROTECTED]) wrote: > Ron, > > Thanks for your help, but my problem still there even when I made the > following changes: > > $sql = "SELECT * FROM \"Remodel\" WHERE Email=\"".$find."\""; You don't have to quote the table name either. You can probably get away with something like this: $sql = "SELECT * FROM Remodel WHERE Email=\"$find\" \n"; I put the \n at the end of my queries, at the end of every line if it's a long query, just makes it easier for me to read if I need to do debugging later, it won't affect your query. Jeff -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] SELECT statement
Ron, Thanks for your help, but my problem still there even when I made the following changes: $sql = "SELECT * FROM \"Remodel\" WHERE Email=\"".$find."\""; By the way, you're correct when saying that "WO" and "Email" are column names and '"$find" is what the user entered when searching for a record. At 03:39 PM 4/4/2001 -0700, you wrote: >At 03:24 PM 4/4/2001 -0700, Julio Cuz, Jr. wrote: >>Every time I run the following code, I get this error if I use a NUMBER >>for the 'Email' case: >>"Warning: Unable to jump to row 0 on PostgreSQL result index 2 in >>/html/rccd/remodel/display2.php on line 35" > >The problem is that your query is equivalent to going: > >SELECT * FROM "Remodel" WHERE "WO" LIKE 1; // or some other numerical value > >The "WO" is a string because it is in quotes so you are saying: > >WHERE string = integer > >Ain't gonna happen so the result is always empty. What you want probably >is (assuming WO is a column name): > >$sql = "SELECT * FROM \"Remodel\" WHERE WO=\"".$find."\""; > >It is equals since you are using an integer instead of a string (like is >for strings). > >>or, if I use a string (i.e. [EMAIL PROTECTED]), I get this error message: >>"Warning: PostgreSQL query failed: ERROR: Attribute 'jcuz' not found in >>/html/rccd/remodel/display2.php on line 33" > >Same problem. "EMAIL" means the string "EMAIL" not the column called "EMAIL". > >Cheers, > >Ron > >- >Island Net AMT Solutions Group Inc. Telephone: 250 383-0096 >1412 Quadra Toll Free:1 800 331-3055 >Victoria, B.C. Fax:250 383-6698 >V8W 2L1 E-Mail:[EMAIL PROTECTED] >Canada WWW: http://www.islandnet.com/ >- > Julio Cuz, Jr. Riverside Community College [EMAIL PROTECTED]
Re: [PHP-DB] SELECT statement
At 03:24 PM 4/4/2001 -0700, Julio Cuz, Jr. wrote: >Every time I run the following code, I get this error if I use a NUMBER >for the 'Email' case: >"Warning: Unable to jump to row 0 on PostgreSQL result index 2 in >/html/rccd/remodel/display2.php on line 35" The problem is that your query is equivalent to going: SELECT * FROM "Remodel" WHERE "WO" LIKE 1; // or some other numerical value The "WO" is a string because it is in quotes so you are saying: WHERE string = integer Ain't gonna happen so the result is always empty. What you want probably is (assuming WO is a column name): $sql = "SELECT * FROM \"Remodel\" WHERE WO=\"".$find."\""; It is equals since you are using an integer instead of a string (like is for strings). >or, if I use a string (i.e. [EMAIL PROTECTED]), I get this error message: >"Warning: PostgreSQL query failed: ERROR: Attribute 'jcuz' not found in >/html/rccd/remodel/display2.php on line 33" Same problem. "EMAIL" means the string "EMAIL" not the column called "EMAIL". Cheers, Ron - Island Net AMT Solutions Group Inc. Telephone: 250 383-0096 1412 Quadra Toll Free:1 800 331-3055 Victoria, B.C. Fax:250 383-6698 V8W 2L1 E-Mail:[EMAIL PROTECTED] Canada WWW: http://www.islandnet.com/ - -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP-DB] SELECT statement
Hi--
Every time I run the following code, I get this error if I use a NUMBER for
the 'Email' case:
"Warning: Unable to jump to row 0 on PostgreSQL result index 2 in
/html/rccd/remodel/display2.php on line 35"
or, if I use a string (i.e. [EMAIL PROTECTED]), I get this error message:
"Warning: PostgreSQL query failed: ERROR: Attribute 'jcuz' not found in
/html/rccd/remodel/display2.php on line 33"
Please help!
switch ($Search) {
case "WO":
settype ($find,integer);
$sql = "SELECT * FROM \"Remodel\" WHERE \"WO\" LIKE $find";
break;
case "Email":
$sql = "SELECT * FROM \"Remodel\" WHERE \"Email\" LIKE $find";
break;
}
--
Julio Cuz, Jr.
Riverside Community College
[EMAIL PROTECTED]
