RE: [PHP-DB] Dynamic SQL + result resource error

2001-07-09 Thread Mats Remman

Look at this snippet :
 $word = current($wordsarray); 
 next($wordsarray);
 $sql=$sql.$word);

if you have 2 words the sql would extend to
word1)word2) .. this is invalid sql syntax.

Change the 
while ($i  count($wordsarray)) 
{ 
$word = current($wordsarray); 
next($wordsarray);
$sql=$sql.$word);
$i++; 
}
bit into something like this
$sep = ;
foreach( $wordsarray as $v ) {
$sql .= $sep. $word ;
$sep =  OR band LIKE ;
}
$sql .= );

Then retry the queries.

oh and yes, include the mysql_query() as Mark Gordon mentioned.


Mats Remman
PHP Developer/Mysql DBA
Coretrek, Norway
+47 51978597 / +47 916 23566

 

 -Original Message-
 From: Matthew Loff [mailto:[EMAIL PROTECTED]]
 Sent: Monday, July 09, 2001 2:20 AM
 To: 'Ben Bleything'; 'Mark Gordon'; [EMAIL PROTECTED]
 Subject: RE: [PHP-DB] Dynamic SQL + result resource error
 
 
 
 Ha ha... I should have meantioned Here's what it -should- be or
 something of the like. :)
 
 I suspect that was the only problem with his code, but perhaps there's
 something else there.
 
 
 -Original Message-
 From: Ben Bleything [mailto:[EMAIL PROTECTED]] 
 Sent: Sunday, July 08, 2001 8:13 PM
 To: 'Matthew Loff'; 'Mark Gordon'; [EMAIL PROTECTED]
 Subject: RE: [PHP-DB] Dynamic SQL + result resource error
 
 
 Guess I'm just a big dumbass then, aren't I =P
 
 Oops.
 
 I suppose that would cause it to fail then, wouldn't it?
 
 =  Ben
 
 -Original Message-
 From: Matthew Loff [mailto:[EMAIL PROTECTED]] 
 Sent: Sunday, July 08, 2001 5:10 PM
 To: 'Ben Bleything'; 'Mark Gordon'; [EMAIL PROTECTED]
 Subject: RE: [PHP-DB] Dynamic SQL + result resource error
 
 
 The code you're referencing is my modification of his original post. :)
 
 
 -Original Message-
 From: Ben Bleything [mailto:[EMAIL PROTECTED]] 
 Sent: Sunday, July 08, 2001 8:04 PM
 To: 'Matthew Loff'; 'Mark Gordon'; [EMAIL PROTECTED]
 Subject: RE: [PHP-DB] Dynamic SQL + result resource error
 
 
 Sure he is.  Right here:
 
 $queryResult = mysql_query($sql);
 
 what exact error is occurring?
 
 -Original Message-
 From: Matthew Loff [mailto:[EMAIL PROTECTED]] 
 Sent: Sunday, July 08, 2001 5:00 PM
 To: 'Mark Gordon'; [EMAIL PROTECTED]
 Subject: RE: [PHP-DB] Dynamic SQL + result resource error
 
 
 You aren't calling mysql_query() to execute the query.
 
 //$find is text box input
 $wordsarray = explode( ,$find); 
 $sql = SELECT bandname FROM bands WHERE (bandname
 LIKE ;
 $i = 0;
 while ($i  count($wordsarray)) 
 { 
 $word = current($wordsarray); 
 next($wordsarray);
 $sql=$sql.$word);
 $i++; 
 }
 print $sqlhr;
 
 $queryResult = mysql_query($sql);
 
 while ($myrow=mysql_fetch_row($queryResult))
 {
 print $myrow[0],p;
 }
 
 
 -Original Message-
 From: Mark Gordon [mailto:[EMAIL PROTECTED]] 
 Sent: Sunday, July 08, 2001 7:54 PM
 To: [EMAIL PROTECTED]
 Subject: [PHP-DB] Dynamic SQL + result resource error
 
 
 Why is this code generating an error when it outputs a
 valid SQL statement?  (there are no parse errors)
 
 //$find is text box input
 $wordsarray = explode( ,$find); 
 $sql = SELECT bandname FROM bands WHERE (bandname
 LIKE ;
 $i = 0;
 while ($i  count($wordsarray)) 
 { 
 $word = current($wordsarray); 
 next($wordsarray);
 $sql=$sql.$word);
 $i++; 
 }
 print $sqlhr;
 while ($myrow=mysql_fetch_row($sql))
 {
 print $myrow[0],p;
 }
 
 =
 Mark
 [EMAIL PROTECTED]
 
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RE: [PHP-DB] Dynamic SQL + result resource error

2001-07-09 Thread Michael Rudel

Hi Mark,

... beside of what was mentioned be4, you have also
a logical error in the way you append to your $sql-
query in the while loop.

Look: letz say, $find is The Fields of the Nephilim,
so your $wordsarray would be:

0 == The
1 == Fields
2 == of
3 == the
4 == Nephilim

in your while-loop, you append every array-entry to
your $sql with $word).

So your whole SQL-Statement would read:

SELECT bandname FROM bands WHERE (bandname LIKE The)Fields)of)the)Nephilim)

This statement will result in error !

Try this 1:

/
//$find is text box input
$wordsarray = explode( ,$find);

$sql = SELECT bandname FROM bands WHERE bandname  '';

while ( list( $key, $word ) = each( $wordsarray ) )
{ 
   $sql .=  OR bandname like '%.$word.%';
}
echo $sql.hr;

$queryResult = mysql_query($sql);

while ( $myrow = mysql_fetch_assoc( $queryResult ) )
{
   echo p .$myrow[bandname]. /p.\n;
}
/

Hope this helps ;)

Greetinx,
  Mike

Michael Rudel 
- Web-Development, Systemadministration -

Besuchen Sie uns am 20. und 21. August 2001 auf der
online-marketing-d├╝sseldorf in Halle 1 Stand E 16
___

Suchtreffer AG 
Bleicherstra├če 20 
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Germany 
fon: +49-(0)7531-89207-17 
fax: +49-(0)7531-89207-13 
e-mail: mailto:[EMAIL PROTECTED] 
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 -Original Message-
 From: Mark Gordon [mailto:[EMAIL PROTECTED]]
 Sent: Monday, July 09, 2001 1:54 AM
 To: [EMAIL PROTECTED]
 Subject: [PHP-DB] Dynamic SQL + result resource error
 
 
 Why is this code generating an error when it outputs a
 valid SQL statement?  (there are no parse errors)
 
 //$find is text box input
 $wordsarray = explode( ,$find); 
 $sql = SELECT bandname FROM bands WHERE (bandname
 LIKE ;
 $i = 0;
 while ($i  count($wordsarray)) 
 { 
 $word = current($wordsarray); 
 next($wordsarray);
 $sql=$sql.$word);
 $i++; 
 }
 print $sqlhr;
 while ($myrow=mysql_fetch_row($sql))
 {
 print $myrow[0],p;
 }
 
 =
 Mark
 [EMAIL PROTECTED]
 
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RE: [PHP-DB] Dynamic SQL + result resource error

2001-07-08 Thread Matthew Loff


You aren't calling mysql_query() to execute the query.

//$find is text box input
$wordsarray = explode( ,$find); 
$sql = SELECT bandname FROM bands WHERE (bandname
LIKE ;
$i = 0;
while ($i  count($wordsarray)) 
{ 
$word = current($wordsarray); 
next($wordsarray);
$sql=$sql.$word);
$i++; 
}
print $sqlhr;

$queryResult = mysql_query($sql);

while ($myrow=mysql_fetch_row($queryResult))
{
print $myrow[0],p;
}


-Original Message-
From: Mark Gordon [mailto:[EMAIL PROTECTED]] 
Sent: Sunday, July 08, 2001 7:54 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Dynamic SQL + result resource error


Why is this code generating an error when it outputs a
valid SQL statement?  (there are no parse errors)

//$find is text box input
$wordsarray = explode( ,$find); 
$sql = SELECT bandname FROM bands WHERE (bandname
LIKE ;
$i = 0;
while ($i  count($wordsarray)) 
{ 
$word = current($wordsarray); 
next($wordsarray);
$sql=$sql.$word);
$i++; 
}
print $sqlhr;
while ($myrow=mysql_fetch_row($sql))
{
print $myrow[0],p;
}

=
Mark
[EMAIL PROTECTED]

__
Do You Yahoo!?
Get personalized email addresses from Yahoo! Mail
http://personal.mail.yahoo.com/

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RE: [PHP-DB] Dynamic SQL + result resource error

2001-07-08 Thread Ben Bleything

Sure he is.  Right here:

$queryResult = mysql_query($sql);

what exact error is occurring?

-Original Message-
From: Matthew Loff [mailto:[EMAIL PROTECTED]] 
Sent: Sunday, July 08, 2001 5:00 PM
To: 'Mark Gordon'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Dynamic SQL + result resource error


You aren't calling mysql_query() to execute the query.

//$find is text box input
$wordsarray = explode( ,$find); 
$sql = SELECT bandname FROM bands WHERE (bandname
LIKE ;
$i = 0;
while ($i  count($wordsarray)) 
{ 
$word = current($wordsarray); 
next($wordsarray);
$sql=$sql.$word);
$i++; 
}
print $sqlhr;

$queryResult = mysql_query($sql);

while ($myrow=mysql_fetch_row($queryResult))
{
print $myrow[0],p;
}


-Original Message-
From: Mark Gordon [mailto:[EMAIL PROTECTED]] 
Sent: Sunday, July 08, 2001 7:54 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Dynamic SQL + result resource error


Why is this code generating an error when it outputs a
valid SQL statement?  (there are no parse errors)

//$find is text box input
$wordsarray = explode( ,$find); 
$sql = SELECT bandname FROM bands WHERE (bandname
LIKE ;
$i = 0;
while ($i  count($wordsarray)) 
{ 
$word = current($wordsarray); 
next($wordsarray);
$sql=$sql.$word);
$i++; 
}
print $sqlhr;
while ($myrow=mysql_fetch_row($sql))
{
print $myrow[0],p;
}

=
Mark
[EMAIL PROTECTED]

__
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RE: [PHP-DB] Dynamic SQL + result resource error

2001-07-08 Thread Matthew Loff


The code you're referencing is my modification of his original post. :)


-Original Message-
From: Ben Bleything [mailto:[EMAIL PROTECTED]] 
Sent: Sunday, July 08, 2001 8:04 PM
To: 'Matthew Loff'; 'Mark Gordon'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Dynamic SQL + result resource error


Sure he is.  Right here:

$queryResult = mysql_query($sql);

what exact error is occurring?

-Original Message-
From: Matthew Loff [mailto:[EMAIL PROTECTED]] 
Sent: Sunday, July 08, 2001 5:00 PM
To: 'Mark Gordon'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Dynamic SQL + result resource error


You aren't calling mysql_query() to execute the query.

//$find is text box input
$wordsarray = explode( ,$find); 
$sql = SELECT bandname FROM bands WHERE (bandname
LIKE ;
$i = 0;
while ($i  count($wordsarray)) 
{ 
$word = current($wordsarray); 
next($wordsarray);
$sql=$sql.$word);
$i++; 
}
print $sqlhr;

$queryResult = mysql_query($sql);

while ($myrow=mysql_fetch_row($queryResult))
{
print $myrow[0],p;
}


-Original Message-
From: Mark Gordon [mailto:[EMAIL PROTECTED]] 
Sent: Sunday, July 08, 2001 7:54 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Dynamic SQL + result resource error


Why is this code generating an error when it outputs a
valid SQL statement?  (there are no parse errors)

//$find is text box input
$wordsarray = explode( ,$find); 
$sql = SELECT bandname FROM bands WHERE (bandname
LIKE ;
$i = 0;
while ($i  count($wordsarray)) 
{ 
$word = current($wordsarray); 
next($wordsarray);
$sql=$sql.$word);
$i++; 
}
print $sqlhr;
while ($myrow=mysql_fetch_row($sql))
{
print $myrow[0],p;
}

=
Mark
[EMAIL PROTECTED]

__
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RE: [PHP-DB] Dynamic SQL + result resource error

2001-07-08 Thread Ben Bleything

Guess I'm just a big dumbass then, aren't I =P

Oops.

I suppose that would cause it to fail then, wouldn't it?

=  Ben

-Original Message-
From: Matthew Loff [mailto:[EMAIL PROTECTED]] 
Sent: Sunday, July 08, 2001 5:10 PM
To: 'Ben Bleything'; 'Mark Gordon'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Dynamic SQL + result resource error


The code you're referencing is my modification of his original post. :)


-Original Message-
From: Ben Bleything [mailto:[EMAIL PROTECTED]] 
Sent: Sunday, July 08, 2001 8:04 PM
To: 'Matthew Loff'; 'Mark Gordon'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Dynamic SQL + result resource error


Sure he is.  Right here:

$queryResult = mysql_query($sql);

what exact error is occurring?

-Original Message-
From: Matthew Loff [mailto:[EMAIL PROTECTED]] 
Sent: Sunday, July 08, 2001 5:00 PM
To: 'Mark Gordon'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Dynamic SQL + result resource error


You aren't calling mysql_query() to execute the query.

//$find is text box input
$wordsarray = explode( ,$find); 
$sql = SELECT bandname FROM bands WHERE (bandname
LIKE ;
$i = 0;
while ($i  count($wordsarray)) 
{ 
$word = current($wordsarray); 
next($wordsarray);
$sql=$sql.$word);
$i++; 
}
print $sqlhr;

$queryResult = mysql_query($sql);

while ($myrow=mysql_fetch_row($queryResult))
{
print $myrow[0],p;
}


-Original Message-
From: Mark Gordon [mailto:[EMAIL PROTECTED]] 
Sent: Sunday, July 08, 2001 7:54 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Dynamic SQL + result resource error


Why is this code generating an error when it outputs a
valid SQL statement?  (there are no parse errors)

//$find is text box input
$wordsarray = explode( ,$find); 
$sql = SELECT bandname FROM bands WHERE (bandname
LIKE ;
$i = 0;
while ($i  count($wordsarray)) 
{ 
$word = current($wordsarray); 
next($wordsarray);
$sql=$sql.$word);
$i++; 
}
print $sqlhr;
while ($myrow=mysql_fetch_row($sql))
{
print $myrow[0],p;
}

=
Mark
[EMAIL PROTECTED]

__
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RE: [PHP-DB] Dynamic SQL + result resource error

2001-07-08 Thread Matthew Loff


Ha ha... I should have meantioned Here's what it -should- be or
something of the like. :)

I suspect that was the only problem with his code, but perhaps there's
something else there.


-Original Message-
From: Ben Bleything [mailto:[EMAIL PROTECTED]] 
Sent: Sunday, July 08, 2001 8:13 PM
To: 'Matthew Loff'; 'Mark Gordon'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Dynamic SQL + result resource error


Guess I'm just a big dumbass then, aren't I =P

Oops.

I suppose that would cause it to fail then, wouldn't it?

=  Ben

-Original Message-
From: Matthew Loff [mailto:[EMAIL PROTECTED]] 
Sent: Sunday, July 08, 2001 5:10 PM
To: 'Ben Bleything'; 'Mark Gordon'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Dynamic SQL + result resource error


The code you're referencing is my modification of his original post. :)


-Original Message-
From: Ben Bleything [mailto:[EMAIL PROTECTED]] 
Sent: Sunday, July 08, 2001 8:04 PM
To: 'Matthew Loff'; 'Mark Gordon'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Dynamic SQL + result resource error


Sure he is.  Right here:

$queryResult = mysql_query($sql);

what exact error is occurring?

-Original Message-
From: Matthew Loff [mailto:[EMAIL PROTECTED]] 
Sent: Sunday, July 08, 2001 5:00 PM
To: 'Mark Gordon'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Dynamic SQL + result resource error


You aren't calling mysql_query() to execute the query.

//$find is text box input
$wordsarray = explode( ,$find); 
$sql = SELECT bandname FROM bands WHERE (bandname
LIKE ;
$i = 0;
while ($i  count($wordsarray)) 
{ 
$word = current($wordsarray); 
next($wordsarray);
$sql=$sql.$word);
$i++; 
}
print $sqlhr;

$queryResult = mysql_query($sql);

while ($myrow=mysql_fetch_row($queryResult))
{
print $myrow[0],p;
}


-Original Message-
From: Mark Gordon [mailto:[EMAIL PROTECTED]] 
Sent: Sunday, July 08, 2001 7:54 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Dynamic SQL + result resource error


Why is this code generating an error when it outputs a
valid SQL statement?  (there are no parse errors)

//$find is text box input
$wordsarray = explode( ,$find); 
$sql = SELECT bandname FROM bands WHERE (bandname
LIKE ;
$i = 0;
while ($i  count($wordsarray)) 
{ 
$word = current($wordsarray); 
next($wordsarray);
$sql=$sql.$word);
$i++; 
}
print $sqlhr;
while ($myrow=mysql_fetch_row($sql))
{
print $myrow[0],p;
}

=
Mark
[EMAIL PROTECTED]

__
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