RE: [PHP-DB] Dynamic SQL + result resource error
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."";
$queryResult = mysql_query($sql);
while ( $myrow = mysql_fetch_assoc( $queryResult ) )
{
echo " ".$myrow["bandname"]." "."\n";
}
/
Hope this helps ;)
Greetinx,
Mike
Michael Rudel
- Web-Development, Systemadministration -
Besuchen Sie uns am 20. und 21. August 2001 auf der
online-marketing-düsseldorf in Halle 1 Stand E 16
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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)";
> $i++;
> }
> print "$sql";
> while ($myrow=mysql_fetch_row($sql))
> {
> print "$myrow[0],";
> }
>
> =
> Mark
> [EMAIL PROTECTED]
>
> __
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RE: [PHP-DB] Dynamic SQL + result resource error
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)";
> $i++;
> }
> print "$sql";
>
> $queryResult = mysql_query($sql);
>
> while ($myrow=mysql_fetch_row($queryResult))
> {
> print "$myrow[0],";
> }
>
>
> -Original Message-
> From: Mark Gordon [mailto:[EMAIL PROTECTED]]
> Sent: Sunday, July 08, 2001 7:54 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] Dynamic SQL + result resource error
>
>
> Why is this code generating an error when it outputs a
> valid SQL statement? (there are no parse errors)
>
> //$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)";
> $i++;
> }
> print "$sql";
> while ($myrow=mysql_fetch_row($sql))
> {
> print "$myrow[0],";
> }
>
> =
> Mark
> [EMAIL PROTECTED]
>
> __
> Do You Yahoo!?
> Get personalized email addresses from Yahoo! Mail
> http://personal.mail.yahoo.com/
>
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>
>
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>
>
>
>
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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)";
$i++;
}
print "$sql";
$queryResult = mysql_query($sql);
while ($myrow=mysql_fetch_row($queryResult))
{
print "$myrow[0],";
}
-Original Message-
From: Mark Gordon [mailto:[EMAIL PROTECTED]]
Sent: Sunday, July 08, 2001 7:54 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Dynamic SQL + result resource error
Why is this code generating an error when it outputs a
valid SQL statement? (there are no parse errors)
//$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)";
$i++;
}
print "$sql";
while ($myrow=mysql_fetch_row($sql))
{
print "$myrow[0],";
}
=
Mark
[EMAIL PROTECTED]
__
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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)";
$i++;
}
print "$sql";
$queryResult = mysql_query($sql);
while ($myrow=mysql_fetch_row($queryResult))
{
print "$myrow[0],";
}
-Original Message-
From: Mark Gordon [mailto:[EMAIL PROTECTED]]
Sent: Sunday, July 08, 2001 7:54 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Dynamic SQL + result resource error
Why is this code generating an error when it outputs a
valid SQL statement? (there are no parse errors)
//$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)";
$i++;
}
print "$sql";
while ($myrow=mysql_fetch_row($sql))
{
print "$myrow[0],";
}
=
Mark
[EMAIL PROTECTED]
__
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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)";
$i++;
}
print "$sql";
$queryResult = mysql_query($sql);
while ($myrow=mysql_fetch_row($queryResult))
{
print "$myrow[0],";
}
-Original Message-
From: Mark Gordon [mailto:[EMAIL PROTECTED]]
Sent: Sunday, July 08, 2001 7:54 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Dynamic SQL + result resource error
Why is this code generating an error when it outputs a
valid SQL statement? (there are no parse errors)
//$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)";
$i++;
}
print "$sql";
while ($myrow=mysql_fetch_row($sql))
{
print "$myrow[0],";
}
=
Mark
[EMAIL PROTECTED]
__
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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)";
$i++;
}
print "$sql";
$queryResult = mysql_query($sql);
while ($myrow=mysql_fetch_row($queryResult))
{
print "$myrow[0],";
}
-Original Message-
From: Mark Gordon [mailto:[EMAIL PROTECTED]]
Sent: Sunday, July 08, 2001 7:54 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Dynamic SQL + result resource error
Why is this code generating an error when it outputs a
valid SQL statement? (there are no parse errors)
//$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)";
$i++;
}
print "$sql";
while ($myrow=mysql_fetch_row($sql))
{
print "$myrow[0],";
}
=
Mark
[EMAIL PROTECTED]
__
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RE: [PHP-DB] Dynamic SQL + result resource error
You aren't calling mysql_query() to execute the query.
//$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)";
$i++;
}
print "$sql";
$queryResult = mysql_query($sql);
while ($myrow=mysql_fetch_row($queryResult))
{
print "$myrow[0],";
}
-Original Message-
From: Mark Gordon [mailto:[EMAIL PROTECTED]]
Sent: Sunday, July 08, 2001 7:54 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Dynamic SQL + result resource error
Why is this code generating an error when it outputs a
valid SQL statement? (there are no parse errors)
//$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)";
$i++;
}
print "$sql";
while ($myrow=mysql_fetch_row($sql))
{
print "$myrow[0],";
}
=
Mark
[EMAIL PROTECTED]
__
Do You Yahoo!?
Get personalized email addresses from Yahoo! Mail
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