newbie wanting advice on search form

Hi
having trouble with the following code. when I try it comes up with the
message "could'nt execute query and returns zero results. It also gives
a warning about the mysql_num_rows line that it is not a valid mysql
resource. Any help given greatfully recieved as I have been struggling
to make this work for about a month now.

<html>
<head>
<title>search08</title>
<
</head>

<body>

<form name="form" action="search08.php" method="get">
<input type="text" name="q" />
<input type="submit" name="Submit" value="Search" />
</form>
</body>
</html>
<?php

// Get the search variable from URL

$var = [at] $_GET['q'] ;
$trimmed = trim($var); //trim whitespace from the stored variable

// rows to return
$limit=10;

// check for an empty string and display a message.
if ($trimmed == "")
{
echo "<p>Please enter a search...</p>";
exit;
}
// check for a search parameter
if (!isset($var))
{
echo "<p>We dont seem to have a search parameter!</p>";
exit ;
}
//connect to your database ** EDIT REQUIRED HERE **
mysql_connect("host","username","passord"); //(host, username,
password)

//specify database ** EDIT REQUIRED HERE **
mysql_select_db("online") or die("Unable to select database"); //select
which database we're using

// Build SQL Query

//$query = 'SELECT * FROM `arrow` WHERE `Surname`LIKE \"%trimmed%\" ;

$results= mysql_query($query);
$numrows= mysql_numrows($results);

// If we have no results, offer a google search as an alternative

if ($numrows == 0)
{
echo "<h4>Results</h4>";
echo "<p>Sorry, your search: "" . $trimmed . "" returned
zero results</p>";

// google
echo "<p><a href=\"http://www.google.com/search?q="
. $trimmed . "\" target=\"_blank\" title=\"Look up
" . $trimmed . " on Google\">Click here</a> to try the
search on google</p>";
}

// next determine if s has been passed to script, if not use 0
if (empty($s)) {
$s=0;
}

// get results
$query = "limit$s,$limit";
$results = mysql_query($query) or die ("Couldnt execute query");

// display what the person searched for
echo "<p>You searched for: "" . $var. ""</p>";

// begin to show results set
echo "Results";
$count = 1 + $s;

// now you can display the results returned
while ($row= mysql_fetch_array($result)){
$title =$row["surname"];

echo "$count.) $title";
$count++;
}

$currPage =(($s/$limit) + 1);

//break before paging
echo "
";

// next we need to do the links to other results
if ($s>=1){// bypass PREV link if s is 0
$prevs=($s-$limit);
print " <a href=\"$PHP_SELF?s=$prevs&q=$var\"><<
Prev 10</a>  ";
}

// calculate number of pages needing links
$pages=intval($numrows/$limit);

// $pages now contains int of pages needed unless there is a remainder
from division

if ($numrows%$limit){
// has remainder so add one page
$pages++;
}

// check to see if last page
if (!((($s+$limit)/$limit)==$pages) && $pages!=1){

// not last page so give NEXT link
$news=$s+$limit;

echo " <a href=\"$PHP_SELF?s=$news&q=$var\">Next 10
>></a>";
}

$a = $s + ($limit);
if ($a >$numrows($a = $numrows));
$b = $s + 1 ;
echo "<p>Showing results $b to $a of $numrows</p>";

?>
wgerry [ Do, 22 September 2005 23:25 ] [ ID #978831 ]

Re: newbie wanting advice on search form

I noticed that Message-ID:
<1127424304.472600.123120 [at] g43g2000cwa.googlegroups.com> from wgerry
contained the following:

>//$query = 'SELECT * FROM `arrow` WHERE `Surname`LIKE \"%trimmed%\" ;

Well it won't work commented out. and trimmed should be $trimmed

--
Geoff Berrow (put thecat out to email)
It's only Usenet, no one dies.
My opinions, not the committee's, mine.
Simple RFDs http://www.ckdog.co.uk/rfdmaker/
Geoff Berrow [ Fr, 23 September 2005 00:09 ] [ ID #978833 ]

Re: newbie wanting advice on search form

wgerry wrote:

> //$query = 'SELECT * FROM `arrow` WHERE `Surname`LIKE \"%trimmed%\" ;

Should be something like:

$query = 'SELECT * FROM arrow WHERE Surname LIKE "%' . $trimmed . '%"';

> $results= mysql_query($query);
> $numrows= mysql_numrows($results);

Should be:

$numrows = mysql_num_rows($results);

HTH.

Peter.


--
http://www.phpforums.nl
Peter van Schie [ Fr, 23 September 2005 01:18 ] [ ID #980513 ]

Re: newbie wanting advice on search form

wgerry wrote:
>
> //$query = 'SELECT * FROM `arrow` WHERE `Surname`LIKE \"%trimmed%\" ;
>

In addition to the other comments, this should be:

$query = "SELECT * FROM arrow WHERE Surname LIKE '%$trimmed%'" ;

The `s around the table and column name are unnecessary unless you have
a conflict with a reserved word. They don't hurt, they're just not
necessary.

However - you need single quotes around the LIKE value - and it should
be $trimmed - you forgot the '$'. Additionally, to get $trimmed to
expand, you need the whole thing in double quotes, not single (or
concatenate the strings together).

--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex [at] attglobal.net
==================
Jerry Stuckle [ Fr, 23 September 2005 05:14 ] [ ID #980527 ]

Re: newbie wanting advice on search form

Thanks for all your kind help guys. I have tried what you have
suggested but I am still having trouble. I have made the following
changes but it still comes up with the warning mysql_num_rows is not a
valid SQL resource and my query does not return a result and it comes
up with couldnt execute query,
<html>
<head>
<title>search</title>

</head>

<body>

<form name="form" action="search09.php" method="get">
<input type="text" name="q" />
<input type="submit" name="Submit" value="Search" />
</form>
</body>
</html>

<?php

// Get the search variable from URL

$var = [at] $_GET['q'] ;
$trimmed = trim($var); //trim whitespace from the stored variable

// rows to return
$limit=10;

// check for an empty string and display a message.
if ($trimmed == "")
{
echo "<p>Please enter a search...</p>";
exit;
}

// check for a search parameter
if (!isset($var))
{
echo "<p>We dont seem to have a search parameter!</p>";
exit;
}

//connect to your database ** EDIT REQUIRED HERE **
mysql_connect("host ","usename",password"); //(host, username,
password)

//specify database ** EDIT REQUIRED HERE **
mysql_select_db("online") or die("Unable to select database"); //select
which database we're using


// Build SQL Query
$query = "SELECT * FROM arrow WHERE surname LIKE %'$trimmed %.'" ;

$results= mysql_query($query);
$numrows= mysql_num_rows($results);

// If we have no results, offer a google search as an alternative

if ($numrows == 0)
{
echo "<h4>Results</h4>";
echo "<p>Sorry, your search: "" . $trimmed . "" returned
zero results</p>";

// google
echo "<p><a href=\"http://www.google.com/search?q="
. $trimmed . "\" target=\"_blank\" title=\"Look up
" . $trimmed . " on Google\">Click here</a> to try the
search on google</p>";
}

// next determine if s has been passed to script, if not use 0
if (empty($s)) {
$s=0;
}
{
// get results
$query = "$limit$s,$limit";
$results = mysql_query($query) or die ("Couldnt execute query");
}
// now you can display the results returned
while ($row = mysql_fetch_array($results)){
$title =$row["surname"];

echo "$count.) $title";
$count++;
}
{
$currPage =(($s/$limit) + 1);

//break before paging
echo "
";
}
// next we need to do the links to other results
if ($s>=1){// bypass PREV link if s is 0
$prevs=($s-$limit);
print " <a href=\"$PHP_SELF?s=$prevs&q=$var\"><<
Prev 10</a>  ";
}

// calculate number of pages needing links
$pages=intval($numrows/$limit);

// $pages now contains int of pages needed unless there is a remainder
from division

if ($numrows%$limit){
// has remainder so add one page
$pages++;
}

// check to see if last page
if (!((($s+$limit)/$limit)==$pages) && $pages!=1){

// not last page so give NEXT link
$news=$s+$limit;

echo " <a href=\"$PHP_SELF?s=$news&q=$var\">Next 10
>></a>";
}

$a = $s + ($limit);
if ($a >$numrows){$a = $numrows;}
$b = $s + 1 ;
echo "<p>Showing results $b to $a of $numrows</p>";
?>

any help give me with the above greatfully recieved and thanks again
for the help already given.

Kind Regards
Gerry
wgerry [ Mo, 26 September 2005 20:07 ] [ ID #984260 ]

Re: newbie wanting advice on search form

I noticed that Message-ID:
<1127758071.691493.182600 [at] f14g2000cwb.googlegroups.com> from wgerry
contained the following:

>$query = "SELECT * FROM arrow WHERE surname LIKE %'$trimmed %.'" ;

Spot the difference.

$query="SELECT * FROM arrow WHERE surname LIKE '%$trimmed%'";

--
Geoff Berrow (put thecat out to email)
It's only Usenet, no one dies.
My opinions, not the committee's, mine.
Simple RFDs http://www.ckdog.co.uk/rfdmaker/
Geoff Berrow [ Mo, 26 September 2005 21:01 ] [ ID #984262 ]

Re: newbie wanting advice on search form

Thanks for your reply so quick Geoff, but I tried it a few different
ways and Ive just tried it your way by copying and pasting (just to
make sure) and I still get the same result.

Kind Regards

Gerry
wgerry [ Mo, 26 September 2005 22:32 ] [ ID #984271 ]

Re: newbie wanting advice on search form

I noticed that Message-ID:
<1127766768.439415.269120 [at] o13g2000cwo.googlegroups.com> from wgerry
contained the following:

>Thanks for your reply so quick Geoff, but I tried it a few different
>ways and Ive just tried it your way by copying and pasting (just to
>make sure) and I still get the same result.

OK. I assume the database connection is ok and that
you have edited all the bits you need to edit to connect to your
database?

//connect to your database ** EDIT REQUIRED HERE **
mysql_connect("host ","usename",password"); //(host, username,
password)

//specify database ** EDIT REQUIRED HERE **
mysql_select_db("online") or die("Unable to select database"); //select
which database we're using

If that's ok...

Have you got phpMyadmin? If so, echo the $query to the screen and then
cut and paste it into PHPMyadmins SQL box. You can then test the query.

--
Geoff Berrow (put thecat out to email)
It's only Usenet, no one dies.
My opinions, not the committee's, mine.
Simple RFDs http://www.ckdog.co.uk/rfdmaker/
Geoff Berrow [ Mo, 26 September 2005 22:50 ] [ ID #984272 ]

Re: newbie wanting advice on search form

wgerry wrote:
> Thanks for all your kind help guys. I have tried what you have
> suggested but I am still having trouble. I have made the following
> changes but it still comes up with the warning mysql_num_rows is not a
> valid SQL resource and my query does not return a result and it comes
> up with couldnt execute query,
> <html>
> <head>
> <title>search</title>
>
> </head>
>
> <body>
>
> <form name="form" action="search09.php" method="get">
> <input type="text" name="q" />
> <input type="submit" name="Submit" value="Search" />
> </form>
> </body>
> </html>
>
> <?php
>
> // Get the search variable from URL
>
> $var = [at] $_GET['q'] ;
> $trimmed = trim($var); //trim whitespace from the stored variable
>
> // rows to return
> $limit=10;
>
> // check for an empty string and display a message.
> if ($trimmed == "")
> {
> echo "<p>Please enter a search...</p>";
> exit;
> }
>
> // check for a search parameter
> if (!isset($var))
> {
> echo "<p>We dont seem to have a search parameter!</p>";
> exit;
> }
>
> //connect to your database ** EDIT REQUIRED HERE **
> mysql_connect("host ","usename",password"); //(host, username,
> password)
>
> //specify database ** EDIT REQUIRED HERE **
> mysql_select_db("online") or die("Unable to select database"); //select
> which database we're using
>
>
> // Build SQL Query
> $query = "SELECT * FROM arrow WHERE surname LIKE %'$trimmed %.'" ;
>
> $results= mysql_query($query);
> $numrows= mysql_num_rows($results);
>
> // If we have no results, offer a google search as an alternative
>
> if ($numrows == 0)
> {
> echo "<h4>Results</h4>";
> echo "<p>Sorry, your search: "" . $trimmed . "" returned
> zero results</p>";
>
> // google
> echo "<p><a href=\"http://www.google.com/search?q="
> . $trimmed . "\" target=\"_blank\" title=\"Look up
> " . $trimmed . " on Google\">Click here</a> to try the
> search on google</p>";
> }
>
> // next determine if s has been passed to script, if not use 0
> if (empty($s)) {
> $s=0;
> }
> {
> // get results
> $query = "$limit$s,$limit";
> $results = mysql_query($query) or die ("Couldnt execute query");
> }
> // now you can display the results returned
> while ($row = mysql_fetch_array($results)){
> $title =$row["surname"];
>
> echo "$count.) $title";
> $count++;
> }
> {
> $currPage =(($s/$limit) + 1);
>
> //break before paging
> echo "
";
> }
> // next we need to do the links to other results
> if ($s>=1){// bypass PREV link if s is 0
> $prevs=($s-$limit);
> print " <a href=\"$PHP_SELF?s=$prevs&q=$var\"><<
> Prev 10</a>  ";
> }
>
> // calculate number of pages needing links
> $pages=intval($numrows/$limit);
>
> // $pages now contains int of pages needed unless there is a remainder
> from division
>
> if ($numrows%$limit){
> // has remainder so add one page
> $pages++;
> }
>
> // check to see if last page
> if (!((($s+$limit)/$limit)==$pages) && $pages!=1){
>
> // not last page so give NEXT link
> $news=$s+$limit;
>
> echo " <a href=\"$PHP_SELF?s=$news&q=$var\">Next 10
> >></a>";
> }
>
> $a = $s + ($limit);
> if ($a >$numrows){$a = $numrows;}
> $b = $s + 1 ;
> echo "<p>Showing results $b to $a of $numrows</p>";
> ?>
>
> any help give me with the above greatfully recieved and thanks again
> for the help already given.
>
> Kind Regards
> Gerry
>

Gerry,

You should ALWAYS check the results of any call to MySQL. If the result
is false, echo mysql_error() to help find the problem.

--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex [at] attglobal.net
==================
Jerry Stuckle [ Di, 27 September 2005 03:16 ] [ ID #986110 ]

Re: newbie wanting advice on search form

Once again thanks for the help guys, sorted out the problem, thanks for
all the instructive device, much appreciated.
Kind Regards
Gerry
wgerry [ Fr, 30 September 2005 20:36 ] [ ID #991987 ]
PHP » comp.lang.php » newbie wanting advice on search form

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