Hi Everyone, I am attempting to setup a search field on a database
application I'm dinking around with and running into problems that
I'm hoping someone might be able to shed some light on.

Here is the code I am using to display the results of the search:

echo ('<table border="1">');
echo "<tr><th>First Name</th><th>Author</th><th>Pages</th></tr>";

$result_row[] = mysql_query($query) or die(mysql_error());
echo "Result_row dump: $result_row <BR>";
$num=mysql_numrows($result);
echo "Num dump: $num <BR>";
$i= 0;

while($i < $num) {
echo "<tr><td>";
echo $result_row[0] . '</td><td>';
echo $result_row[1] . '</td><td>';
echo $result_row[2] . '</td></tr>';
$i++;
}

echo ("</table>");


the problem is instead of printing out the actual results it prints
out 6 fields that say: "Resource id #5"

the query I'm using is: SELECT FName, LName, Add1, Add2 FROM current
WHERE FName like '%jason%' which works in MySQL

I can't find any errors in my log files to even give me a hint as to
what is going on.

If someone could take a look I would greatly appreciate it.

Jason Pruim

Em Quinta 31 Maio 2007 16:25, Jason Pruim escreveu:
> Hi Everyone, I am attempting to setup a search field on a database
> application I'm dinking around with and running into problems that
> I'm hoping someone might be able to shed some light on.
>
> Here is the code I am using to display the results of the search:
>
> echo ('<table border="1">');
> echo "<tr><th>First Name</th><th>Author</th><th>Pages</th></tr>";
>
> $result_row[] = mysql_query($query) or die(mysql_error());
> echo "Result_row dump: $result_row <BR>";
> $num=mysql_numrows($result);
> echo "Num dump: $num <BR>";
> $i= 0;
>
/*
> while($i < $num) {
*/

$result=$result_row[0];

while($result_row = mysql_fetch_array($result) {

> echo "<tr><td>";
> echo $result_row[0] . '</td><td>';
> echo $result_row[1] . '</td><td>';
> echo $result_row[2] . '</td></tr>';
> $i++;
> }
>
> echo ("</table>");
>
>
> the problem is instead of printing out the actual results it prints
> out 6 fields that say: "Resource id #5"
>
> the query I'm using is: SELECT FName, LName, Add1, Add2 FROM current
> WHERE FName like '%jason%' which works in MySQL
>
> I can't find any errors in my log files to even give me a hint as to
> what is going on.
>
> If someone could take a look I would greatly appreciate it.
>

HTH

--
Davi Vidal
EMAIL REMOVED
EMAIL REMOVED
--
"Religion, ideology, resources, land,
spite, love or "just because"...
No matter how pathetic the reason,
it's enough to start a war. "
--------------------------------------------------------
Por favor não faça top-posting, coloque a sua resposta abaixo desta linha.
Please don't do top-posting, put your reply below the following line.
--------------------------------------------------------

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On Thu, 2007-05-31 at 15:25 -0400, Jason Pruim wrote:
> Hi Everyone, I am attempting to setup a search field on a database
> application I'm dinking around with and running into problems that
> I'm hoping someone might be able to shed some light on.
>
> Here is the code I am using to display the results of the search:
>
> echo ('<table border="1">');
> echo "<tr><th>First Name</th><th>Author</th><th>Pages</th></tr>";
>
> $result_row[] = mysql_query($query) or die(mysql_error());

mysql_query() doesn't return the rows. Go read the manual.

Cheers,
Rob.
--
..------------------------------------------------------------.
| InterJinn Application Framework - http://www.interjinn.com |
:------------------------------------------------------------:
| An application and templating framework for PHP. Boasting |
| a powerful, scalable system for accessing system services |
| such as forms, properties, sessions, and caches. InterJinn |
| also provides an extremely flexible architecture for |
| creating re-usable components quickly and easily. |
`------------------------------------------------------------'

Your problem is this:

>
> $result_row[] = mysql_query($query) or die(mysql_error());


....you are ***igning a query to a variable. What you need to do is
something like this:

$result = mysql_query($query) or die(mysql_error());
while ($result_row = mysql_fetch_array($result)) {
......
}

On May 31, 2007, at 3:27 PM, Davi wrote:

> Em Quinta 31 Maio 2007 16:25, Jason Pruim escreveu:
>> Hi Everyone, I am attempting to setup a search field on a database
>> application I'm dinking around with and running into problems that
>> I'm hoping someone might be able to shed some light on.
>>
>> Here is the code I am using to display the results of the search:
>>
>> echo ('<table border="1">');
>> echo "<tr><th>First Name</th><th>Author</th><th>Pages</th></tr>";
>>
>> $result_row[] = mysql_query($query) or die(mysql_error());
>> echo "Result_row dump: $result_row <BR>";
>> $num=mysql_numrows($result);
>> echo "Num dump: $num <BR>";
>> $i= 0;
>>
> /*
>> while($i < $num) {
> */
>
> $result=$result_row[0];
>
> while($result_row = mysql_fetch_array($result) {

Worked perfectly after adding a closing ) Thanks for the tip!

>
>> echo "<tr><td>";
>> echo $result_row[0] . '</td><td>';
>> echo $result_row[1] . '</td><td>';
>> echo $result_row[2] . '</td></tr>';
>> $i++;
>> }
>>
>> echo ("</table>");
>>
>>
>> the problem is instead of printing out the actual results it prints
>> out 6 fields that say: "Resource id #5"
>>
>> the query I'm using is: SELECT FName, LName, Add1, Add2 FROM current
>> WHERE FName like '%jason%' which works in MySQL
>>
>> I can't find any errors in my log files to even give me a hint as to
>> what is going on.
>>
>> If someone could take a look I would greatly appreciate it.
>>
>
> HTH
>
> --
> Davi Vidal
> EMAIL REMOVED
> EMAIL REMOVED
> --
> "Religion, ideology, resources, land,
> spite, love or "just because"...
> No matter how pathetic the reason,
> it's enough to start a war. "
> --------------------------------------------------------
> Por favor não faça top-posting, coloque a sua resposta abaixo desta
> linha.
> Please don't do top-posting, put your reply below the following line.
> --------------------------------------------------------

Em Quinta 31 Maio 2007 16:38, Jason Pruim escreveu:
> >
> > while($result_row = mysql_fetch_array($result) {
>
> Worked perfectly after adding a closing ) Thanks for the tip!
>

Forgive me for that!!! #'_'#
I *always* forget the closing )... =P


--
Davi Vidal
EMAIL REMOVED
EMAIL REMOVED
--
"Religion, ideology, resources, land,
spite, love or "just because"...
No matter how pathetic the reason,
it's enough to start a war. "
--------------------------------------------------------
Por favor não faça top-posting, coloque a sua resposta abaixo desta linha.
Please don't do top-posting, put your reply below the following line.
--------------------------------------------------------

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On Thu, May 31, 2007 2:25 pm, Jason Pruim wrote:
> Hi Everyone, I am attempting to setup a search field on a database
> application I'm dinking around with and running into problems that
> I'm hoping someone might be able to shed some light on.
>
> Here is the code I am using to display the results of the search:
>
> echo ('<table border="1">');
> echo "<tr><th>First Name</th><th>Author</th><th>Pages</th></tr>";
>
> $result_row[] = mysql_query($query) or die(mysql_error());

I also have to wonder why you are building an array of query result
resources...

You probably could do this with just one SQL query and some order by
clauses to get what you want, without hitting the DB so much.

--
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Know what I want?
I want you to buy a CD from some indie artist.
http://cdbaby.com/browse/from/lynch
Yeah, I get a buck. So?