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You have an error in your SQL syntax;

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  You have an error in your SQL syntax; Co <vonclausowitz@gmail.com> - 2011-05-22 07:01 -0700
    Re: You have an error in your SQL syntax; Luuk <Luuk@invalid.lan> - 2011-05-22 19:11 +0200
      Re: You have an error in your SQL syntax; Co <vonclausowitz@gmail.com> - 2011-05-22 13:11 -0700
        Re: You have an error in your SQL syntax; The Natural Philosopher <tnp@invalid.invalid> - 2011-05-22 21:21 +0100
        Re: You have an error in your SQL syntax; Jerry Stuckle <jstucklex@attglobal.net> - 2011-05-22 16:21 -0400
        Re: You have an error in your SQL syntax; Luuk <Luuk@invalid.lan> - 2011-05-22 22:35 +0200
          Re: You have an error in your SQL syntax; Co <vonclausowitz@gmail.com> - 2011-05-22 13:56 -0700
            Re: You have an error in your SQL syntax; Jerry Stuckle <jstucklex@attglobal.net> - 2011-05-22 17:16 -0400
              Re: You have an error in your SQL syntax; Co <vonclausowitz@gmail.com> - 2011-05-22 14:22 -0700

#1768 — You have an error in your SQL syntax;

Hi all,

I run a query based on the input in a listbox.
The query looks for users from a certain country.
When no users are from the chosen country I get a error message.
Is there no way to check if there are records in the query before
trying to output
so we only get a message saying: No records for this search.... or
something like it.

Regards
Marco

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#1772

On 22-05-2011 16:01, Co wrote:
> Hi all,
> 
> I run a query based on the input in a listbox.
> The query looks for users from a certain country.
> When no users are from the chosen country I get a error message.
> Is there no way to check if there are records in the query before
> trying to output
> so we only get a message saying: No records for this search.... or
> something like it.
> 
> Regards
> Marco

I hope you know that there is a great MANUAL online at:
http://www.php.net

It has this great info:
http://php.net/manual/en/function.mysql-num-rows.php

-- 
Luuk

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#1777

On 22 mei, 19:11, Luuk <L...@invalid.lan> wrote:
> On 22-05-2011 16:01, Co wrote:
>
> > Hi all,
>
> > I run a query based on the input in a listbox.
> > The query looks for users from a certain country.
> > When no users are from the chosen country I get a error message.
> > Is there no way to check if there are records in the query before
> > trying to output
> > so we only get a message saying: No records for this search.... or
> > something like it.
>
> > Regards
> > Marco
>
> I hope you know that there is a great MANUAL online at:http://www.php.net
>
> It has this great info:http://php.net/manual/en/function.mysql-num-rows.php
>
> --
> Luuk

Thanks Luuk,
I didn't know.
Anyways I tried an example from the page:

$num_rows = mysql_num_rows($sql2);
if($num_rows <> 0) {
while($row = mysql_fetch_array($sql2)) { ......

else {
print ("<p>No records for this search were found.</p>");
}

But I still get the error line saying:
Warning: mysql_num_rows(): supplied argument is not a valid MySQL
result resource

Marco

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#1778

Co wrote:
> On 22 mei, 19:11, Luuk <L...@invalid.lan> wrote:
>> On 22-05-2011 16:01, Co wrote:
>>
>>> Hi all,
>>> I run a query based on the input in a listbox.
>>> The query looks for users from a certain country.
>>> When no users are from the chosen country I get a error message.
>>> Is there no way to check if there are records in the query before
>>> trying to output
>>> so we only get a message saying: No records for this search.... or
>>> something like it.
>>> Regards
>>> Marco
>> I hope you know that there is a great MANUAL online at:http://www.php.net
>>
>> It has this great info:http://php.net/manual/en/function.mysql-num-rows.php
>>
>> --
>> Luuk
> 
> Thanks Luuk,
> I didn't know.
> Anyways I tried an example from the page:
> 
> $num_rows = mysql_num_rows($sql2);
> if($num_rows <> 0) {
> while($row = mysql_fetch_array($sql2)) { ......
> 
> else {
> print ("<p>No records for this search were found.</p>");
> }
> 
> But I still get the error line saying:
> Warning: mysql_num_rows(): supplied argument is not a valid MySQL
> result resource
> 
> Marco

Then the original query failed.

If you stick in some debug like $query=(whatever your query is);
echo $query;

then you can cut and paste that into an interactive mysql session and 
test it.

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#1779

On 5/22/2011 4:11 PM, Co wrote:
> On 22 mei, 19:11, Luuk<L...@invalid.lan>  wrote:
>> On 22-05-2011 16:01, Co wrote:
>>
>>> Hi all,
>>
>>> I run a query based on the input in a listbox.
>>> The query looks for users from a certain country.
>>> When no users are from the chosen country I get a error message.
>>> Is there no way to check if there are records in the query before
>>> trying to output
>>> so we only get a message saying: No records for this search.... or
>>> something like it.
>>
>>> Regards
>>> Marco
>>
>> I hope you know that there is a great MANUAL online at:http://www.php.net
>>
>> It has this great info:http://php.net/manual/en/function.mysql-num-rows.php
>>
>> --
>> Luuk
>
> Thanks Luuk,
> I didn't know.
> Anyways I tried an example from the page:
>
> $num_rows = mysql_num_rows($sql2);
> if($num_rows<>  0) {
> while($row = mysql_fetch_array($sql2)) { ......
>
> else {
> print ("<p>No records for this search were found.</p>");
> }
>
> But I still get the error line saying:
> Warning: mysql_num_rows(): supplied argument is not a valid MySQL
> result resource
>
> Marco

It means there was a problem with your SQL statement.

You need to check the results of mysql_query(), and if it is false, 
determine why the SQL statement was in error.

You should ALWAYS check the return value of mysql_query() to ensure 
worked correctly.

-- 
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attglobal.net
==================

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#1781

On 22-05-2011 22:11, Co wrote:
> On 22 mei, 19:11, Luuk <L...@invalid.lan> wrote:
>> On 22-05-2011 16:01, Co wrote:
>>
>>> Hi all,
>>
>>> I run a query based on the input in a listbox.
>>> The query looks for users from a certain country.
>>> When no users are from the chosen country I get a error message.
>>> Is there no way to check if there are records in the query before
>>> trying to output
>>> so we only get a message saying: No records for this search.... or
>>> something like it.
>>
>>> Regards
>>> Marco
>>
>> I hope you know that there is a great MANUAL online at:http://www.php.net
>>
>> It has this great info:http://php.net/manual/en/function.mysql-num-rows.php
>>
>> --
>> Luuk
> 
> Thanks Luuk,
> I didn't know.
> Anyways I tried an example from the page:
> 
> $num_rows = mysql_num_rows($sql2);
> if($num_rows <> 0) {
> while($row = mysql_fetch_array($sql2)) { ......
> 
> else {
> print ("<p>No records for this search were found.</p>");
> }
> 
> But I still get the error line saying:
> Warning: mysql_num_rows(): supplied argument is not a valid MySQL
> result resource
> 
> Marco

The argument to mysql_num_rows() should not be a sql-statement, so
sending a parameter with the name $sql2 seems confusing

The argument you need to pass is the result from mysql_query();

like the example:
$result = mysql_query("SELECT * FROM table1", $link);
$num_rows = mysql_num_rows($result);




-- 
Luuk

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#1782

On 22 mei, 22:35, Luuk <L...@invalid.lan> wrote:
> On 22-05-2011 22:11, Co wrote:
>
>
>
>
>
>
>
>
>
> > On 22 mei, 19:11, Luuk <L...@invalid.lan> wrote:
> >> On 22-05-2011 16:01, Co wrote:
>
> >>> Hi all,
>
> >>> I run a query based on the input in a listbox.
> >>> The query looks for users from a certain country.
> >>> When no users are from the chosen country I get a error message.
> >>> Is there no way to check if there are records in the query before
> >>> trying to output
> >>> so we only get a message saying: No records for this search.... or
> >>> something like it.
>
> >>> Regards
> >>> Marco
>
> >> I hope you know that there is a great MANUAL online at:http://www.php.net
>
> >> It has this great info:http://php.net/manual/en/function.mysql-num-rows.php
>
> >> --
> >> Luuk
>
> > Thanks Luuk,
> > I didn't know.
> > Anyways I tried an example from the page:
>
> > $num_rows = mysql_num_rows($sql2);
> > if($num_rows <> 0) {
> > while($row = mysql_fetch_array($sql2)) { ......
>
> > else {
> > print ("<p>No records for this search were found.</p>");
> > }
>
> > But I still get the error line saying:
> > Warning: mysql_num_rows(): supplied argument is not a valid MySQL
> > result resource
>
> > Marco
>
> The argument to mysql_num_rows() should not be a sql-statement, so
> sending a parameter with the name $sql2 seems confusing
>
> The argument you need to pass is the result from mysql_query();
>
> like the example:
> $result = mysql_query("SELECT * FROM table1", $link);
> $num_rows = mysql_num_rows($result);
>
> --
> Luuk

Guys,

I solved it.
I put this in the beginning:
$nr = mysql_num_rows($sql); // Get total of Num rows from the database
query
if ($nr){  // if we found any records we will proceed

If the query returns 0 records we pass all the code and just say "No
records found".

The only problem I still have is when I load the page first time I get
an error saying:
Notice: Undefined index
Somehow on this line:

if (($_POST['listByq'] == "newest_members")) {

listByq has not been defined.
Can I declare it at the top of the page?

Marco

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#1783

On 5/22/2011 4:56 PM, Co wrote:
> On 22 mei, 22:35, Luuk<L...@invalid.lan>  wrote:
>> On 22-05-2011 22:11, Co wrote:
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>> On 22 mei, 19:11, Luuk<L...@invalid.lan>  wrote:
>>>> On 22-05-2011 16:01, Co wrote:
>>
>>>>> Hi all,
>>
>>>>> I run a query based on the input in a listbox.
>>>>> The query looks for users from a certain country.
>>>>> When no users are from the chosen country I get a error message.
>>>>> Is there no way to check if there are records in the query before
>>>>> trying to output
>>>>> so we only get a message saying: No records for this search.... or
>>>>> something like it.
>>
>>>>> Regards
>>>>> Marco
>>
>>>> I hope you know that there is a great MANUAL online at:http://www.php.net
>>
>>>> It has this great info:http://php.net/manual/en/function.mysql-num-rows.php
>>
>>>> --
>>>> Luuk
>>
>>> Thanks Luuk,
>>> I didn't know.
>>> Anyways I tried an example from the page:
>>
>>> $num_rows = mysql_num_rows($sql2);
>>> if($num_rows<>  0) {
>>> while($row = mysql_fetch_array($sql2)) { ......
>>
>>> else {
>>> print ("<p>No records for this search were found.</p>");
>>> }
>>
>>> But I still get the error line saying:
>>> Warning: mysql_num_rows(): supplied argument is not a valid MySQL
>>> result resource
>>
>>> Marco
>>
>> The argument to mysql_num_rows() should not be a sql-statement, so
>> sending a parameter with the name $sql2 seems confusing
>>
>> The argument you need to pass is the result from mysql_query();
>>
>> like the example:
>> $result = mysql_query("SELECT * FROM table1", $link);
>> $num_rows = mysql_num_rows($result);
>>
>> --
>> Luuk
>
> Guys,
>
> I solved it.
> I put this in the beginning:
> $nr = mysql_num_rows($sql); // Get total of Num rows from the database
> query
> if ($nr){  // if we found any records we will proceed
>
> If the query returns 0 records we pass all the code and just say "No
> records found".
>
> The only problem I still have is when I load the page first time I get
> an error saying:
> Notice: Undefined index
> Somehow on this line:
>
> if (($_POST['listByq'] == "newest_members")) {
>
> listByq has not been defined.
> Can I declare it at the top of the page?
>
> Marco

You can declare it, but rather you should use:

if (isset($_POST['listByq']))
   $listByq = $_POST['listByq'];
else
   $listByq = 'Some default value'; // Use an appropriate value

Which can also be shortened to:

$listByq = isset($_POST['listByq']) ? $_POST['listByq'] : 'Some default 
value';

Then use $listByq from there on.

And BTW - ALWAYS validate any input from the user, including $_POST 
values.  There is no guarantee these values came from your form, for 
instance.  A hacker can easily provide whatever value he wants.

-- 
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attglobal.net
==================

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#1785

On 22 mei, 23:16, Jerry Stuckle <jstuck...@attglobal.net> wrote:
> On 5/22/2011 4:56 PM, Co wrote:
>
>
>
>
>
>
>
>
>
> > On 22 mei, 22:35, Luuk<L...@invalid.lan>  wrote:
> >> On 22-05-2011 22:11, Co wrote:
>
> >>> On 22 mei, 19:11, Luuk<L...@invalid.lan>  wrote:
> >>>> On 22-05-2011 16:01, Co wrote:
>
> >>>>> Hi all,
>
> >>>>> I run a query based on the input in a listbox.
> >>>>> The query looks for users from a certain country.
> >>>>> When no users are from the chosen country I get a error message.
> >>>>> Is there no way to check if there are records in the query before
> >>>>> trying to output
> >>>>> so we only get a message saying: No records for this search.... or
> >>>>> something like it.
>
> >>>>> Regards
> >>>>> Marco
>
> >>>> I hope you know that there is a great MANUAL online at:http://www.php.net
>
> >>>> It has this great info:http://php.net/manual/en/function.mysql-num-rows.php
>
> >>>> --
> >>>> Luuk
>
> >>> Thanks Luuk,
> >>> I didn't know.
> >>> Anyways I tried an example from the page:
>
> >>> $num_rows = mysql_num_rows($sql2);
> >>> if($num_rows<>  0) {
> >>> while($row = mysql_fetch_array($sql2)) { ......
>
> >>> else {
> >>> print ("<p>No records for this search were found.</p>");
> >>> }
>
> >>> But I still get the error line saying:
> >>> Warning: mysql_num_rows(): supplied argument is not a valid MySQL
> >>> result resource
>
> >>> Marco
>
> >> The argument to mysql_num_rows() should not be a sql-statement, so
> >> sending a parameter with the name $sql2 seems confusing
>
> >> The argument you need to pass is the result from mysql_query();
>
> >> like the example:
> >> $result = mysql_query("SELECT * FROM table1", $link);
> >> $num_rows = mysql_num_rows($result);
>
> >> --
> >> Luuk
>
> > Guys,
>
> > I solved it.
> > I put this in the beginning:
> > $nr = mysql_num_rows($sql); // Get total of Num rows from the database
> > query
> > if ($nr){  // if we found any records we will proceed
>
> > If the query returns 0 records we pass all the code and just say "No
> > records found".
>
> > The only problem I still have is when I load the page first time I get
> > an error saying:
> > Notice: Undefined index
> > Somehow on this line:
>
> > if (($_POST['listByq'] == "newest_members")) {
>
> > listByq has not been defined.
> > Can I declare it at the top of the page?
>
> > Marco
>
> You can declare it, but rather you should use:
>
> if (isset($_POST['listByq']))
>    $listByq = $_POST['listByq'];
> else
>    $listByq = 'Some default value'; // Use an appropriate value
>
> Which can also be shortened to:
>
> $listByq = isset($_POST['listByq']) ? $_POST['listByq'] : 'Some default
> value';
>
> Then use $listByq from there on.
>
> And BTW - ALWAYS validate any input from the user, including $_POST
> values.  There is no guarantee these values came from your form, for
> instance.  A hacker can easily provide whatever value he wants.
>
> --
> ==================
> Remove the "x" from my email address
> Jerry Stuckle
> JDS Computer Training Corp.
> jstuck...@attglobal.net
> ==================

Jerry,

thanks this works,
and thanks for the advises.

Marco

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