[PHP] Re: mysql_num_rows()
On 22/02/11 14:40, Gary wrote:
Pete Fordp...@justcroft.com wrote in message
news:76.48.39221.054c3...@pb1.pair.com...
On 22/02/11 13:59, Gary wrote:
Pete Fordp...@justcroft.com wrote in message
news:a4.c0.39221.b3ca3...@pb1.pair.com...
On 22/02/11 05:40, Gary wrote:
Can someone tell me why this is not working? I do not get an error
message,
the results are called and echo'd to screen, the count does not work, I
get
a 0 for a result...
$result = mysql_query(SELECT * FROM `counties` WHERE name =
'checked')
or
die(mysql_error());
if ( isset($_POST['submit']) ) {
for($i=1; $i=$_POST['counties']; $i++) {
if ( isset($_POST[county$i] ) ) {
echo You have chosen . $_POST[county$i].br/;
}
}
}
$county_total=mysql_num_rows($result);
while($row = mysql_fetch_array($result))
{$i++;
echo $row['name'];
}
$county_total=mysql_num_rows($result);
echo $county_total;
echo 'You Have ' . $county_total . ' Counties ';
?
The first thing I see is that you do
$county_total=mysql_num_rows($result)
twice. Now I'm not to sure, but it is possible that looping over
mysql_fetch_array($result) moves an array pointer so the count is not
valid after the operation... that's a long shot.
But since you are looping over the database records anyway, why not just
count them as you go? Isn't $i the number of counties after all the
looping?
--
Peter Ford, Developer phone: 01580 89 fax: 01580
893399
Justcroft International Ltd.
www.justcroft.com
Justcroft House, High Street, Staplehurst, Kent TN12 0AH United
Kingdom
Registered in England and Wales: 2297906
Registered office: Stag Gates House, 63/64 The Avenue, Southampton SO17
1XS
Peter
Thank you for your reply.
I did notice that I had the $county_total=mysql_num_rows($result) twice.
I
had moved and removed it, always getting either a 0 or 1 result.
I put up another test page with
$result = mysql_query(SELECT * FROM `counties` ) or die(mysql_error());
$tot=mysql_num_rows($result);
echo $tot;
?
And it worked fine.
I have tried count() as well as mysql_num_rows() but am getting the same
result.
Could you explain what you mean by count them as I go?
Again, Thank you for your reply.
Gary
__ Information from ESET Smart Security, version of virus
signature database 5895 (20110222) __
The message was checked by ESET Smart Security.
http://www.eset.com
Well,
Lets go back to your original code and cut out the decoration:
$result = mysql_query(SELECT * FROM `counties` WHERE name = 'checked')
$county_total=mysql_num_rows($result);
while($row = mysql_fetch_array($result))
{
echo $row['name'];
}
echo $county_total;
That code should do pretty much what you want...
But, because you only show the number of records AFTER the loop, you could
do:
$result = mysql_query(SELECT * FROM `counties` WHERE name = 'checked')
$county_total = 0;
while($row = mysql_fetch_array($result))
{
echo $row['name'];
$county_total++;
}
echo $county_total;
Peter
Thank you for your suggestion...btw cut out the decoration LOL
I'm sorry to report it did not work. I tried various configurations of it
but to no avail.
Any other suggestions?
Again, thank you for your help.
Gary
__ Information from ESET Smart Security, version of virus signature
database 5895 (20110222) __
The message was checked by ESET Smart Security.
http://www.eset.com
Gary,
I'd probably need a bit more detail on the it did not work to go much further!
Actually, it looks like I missed a semicolon on the mysql_query line in *both*
versions - maybe that was the problem...
Do you have access to your server logs, to see if any errors are reported when
you run this?
Cheers
Peter
--
Peter Ford, Developer phone: 01580 89 fax: 01580 893399
Justcroft International Ltd. www.justcroft.com
Justcroft House, High Street, Staplehurst, Kent TN12 0AH United Kingdom
Registered in England and Wales: 2297906
Registered office: Stag Gates House, 63/64 The Avenue, Southampton SO17 1XS
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[PHP] Re: mysql_num_rows()
On 22/02/11 05:40, Gary wrote:
Can someone tell me why this is not working? I do not get an error message,
the results are called and echo'd to screen, the count does not work, I get
a 0 for a result...
$result = mysql_query(SELECT * FROM `counties` WHERE name = 'checked') or
die(mysql_error());
if ( isset($_POST['submit']) ) {
for($i=1; $i=$_POST['counties']; $i++) {
if ( isset($_POST[county$i] ) ) {
echo You have chosen . $_POST[county$i].br/;
}
}
}
$county_total=mysql_num_rows($result);
while($row = mysql_fetch_array($result))
{$i++;
echo $row['name'];
}
$county_total=mysql_num_rows($result);
echo $county_total;
echo 'You Have ' . $county_total . ' Counties ';
?
The first thing I see is that you do
$county_total=mysql_num_rows($result)
twice. Now I'm not to sure, but it is possible that looping over
mysql_fetch_array($result) moves an array pointer so the count is not valid
after the operation... that's a long shot.
But since you are looping over the database records anyway, why not just count
them as you go? Isn't $i the number of counties after all the looping?
--
Peter Ford, Developer phone: 01580 89 fax: 01580 893399
Justcroft International Ltd. www.justcroft.com
Justcroft House, High Street, Staplehurst, Kent TN12 0AH United Kingdom
Registered in England and Wales: 2297906
Registered office: Stag Gates House, 63/64 The Avenue, Southampton SO17 1XS
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Re: mysql_num_rows()
Pete Ford p...@justcroft.com wrote in message
news:a4.c0.39221.b3ca3...@pb1.pair.com...
On 22/02/11 05:40, Gary wrote:
Can someone tell me why this is not working? I do not get an error
message,
the results are called and echo'd to screen, the count does not work, I
get
a 0 for a result...
$result = mysql_query(SELECT * FROM `counties` WHERE name = 'checked')
or
die(mysql_error());
if ( isset($_POST['submit']) ) {
for($i=1; $i=$_POST['counties']; $i++) {
if ( isset($_POST[county$i] ) ) {
echo You have chosen . $_POST[county$i].br/;
}
}
}
$county_total=mysql_num_rows($result);
while($row = mysql_fetch_array($result))
{$i++;
echo $row['name'];
}
$county_total=mysql_num_rows($result);
echo $county_total;
echo 'You Have ' . $county_total . ' Counties ';
?
The first thing I see is that you do
$county_total=mysql_num_rows($result)
twice. Now I'm not to sure, but it is possible that looping over
mysql_fetch_array($result) moves an array pointer so the count is not
valid after the operation... that's a long shot.
But since you are looping over the database records anyway, why not just
count them as you go? Isn't $i the number of counties after all the
looping?
--
Peter Ford, Developer phone: 01580 89 fax: 01580
893399
Justcroft International Ltd.
www.justcroft.com
Justcroft House, High Street, Staplehurst, Kent TN12 0AH United
Kingdom
Registered in England and Wales: 2297906
Registered office: Stag Gates House, 63/64 The Avenue, Southampton SO17
1XS
Peter
Thank you for your reply.
I did notice that I had the $county_total=mysql_num_rows($result) twice. I
had moved and removed it, always getting either a 0 or 1 result.
I put up another test page with
$result = mysql_query(SELECT * FROM `counties` ) or die(mysql_error());
$tot=mysql_num_rows($result);
echo $tot;
?
And it worked fine.
I have tried count() as well as mysql_num_rows() but am getting the same
result.
Could you explain what you mean by count them as I go?
Again, Thank you for your reply.
Gary
__ Information from ESET Smart Security, version of virus signature
database 5895 (20110222) __
The message was checked by ESET Smart Security.
http://www.eset.com
--
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[PHP] Re: mysql_num_rows()
On 22/02/11 13:59, Gary wrote:
Pete Fordp...@justcroft.com wrote in message
news:a4.c0.39221.b3ca3...@pb1.pair.com...
On 22/02/11 05:40, Gary wrote:
Can someone tell me why this is not working? I do not get an error
message,
the results are called and echo'd to screen, the count does not work, I
get
a 0 for a result...
$result = mysql_query(SELECT * FROM `counties` WHERE name = 'checked')
or
die(mysql_error());
if ( isset($_POST['submit']) ) {
for($i=1; $i=$_POST['counties']; $i++) {
if ( isset($_POST[county$i] ) ) {
echo You have chosen . $_POST[county$i].br/;
}
}
}
$county_total=mysql_num_rows($result);
while($row = mysql_fetch_array($result))
{$i++;
echo $row['name'];
}
$county_total=mysql_num_rows($result);
echo $county_total;
echo 'You Have ' . $county_total . ' Counties ';
?
The first thing I see is that you do
$county_total=mysql_num_rows($result)
twice. Now I'm not to sure, but it is possible that looping over
mysql_fetch_array($result) moves an array pointer so the count is not
valid after the operation... that's a long shot.
But since you are looping over the database records anyway, why not just
count them as you go? Isn't $i the number of counties after all the
looping?
--
Peter Ford, Developer phone: 01580 89 fax: 01580
893399
Justcroft International Ltd.
www.justcroft.com
Justcroft House, High Street, Staplehurst, Kent TN12 0AH United
Kingdom
Registered in England and Wales: 2297906
Registered office: Stag Gates House, 63/64 The Avenue, Southampton SO17
1XS
Peter
Thank you for your reply.
I did notice that I had the $county_total=mysql_num_rows($result) twice. I
had moved and removed it, always getting either a 0 or 1 result.
I put up another test page with
$result = mysql_query(SELECT * FROM `counties` ) or die(mysql_error());
$tot=mysql_num_rows($result);
echo $tot;
?
And it worked fine.
I have tried count() as well as mysql_num_rows() but am getting the same
result.
Could you explain what you mean by count them as I go?
Again, Thank you for your reply.
Gary
__ Information from ESET Smart Security, version of virus signature
database 5895 (20110222) __
The message was checked by ESET Smart Security.
http://www.eset.com
Well,
Lets go back to your original code and cut out the decoration:
$result = mysql_query(SELECT * FROM `counties` WHERE name = 'checked')
$county_total=mysql_num_rows($result);
while($row = mysql_fetch_array($result))
{
echo $row['name'];
}
echo $county_total;
That code should do pretty much what you want...
But, because you only show the number of records AFTER the loop, you could do:
$result = mysql_query(SELECT * FROM `counties` WHERE name = 'checked')
$county_total = 0;
while($row = mysql_fetch_array($result))
{
echo $row['name'];
$county_total++;
}
echo $county_total;
--
Peter Ford, Developer phone: 01580 89 fax: 01580 893399
Justcroft International Ltd. www.justcroft.com
Justcroft House, High Street, Staplehurst, Kent TN12 0AH United Kingdom
Registered in England and Wales: 2297906
Registered office: Stag Gates House, 63/64 The Avenue, Southampton SO17 1XS
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Re: mysql_num_rows()
Pete Ford p...@justcroft.com wrote in message
news:76.48.39221.054c3...@pb1.pair.com...
On 22/02/11 13:59, Gary wrote:
Pete Fordp...@justcroft.com wrote in message
news:a4.c0.39221.b3ca3...@pb1.pair.com...
On 22/02/11 05:40, Gary wrote:
Can someone tell me why this is not working? I do not get an error
message,
the results are called and echo'd to screen, the count does not work, I
get
a 0 for a result...
$result = mysql_query(SELECT * FROM `counties` WHERE name =
'checked')
or
die(mysql_error());
if ( isset($_POST['submit']) ) {
for($i=1; $i=$_POST['counties']; $i++) {
if ( isset($_POST[county$i] ) ) {
echo You have chosen . $_POST[county$i].br/;
}
}
}
$county_total=mysql_num_rows($result);
while($row = mysql_fetch_array($result))
{$i++;
echo $row['name'];
}
$county_total=mysql_num_rows($result);
echo $county_total;
echo 'You Have ' . $county_total . ' Counties ';
?
The first thing I see is that you do
$county_total=mysql_num_rows($result)
twice. Now I'm not to sure, but it is possible that looping over
mysql_fetch_array($result) moves an array pointer so the count is not
valid after the operation... that's a long shot.
But since you are looping over the database records anyway, why not just
count them as you go? Isn't $i the number of counties after all the
looping?
--
Peter Ford, Developer phone: 01580 89 fax: 01580
893399
Justcroft International Ltd.
www.justcroft.com
Justcroft House, High Street, Staplehurst, Kent TN12 0AH United
Kingdom
Registered in England and Wales: 2297906
Registered office: Stag Gates House, 63/64 The Avenue, Southampton SO17
1XS
Peter
Thank you for your reply.
I did notice that I had the $county_total=mysql_num_rows($result) twice.
I
had moved and removed it, always getting either a 0 or 1 result.
I put up another test page with
$result = mysql_query(SELECT * FROM `counties` ) or die(mysql_error());
$tot=mysql_num_rows($result);
echo $tot;
?
And it worked fine.
I have tried count() as well as mysql_num_rows() but am getting the same
result.
Could you explain what you mean by count them as I go?
Again, Thank you for your reply.
Gary
__ Information from ESET Smart Security, version of virus
signature database 5895 (20110222) __
The message was checked by ESET Smart Security.
http://www.eset.com
Well,
Lets go back to your original code and cut out the decoration:
$result = mysql_query(SELECT * FROM `counties` WHERE name = 'checked')
$county_total=mysql_num_rows($result);
while($row = mysql_fetch_array($result))
{
echo $row['name'];
}
echo $county_total;
That code should do pretty much what you want...
But, because you only show the number of records AFTER the loop, you could
do:
$result = mysql_query(SELECT * FROM `counties` WHERE name = 'checked')
$county_total = 0;
while($row = mysql_fetch_array($result))
{
echo $row['name'];
$county_total++;
}
echo $county_total;
Peter
Thank you for your suggestion...btw cut out the decoration LOL
I'm sorry to report it did not work. I tried various configurations of it
but to no avail.
Any other suggestions?
Again, thank you for your help.
Gary
__ Information from ESET Smart Security, version of virus signature
database 5895 (20110222) __
The message was checked by ESET Smart Security.
http://www.eset.com
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[PHP] Re: mysql_num_rows
Hi John,
John Taylor-Johnston wrote in message:
I use $mycounter, not saying it is pretty. But if you have a whole bunch
of stuff deleted, your last id might be worth a lot more than the actual
number of rows.
$myconnection = mysql_connect($server,$user,$pass);
mysql_select_db($db,$myconnection);
$news = mysql_query('select * from '.$table.' where '.$where.' order by id
asc');
$mycounter = 0;
while ($mydata = mysql_fetch_object($news))
{
$mycounter++;
}
Have you ever considered just doing a count()?
$count = @mysql_query(select count(*) from [table(s)] where [criteria
[group by something]]);
$total = mysql_result($count, 0);
That will return the number of rows in your table(s). It's also much quicker
and less resource intensive, particularly with large datasets. :)
James
--
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Re: [PHP] Re: mysql_num_rows
First, there is no reason to do mycounter like this
when mysql_num_rows() will work perfectly. It's not
based on any certain column numbers or primary keys.
Although if you're suggesting this method to simply
print 0-n in the loop (as users may want to know the
row number of output not any id number) then that's
good altough it's not really a counter in this case.
Second, we assume this person wants a count for
informational purposes, to go along with the
data in which case mysql_num_rows() is your best
bet. It means having a count before the data is
printed/used. And checking the number or rows
returned before fetching is a form of error handling
as well as if it equals 0 than there is no data to
fetch. But if one ONLY wants a number of rows count,
doing SELECT count(*)... works great as suggested
below.
Regards,
Philip Olson
On Wed, 18 Dec 2002, liljim wrote:
Hi John,
John Taylor-Johnston wrote in message:
I use $mycounter, not saying it is pretty. But if you have a whole bunch
of stuff deleted, your last id might be worth a lot more than the actual
number of rows.
$myconnection = mysql_connect($server,$user,$pass);
mysql_select_db($db,$myconnection);
$news = mysql_query('select * from '.$table.' where '.$where.' order by id
asc');
$mycounter = 0;
while ($mydata = mysql_fetch_object($news))
{
$mycounter++;
}
Have you ever considered just doing a count()?
$count = @mysql_query(select count(*) from [table(s)] where [criteria
[group by something]]);
$total = mysql_result($count, 0);
That will return the number of rows in your table(s). It's also much quicker
and less resource intensive, particularly with large datasets. :)
James
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Re: [PHP] Re: mysql_num_rows
I would also tend to do a count(*) as well as the main query if I intend to
use pagination (with the help of LIMIT in the main query).
Suppose, for example, you're limiting to 30 news posts per page, and from
calculations there are only 2 pages that fall into the criteria, then trying
to bring the result set back on page 3 when it doesn't exist will result
in a warning. Also, you'll want to know how many pages of the news post
there are...
Brief Example (assuming $page has already been validated as an integer):
$num = 30; // per page
$total = 49; // result from count(*)
$total_pages = ceil($total/$num);
if($page $total_pages)
{
$page = $total_pages;
}
// Main query.
// select [whatever] from [table(s)] where [criteria] [whateverelse] limit
. (($page*$num)-$num) . ,$num;
James
Philip Olson [EMAIL PROTECTED] wrote in message
Pine.BSF.4.10.10212181637090.4483-10@localhost">news:Pine.BSF.4.10.10212181637090.4483-10@localhost...
First, there is no reason to do mycounter like this
when mysql_num_rows() will work perfectly. It's not
based on any certain column numbers or primary keys.
Although if you're suggesting this method to simply
print 0-n in the loop (as users may want to know the
row number of output not any id number) then that's
good altough it's not really a counter in this case.
Second, we assume this person wants a count for
informational purposes, to go along with the
data in which case mysql_num_rows() is your best
bet. It means having a count before the data is
printed/used. And checking the number or rows
returned before fetching is a form of error handling
as well as if it equals 0 than there is no data to
fetch. But if one ONLY wants a number of rows count,
doing SELECT count(*)... works great as suggested
below.
Regards,
Philip Olson
On Wed, 18 Dec 2002, liljim wrote:
Hi John,
John Taylor-Johnston wrote in message:
I use $mycounter, not saying it is pretty. But if you have a whole
bunch
of stuff deleted, your last id might be worth a lot more than the actual
number of rows.
$myconnection = mysql_connect($server,$user,$pass);
mysql_select_db($db,$myconnection);
$news = mysql_query('select * from '.$table.' where '.$where.' order
by id
asc');
$mycounter = 0;
while ($mydata = mysql_fetch_object($news))
{
$mycounter++;
}
Have you ever considered just doing a count()?
$count = @mysql_query(select count(*) from [table(s)] where [criteria
[group by something]]);
$total = mysql_result($count, 0);
That will return the number of rows in your table(s). It's also much
quicker
and less resource intensive, particularly with large datasets. :)
James
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Re: [PHP] Re: mysql_num_rows
And now mysql4 allows doing this another way, I *assume*
this new way is slightly faster then the additional
count() although I haven't tested it and wonder if it
affects mysql_num_rows()...
See: SQL_CALC_FOUND_ROWS()
http://www.mysql.com/doc/en/Miscellaneous_functions.html
Regards,
Philip Olson
On Wed, 18 Dec 2002, liljim wrote:
I would also tend to do a count(*) as well as the main query if I intend to
use pagination (with the help of LIMIT in the main query).
Suppose, for example, you're limiting to 30 news posts per page, and from
calculations there are only 2 pages that fall into the criteria, then trying
to bring the result set back on page 3 when it doesn't exist will result
in a warning. Also, you'll want to know how many pages of the news post
there are...
Brief Example (assuming $page has already been validated as an integer):
$num = 30; // per page
$total = 49; // result from count(*)
$total_pages = ceil($total/$num);
if($page $total_pages)
{
$page = $total_pages;
}
// Main query.
// select [whatever] from [table(s)] where [criteria] [whateverelse] limit
. (($page*$num)-$num) . ,$num;
James
Philip Olson [EMAIL PROTECTED] wrote in message
Pine.BSF.4.10.10212181637090.4483-10@localhost">news:Pine.BSF.4.10.10212181637090.4483-10@localhost...
First, there is no reason to do mycounter like this
when mysql_num_rows() will work perfectly. It's not
based on any certain column numbers or primary keys.
Although if you're suggesting this method to simply
print 0-n in the loop (as users may want to know the
row number of output not any id number) then that's
good altough it's not really a counter in this case.
Second, we assume this person wants a count for
informational purposes, to go along with the
data in which case mysql_num_rows() is your best
bet. It means having a count before the data is
printed/used. And checking the number or rows
returned before fetching is a form of error handling
as well as if it equals 0 than there is no data to
fetch. But if one ONLY wants a number of rows count,
doing SELECT count(*)... works great as suggested
below.
Regards,
Philip Olson
On Wed, 18 Dec 2002, liljim wrote:
Hi John,
John Taylor-Johnston wrote in message:
I use $mycounter, not saying it is pretty. But if you have a whole
bunch
of stuff deleted, your last id might be worth a lot more than the actual
number of rows.
$myconnection = mysql_connect($server,$user,$pass);
mysql_select_db($db,$myconnection);
$news = mysql_query('select * from '.$table.' where '.$where.' order
by id
asc');
$mycounter = 0;
while ($mydata = mysql_fetch_object($news))
{
$mycounter++;
}
Have you ever considered just doing a count()?
$count = @mysql_query(select count(*) from [table(s)] where [criteria
[group by something]]);
$total = mysql_result($count, 0);
That will return the number of rows in your table(s). It's also much
quicker
and less resource intensive, particularly with large datasets. :)
James
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[PHP] RE: mysql_num_rows
I found the problem! It was a space before $ in ' $name' in line two of the query. Sorry for the trouble. Roger -Original Message- From: Roger Lewis [mailto:[EMAIL PROTECTED]] Sent: Tuesday, December 17, 2002 4:00 PM To: Php-General Subject: mysql_num_rows Would someone be kind enough to explain why I'm not getting the correct result from the following query. If I select a valid member no. and name, the following query should return 1 row. This is not the case, however. It returns zero rows. $sql = SELECT * FROM users WHERE member_no = '$member_no' and name = ' $name' ; $result = mysql_query($sql) or die (Cannot verify the member); $rows = mysql_num_rows($result); echo rows = $rowsbr; Furthermore when trying the following query where the table users contains 13 rows, the query returns only 12 rows. $sql = SELECT * FROM users; $result = mysql_query($sql) or die (Cannot verify the member); $rows = mysql_num_rows($result); echo rows = $rowsbr; Conclusion: mysql_num_rows seems to be returning one less row than exists. Is this correct? Thanks in advance Roger Lewis -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Re: mysql_num_rows
I use $mycounter, not saying it is pretty. But if you have a whole bunch of stuff
deleted, your last id might be worth a lot more than the actual number of rows.
$myconnection = mysql_connect($server,$user,$pass);
mysql_select_db($db,$myconnection);
$news = mysql_query('select * from '.$table.' where '.$where.' order by id asc');
$mycounter = 0;
while ($mydata = mysql_fetch_object($news))
{
$mycounter++;
}
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[PHP] Re: mysql_num_rows error
Omar,
I would first take the select statement and try running it in mysql to make
sure it actually runs (substituting a valid value for the variable of
course). Then if it has problems you can see what error messages mysql
gives you.
Good Luck,
Jonathan Duncan
Omar Campos [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
I have a file. I'll paste from line 32 to 34
?
.
$sql = select USUARIO from docente where USUARIO = '$usuario';
$result = mysql_query($sql, $link);
if (mysql_num_rows($result) == 1) { // line 34
?
I get the next warning:
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result
resource in /home/olimpiad/public_html/base/alta1.php on line 34
I'm running the script on a linux server.
could someone help me.
Thank you.
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[PHP] Re: mysql_num_rows error
In article [EMAIL PROTECTED],
[EMAIL PROTECTED] says...
I have a file. I'll paste from line 32 to 34
?
.
$sql = select USUARIO from docente where USUARIO = '$usuario';
$result = mysql_query($sql, $link);
Replace the above line with (all on one line)
$result = mysql_query($sql, $link) or die(Error: .
mysql_error().BRQuery: .$sql);
if (mysql_num_rows($result) == 1) { // line 34
?
I get the next warning:
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result
resource in /home/olimpiad/public_html/base/alta1.php on line 34
I'm running the script on a linux server.
could someone help me.
Thank you.
And when you get the error message, you will have the actual query to
check.
Cheers
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David Robley
Temporary Kiwi!
Quod subigo farinam
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[PHP] RE: mysql_num_rows
Thanks to everyone that replied. After trying solutions from everyone and still not getting it to work I finally figured out why my code: $result = mysql_query($query,$connection) or die(Error in Query); $num = mysql_num_rows($result); if ($num == 0) echo something; else echo something else; was not working. I had that piece of code in a while ($row = mysql_fetch_array($result))loop. It seems that if there is nothing to get the condition inside the while loop is false so my code above was never being executed. Even when I tried printing out the contents of $num it would always be blank. The clue came when I deliberately changed the vale of $num right above my code above and it still didn't work. Thanks again to all who replied. It was a good chance to learn about isset(), empty() and other functions. Frank Frank Miller Computer Specialist and Webmaster Technology and Distance Education Texas AM University-Texarkana 2600 North Robison Rd Texarkana, Texas 75501 Phone: 903-223-3156 Fax: 903-223-3139 Office: 165 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Re: Mysql_num_rows
Hi Frank,
I'm not exactly sure what's up with this, I've definitely used something
similar before without any problems. However, maybe you can work around it
by switching the logic...
if($num)
echo blah, blah, blah;
else
echo There are no events scheduled today!;
Wish I had a better suggestion.. Good luck!
Corey Eiseman
Infinite Orange Incorporated
http://infiniteorange.com/
Frank Miller [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Thanks to everyone that helped me with my last problem I've run into
another problem and was wondering if someone here could offer any help or
suggestions. The project I'm working on is an event calender. It does
several things but the problem I've run into is when I click a date on the
calendar if there are no events for that day I want to print a message on
the web page saying as much. I've read the documentaion that said if I'm
using a select statement to use mysql_num_rows and that it returns an
integer. Here is a snippet of my code
$query = SELECT *,TIME_FORMAT(eventtime, '%l:%i %p')AS eventtime,
DATE_FORMAT(dateofevent,'%M %e, %Y') AS fdateofevent FROM tamutevents
where refid=$refid;
$result = mysql_query($query,$connection) or die(Error in Query);
$num = mysql_num_rows($result);
Next I say if ($num == 0)
{
echo There are no events scheduled today!;
}
else
{
echo blah, blah, blah;
}
The problem is if there are no records that match the select then it
always
goes to the else part. I've tried printing the value of $num and it
works
if there is something scheduled but when there is nothing scheduled $num
shows nothing on the screen.
I'm using Mysql 3.23.38 and php 4.06 on a windows test machine but it
works
the same on php 4.06 and Mysql 3.23.46 on my Linux server.
Has anyone else run into this and if so can you tell me what to do about
it.
Thanks in advance - Frank
Frank Miller
Computer Specialist and Webmaster
Technology and Distance Education
Texas AM University-Texarkana
2600 North Robison Rd
Texarkana, Texas 75501
Phone: 903-223-3156
Fax: 903-223-3139
Office: 165
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