[PHP] Maximum execution time exceeded!
hi all,
when i am trying to calculate the number of rows with a '0' in a
particular column using the following code, it is giving the 'Maximum
execution time exceeded' message, eventhough after i gave
set_time_limit(60), it didn't worked, infact now it is even worse the
browser is getting siezed.
$result=mysql_query("select * from mytable");
while ($avgrow=mysql_fetch_array($result))
{
$prescene2 = $avgrow["scene2"];
$prescene4 = $avgrow["scene4"];
$prescene5 = $avgrow["scene5"];
$prescene7 = $avgrow["scene7"];
$prescene8 = $avgrow["scene8"];
$prescene9 = $avgrow["scene9"];
$i = 1;
while ($prescene2 == 0) {
$i=$i+1;
}
}
printf ("Number of zeros=%d",$i);
Can any one please help me with this.
TIA
kalyan
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[PHP] if ... else ...
Hi all,
I am having 5 rows in a mysql database related to five
different users. these rows contains their firstname,
lastname and details of scenes 1 to 5(these details
are numeric values from 1 to 3). if any of these
fields from scenes 1 to 5 are empty that will be
detected from the firstname of the user(collected from
a form with a filed name 'fname') by opening the
database and will be indicated to that user from which
scene he got empty fields. here is the code that i
wrote works like a champ.
$db = mysql_connect ("localhost","myname","password")
or die("bUnable to connect to database./b");
mysql_select_db ("nimitztest",$db) or die("bUnable
to select database./b");
$result = mysql_query ("select * from mytable order by
id",$db);
if ($row=mysql_fetch_array($result))
{
while ($row=mysql_fetch_array($result))
{
$FirstName = $row["firstname"];
if ($fname == $FirstName)
{
for ($i=1; $i=5; $i++)
{
$Num[$i] = $row["num$i"];
if ($Num[$i] == '') {
printf ("You have to goto Scene$ip");
break;
}
}
}
}
}
But when i include the else statement like below, to
indicate an user with all the fields filled up as 'You
don't have any thing to fill', this code is giving
weird results such as printing like this onto the
screen:
You don't have anything to fill
You have to goto Scene2
You have to goto Scene3
You have to goto Scene4
You have to goto Scene5
for a user with scene2 onwards empty fields.
This may be, because of fault in my code or because of
misinterpretation of the mysql functions by me.
-
for ($i=1; $i=5; $i++)
{
$Num[$i] = $row["num$i"];
if ($Num[$i] == '') {
printf ("You have to goto Scene$ip");
}
else {
printf ("You don't have anything to fillp");
}
}
--
Please correct me about the mistake i am making over
above.
TIA
Chakravarthy K Sannedhi
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[PHP] Selecting max(id) of a table
Hello all,
I am having a table named 'mytable' of 5 rows in a
MySQL database. I tried to print '5' which is the
highest row onto the screen. I used the following code
for that:
$result=mysql_query('select max(id) from mytable');
$max=mysql_result($result);
print ('Max id in the table=$max');
But what all i am getting is an error like below:
Warning:Wrong parameter count for mysql_result() in
/home/httpd/info.php on line xx
Max id in the table=$max
i am confused about which function to use instead of
mysql_result().
please help
Chakravarthy K Sannedhi
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[PHP] How to avoid submitting twice?
hello all,
I am having a form which contains a dropdown listbox
and a submit button. The user will choose any number
from 1 to 5 and hits the submit button. After this the
data that he had selected (1,2,3,4 or 5) enters into a
mysql database(this data enters into the row with the
latest userid) and a link appears, so that by click
that link he can go to next page.
I want to avoid the user from submitting his opinion
once again if he does it already.
Here is the code that i have wrote, which is going
logically some where wrong.
FYI : opinion and submit are the names of the dropdown
listbox and submit buttons. scene1 is the name of the
field in the table named testtable into which the data
enters(one of 1 to 5)
mysql_connect ('localhost','myname','password') or die
('bunable to connect to the database./b');
mysql_select_db ('usertest') or die ('bunable to
select the database./b');
if ($submit) {
$result = mysql_query("select * from testtable order
by userid desc limit 1");
$myrow = mysql_fetch_array ($result);
$id = $myrow["userid"];
$query = "update testtable set scene1='$opinion' where
userid=$id";
mysql_query ($query);
echo "a href=next.phpGo to next page/a";
/* this is the part i am trying to use to avoid him
from doing submission second time */
$result=mysql_query("select * from testtable order by
userid desc limit 1");
$myrow = mysql_fetch_array ($result);
$scene1 = $myrow["scene1"];
if ($scene1 == 1||2||3||4||5) {
echo "script language=\"javascript\"";
echo "alert(\"You can choose the rating only
once\");";
echo "/script";
}
}
Thanks
Chakravarthy K Sannedhi
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