Re: [PHP-DB] want to execute comments on new pages after clicking submit.
hallo,
i wrote this
echo Submit = $submit; before 'if statement'
but , the sentences in echo statements are executing on same pages , showing
original page with these echo statement.
i want to execute , some comments and tables on same page(without original
contents) or on new pages , after clicking on submit button
what to do for it.
thank you.
Mike S. [EMAIL PROTECTED] wrote:
hallo ,
i have wriiten simple php script , below.
created databse dollar1_allinfo with table 'totalinfo'
but on clicking submit button ,
information entered is not entering in database.
also, echo statements are also not displaying after submit
if ($submit)
{
$db = mysql_connect(localhost,root);
mysql_select_db(dollar1_allinfo,$db);
mysql_query(INSERT INTO totalinfo (username,password,) VALUES
('$loginusername','$loginpassword'));
echo 'thank you';
echo 'So Hurry Up';
}
?
thank you.
Because your query and echoes are within an if block, I'd say that
$submit is evaluating to FALSE, and thus not executing the query and
echoes.
Add (before the if statement):
echo Submit = $submit;
Make sure that $submit is evaluating to TRUE.
To All Beginning Programmers (Newbies), or just people who want to
easily debug their code:
Try adding debugging statements to help you find problems. A few echoes
or prints in selected locations will help identify the issue. Make sure
your variables are evaluating to values that you think they should be.
If you want to keep the debugging statements in your code and just
activate them when you need to see them, do something like:
// near the top of your code
$debug=TRUE;
// and anywehere in your code
if ($debug) {
echo DEBUG: Submit = $submit
\n;
}
// or whatever echo might help you nail down the problem. Put as many of
these in your code and you can turm them ON and OFF very easily, by
changing one line of code.
OR for even quicker debugging, replace $debug=TRUE; with
$debug=$_GET['debug'];
and then you can turn it on in by adding ?debug=TRUE at the end of the
URL. Example: http://www.php.net/path/my.html?debug=TRUE
or: http://www.php.net/path/my.html?debug=1
**DISCLAIMER** Using the URL method is NOT a secure way to do this!!!
And I'm sure there's plenty of folks groaning out there because it is a
BIG security hole. You really would want to do this on a development
machine and network, and remove the debug code for production. That
really should go without saying.
Good luck.
:Mike S.
:Austin TX USA
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Re: [PHP-DB] want to execute comments on new pages after clicking submit.
The purpose of putting the echo Submit = $submit; is to find out what
is happening with your code. Once everything is working correctly they
will be removed. What you need to find out is what is being held in the
variable submit when you move from one page to the next, a point when
you are expecting to enter the if($submit) clause.
I hope that clarifies it a bit.
all the best,
graeme.
amol patil wrote:
hallo,
i wrote this
echo Submit = $submit; before 'if statement'
but , the sentences in echo statements are executing on same pages , showing
original page with these echo statement.
i want to execute , some comments and tables on same page(without original
contents) or on new pages , after clicking on submit button
what to do for it.
thank you.
Mike S. [EMAIL PROTECTED] wrote:
hallo ,
i have wriiten simple php script , below.
created databse dollar1_allinfo with table 'totalinfo'
but on clicking submit button ,
information entered is not entering in database.
also, echo statements are also not displaying after submit
if ($submit)
{
$db = mysql_connect(localhost,root);
mysql_select_db(dollar1_allinfo,$db);
mysql_query(INSERT INTO totalinfo (username,password,) VALUES
('$loginusername','$loginpassword'));
echo 'thank you';
echo 'So Hurry Up';
}
?
thank you.
Because your query and echoes are within an if block, I'd say that
$submit is evaluating to FALSE, and thus not executing the query and
echoes.
Add (before the if statement):
echo Submit = $submit;
Make sure that $submit is evaluating to TRUE.
To All Beginning Programmers (Newbies), or just people who want to
easily debug their code:
Try adding debugging statements to help you find problems. A few echoes
or prints in selected locations will help identify the issue. Make sure
your variables are evaluating to values that you think they should be.
If you want to keep the debugging statements in your code and just
activate them when you need to see them, do something like:
// near the top of your code
$debug=TRUE;
// and anywehere in your code
if ($debug) {
echo DEBUG: Submit = $submit
\n;
}
// or whatever echo might help you nail down the problem. Put as many of
these in your code and you can turm them ON and OFF very easily, by
changing one line of code.
OR for even quicker debugging, replace $debug=TRUE; with
$debug=$_GET['debug'];
and then you can turn it on in by adding ?debug=TRUE at the end of the
URL. Example: http://www.php.net/path/my.html?debug=TRUE
or: http://www.php.net/path/my.html?debug=1
**DISCLAIMER** Using the URL method is NOT a secure way to do this!!!
And I'm sure there's plenty of folks groaning out there because it is a
BIG security hole. You really would want to do this on a development
machine and network, and remove the debug code for production. That
really should go without saying.
Good luck.
:Mike S.
:Austin TX USA
-
Do you Yahoo!?
Yahoo! Mail - Helps protect you from nasty viruses.
[PHP-DB] want to execute comments on new pages after clicking submit.
hallo,
thanks for reply, but
code is like below , i want comment in echo statemets to executed on
same page(without pages original contenti.e only echo comments )or on new pages
how to link new pages in php, after clicking submit
also.
[these first two echo statements are giving parse errors errors,unexpected '/']
?php
echo --$submit--;echo --$_POST[submit]--;
if ($submit)
{$db = mysql_connect(localhost,root);mysql_select_db(dollar1_allinfo,$db);
mysql_query(INSERT INTO totalinfo (username,password,)
VALUES('$loginusername','$loginpassword'));echo 'thank you';echo 'So Hurry
Up';}?
thank you.
[EMAIL PROTECTED] wrote:
The purpose of putting the echo Submit = $submit; is to find out what is
happening with your code. Once everything is working correctly they will be
removed. What you need to find out is what is being held in the variable submit
when you move from one page to the next, a point when you are expecting to
enter the if($submit) clause.
I hope that clarifies it a bit.
all the best,
graeme.
amol patil wrote:
hallo, i wrote this echo Submit = $submit; before 'if statement' but ,
the sentences in echo statements are executing on same pages , showing
original page with these echo statement. i want to execute , some comments and
tables on same page(without original contents) or on new pages , after clicking
on submit buttonwhat to do for it. thank you.Mike S. [EMAIL PROTECTED]
wrote:
hallo ,i have wriiten simple php script , below.created databse
dollar1_allinfo with table 'totalinfo'but on clicking submit button
,information entered is not entering in database.also, echo statements are also
not displaying after submit
if ($submit)
{$db =
mysql_connect(localhost,root);mysql_select_db(dollar1_allinfo,$db);mysql_query(INSERT
INTO totalinfo (username,password,)
VALUES('$loginusername','$loginpassword'));echo 'thank you';echo 'So Hurry
Up';}?thank you.
Because your query and echoes are within an if block, I'd say that$submit is
evaluating to FALSE, and thus not executing the query andechoes.Add (before the
if statement):echo Submit = $submit;Make sure that $submit is evaluating to
TRUE.To All Beginning Programmers (Newbies), or just people who want toeasily
debug their code:Try adding debugging statements to help you find problems. A
few echoesor prints in selected locations will help identify the issue. Make
sureyour variables are evaluating to values that you think they should be.If
you want to keep the debugging statements in your code and justactivate them
when you need to see them, do something like:// near the top of your
code$debug=TRUE;// and anywehere in your codeif ($debug) {echo DEBUG: Submit =
$submit\n;}// or whatever echo might help you nail down the problem. Put as
many ofthese in your code and you can turm them ON and OFF very easily,
bychanging one line of code.OR for even quicker debugging, replace
$debug=TRUE; with$debug=$_GET['debug'];and then you can turn it on in by
adding ?debug=TRUE at the end of theURL. Example:
http://www.php.net/path/my.html?debug=TRUEor:
http://www.php.net/path/my.html?debug=1**DISCLAIMER** Using the URL method is
NOT a secure way to do this!!! And I'm sure there's plenty of folks groaning
out there because it is aBIG security hole. You really would want to do this on
a developmentmachine and network, and remove the debug code for production.
Thatreally should go without saying.Good luck.:Mike S.:Austin TX USA
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