[PHP] filling an array(2)
h. Ok. Can somebody explain this one? Why won't it work correctly?
for($m=1;$m=5;$m++){
$div_idd[$m]=${'row-sub' . $m . 'd'};
}
Can it not be done with a 3 parter? The columns in the table that $row
gets, are sub1d, sub2d, sub3d, sub4d and sub5d. Or is it the - that is
messing it up? I have tried escaping them row\-\sub, but that didn't work.
What would I search for on the PHP site or where are directions located
that tells me how to use this type of putting a variable together. It
makes it hard to search for it if I don't know what it is called.
Thanks
Steve
--
This message has been scanned for viruses and
dangerous content by MailScanner, and is
believed to be clean.
ow3
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] filling an array
Thank You Andrey, Matt and Martin for your answers. To summarize for
everyone, here are the answers:
From Andrey (tested, works):
for($m=1;$m=5;$m++){
$div_id[$m]=${'divid'.$m};
}
From Matt:
for($m=1;$m=5;$m++){
$varName = 'divId' . $m;
$div_id[$m]=${$varName}
}
From Martin:
for($i =0; $i 5; $i++) {
$offset = $m + 1;
$divid[$m] = $div_id{$offset};
}
At 10:55 PM 7/26/2002 +0300, you wrote:
- Original Message -
From: Steve Buehler [EMAIL PROTECTED]
To: PHP [EMAIL PROTECTED]
Sent: Friday, July 26, 2002 10:48 PM
Subject: [PHP] filling an array
Can anyone tell me what I am doing wrong?
I am essentially trying to do this:
$divid[1] = $div_id1;
$divid[2] = $div_id2;
$divid[3] = $div_id3;
$divid[4] = $div_id4;
$divid[5] = $div_id5;
But I was looking for a tighter way, like the following (which does not
work):
for($m=1;$m=5;$m++){
$div_id[$m]=$divid$m;
}
Can anybody tell me how to do this in a for statement?
Thanks In Advance
Steve
--
This message has been scanned for viruses and
dangerous content by MailScanner, and is
believed to be clean.
ow3
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
This message has been scanned for viruses and
dangerous content by MailScanner, and is
believed to be clean.
ow3
--
This message has been scanned for viruses and
dangerous content by MailScanner, and is
believed to be clean.
ow3
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] filling an array
for($m=1;$m=5;$m++){
$div_id[$m]=${'divid'.$m};
}
Regards,
Andrey
- Original Message -
From: Steve Buehler [EMAIL PROTECTED]
To: PHP [EMAIL PROTECTED]
Sent: Friday, July 26, 2002 10:48 PM
Subject: [PHP] filling an array
Can anyone tell me what I am doing wrong?
I am essentially trying to do this:
$divid[1] = $div_id1;
$divid[2] = $div_id2;
$divid[3] = $div_id3;
$divid[4] = $div_id4;
$divid[5] = $div_id5;
But I was looking for a tighter way, like the following (which does not
work):
for($m=1;$m=5;$m++){
$div_id[$m]=$divid$m;
}
Can anybody tell me how to do this in a for statement?
Thanks In Advance
Steve
--
This message has been scanned for viruses and
dangerous content by MailScanner, and is
believed to be clean.
ow3
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] filling an array(2)
Ok. That makes since.
Thanks
Steve
At 04:20 PM 7/26/2002 -0500, you wrote:
var names can only be letters, numbers, and underscores.
Jim Grill
Support
Web-1 Hosting
http://www.web-1hosting.net
- Original Message -
From: Steve Buehler [EMAIL PROTECTED]
To: PHP [EMAIL PROTECTED]
Sent: Friday, July 26, 2002 3:53 PM
Subject: [PHP] filling an array(2)
h. Ok. Can somebody explain this one? Why won't it work correctly?
for($m=1;$m=5;$m++){
$div_idd[$m]=${'row-sub' . $m . 'd'};
}
Can it not be done with a 3 parter? The columns in the table that $row
gets, are sub1d, sub2d, sub3d, sub4d and sub5d. Or is it the - that is
messing it up? I have tried escaping them row\-\sub, but that didn't
work.
What would I search for on the PHP site or where are directions located
that tells me how to use this type of putting a variable together. It
makes it hard to search for it if I don't know what it is called.
Thanks
Steve
--
This message has been scanned for viruses and
dangerous content by MailScanner, and is
believed to be clean.
ow3
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
This message has been scanned for viruses and
dangerous content by MailScanner, and is
believed to be clean.
ow3
--
This message has been scanned for viruses and
dangerous content by MailScanner, and is
believed to be clean.
ow3
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] filling an array
Can anyone tell me what I am doing wrong?
I am essentially trying to do this:
$divid[1] = $div_id1;
$divid[2] = $div_id2;
$divid[3] = $div_id3;
$divid[4] = $div_id4;
$divid[5] = $div_id5;
But I was looking for a tighter way, like the following (which does not work):
for($m=1;$m=5;$m++){
$div_id[$m]=$divid$m;
}
Can anybody tell me how to do this in a for statement?
Thanks In Advance
Steve
--
This message has been scanned for viruses and
dangerous content by MailScanner, and is
believed to be clean.
ow3
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] filling an array(2)
for($m=1;$m=5;$m++){
$div_idd[$m]=${'row-sub' . $m . 'd'};
}
I'm not sure if it will work, but you might try either using the
mysql_fetch_array($result); function and then refer to them by
$row[src1d]; or try $div_idd[$m] = $row-$$name; where $name = sub . $m
. d; I'm not sure if that will work, but it might be worth a try.
--
Steve Buehler [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Ok. That makes since.
Thanks
Steve
At 04:20 PM 7/26/2002 -0500, you wrote:
var names can only be letters, numbers, and underscores.
Jim Grill
Support
Web-1 Hosting
http://www.web-1hosting.net
- Original Message -
From: Steve Buehler [EMAIL PROTECTED]
To: PHP [EMAIL PROTECTED]
Sent: Friday, July 26, 2002 3:53 PM
Subject: [PHP] filling an array(2)
h. Ok. Can somebody explain this one? Why won't it work
correctly?
for($m=1;$m=5;$m++){
$div_idd[$m]=${'row-sub' . $m . 'd'};
}
Can it not be done with a 3 parter? The columns in the table that
$row
gets, are sub1d, sub2d, sub3d, sub4d and sub5d. Or is it the -
that is
messing it up? I have tried escaping them row\-\sub, but that
didn't
work.
What would I search for on the PHP site or where are directions
located
that tells me how to use this type of putting a variable together. It
makes it hard to search for it if I don't know what it is called.
Thanks
Steve
--
This message has been scanned for viruses and
dangerous content by MailScanner, and is
believed to be clean.
ow3
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
This message has been scanned for viruses and
dangerous content by MailScanner, and is
believed to be clean.
ow3
--
This message has been scanned for viruses and
dangerous content by MailScanner, and is
believed to be clean.
ow3
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
