[PHP] Variables in Variables?
Greetings. Does anyone know how to do this? I have, $var $var2 In a field called two_vars in a MySQL db. I am calling the variables inside PHP document. In that document I am saying: $var = time $var2 = clock !-- I do the query in MySQL here -- echo $two_vars; But the what prints out is $var $var2 not time and clock. I know that is what is in the database, but I want it to replace the variables when printed in the PHP file. Does this make sense to anyone? Does anyone know how to do this? -- Steve -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Variables in variables?
This is a little more specific:
I have a template in the database which looks something like this:
html
title$page_title/title
body
Image: $image p /
img src = \$image\
Main Text: $mtext br /
Text2: $text2 br /
Text3: $text3 br /
/body
html
The variable name is $somecontent
When I call $somecontent in the PHP file, it displays exactly like it is
above. It does not replace the variables.
Is there a reason for this?
Here is my PHP file:
?php
include cnx.php;
include includes/header.php;
//Include the content
$select_data = SELECT * FROM template WHERE tmpl_name = \my_template\;;
$response = mysql_query( $select_data, $cnx );
//now print it out for the user.
if ( $one_line_of_data = mysql_fetch_array( $response ) ) {
extract ( $one_line_of_data );
}
$file = $_POST['file'];
$page_title = $_POST['page_title'];
$image = $_POST['image'];
$mtext = $_POST['mtext'];
$text2 = $_POST['text2'];
$text3 = $_POST['text3'];
if (file_exists($dir . $file)) {
echo centerp /p class=\NormalBold\Success!p /p /;
} else {
echo The file $file does not exist;
}
$filename = $file;
//Update the table in MySQL
$update_data = UPDATE cms_pages SET page_title = '$page_title', image =
'$image', mtext = '$mtext', text2 = '$text2', text3 = '$text2' WHERE
filename = '$file';
$response = mysql_query( $update_data, $cnx );
if(mysql_error()) die ('database errorbr'. mysql_error());
//Begining of Template Construction
$mtext = nl2br($mtext);
$text2 = nl2br($text2);
$text3 = nl2br($text3);
//include content.php; This works with an external page included, but I
would like it to be in the database for easy changes.
// Is the file writable?
if (is_writable($dir. $filename)) {
if (!$handle = fopen($dir . $filename, 'w')) {
echo Cannot open file ($filename);
exit;
}
// Write $somecontent to our opened file.
if (fwrite($handle, $somecontent) === FALSE) {
echo Cannot write to file ($filename);
exit;
}
echo centerp /p class=\NormalText\$filename has been
updated./pp //center;
fclose($handle);
} else {
echo The file $filename is not writable;
}
include includes/footer.php;
?
/html
on 11/18/05 7:14 PM Lists ([EMAIL PROTECTED]) wrote:
?
$var = time;
$var2 = clock;
echo $var$var2;
//outputs timeclock
$two_vars = $var$var2;
echo $two_vars;
//outputs timeclock
?
Why put two variables in one field? But if you want to store them,
do: $var, $var2. After pulling this back from the database,
explode into an array at the comma, using list to call them var and
var2.
On Nov 18, 2005, at 7:54 PM, Marquez Design wrote:
Greetings.
Does anyone know how to do this?
I have,
$var
$var2
In a field called two_vars in a MySQL db.
I am calling the variables inside PHP document.
In that document I am saying:
$var = time
$var2 = clock
!-- I do the query in MySQL here --
echo $two_vars;
But the what prints out is
$var
$var2
not time and clock. I know that is what is in the database,
but I want
it to replace the variables when printed in the PHP file.
Does this make sense to anyone? Does anyone know how to do this?
--
Steve
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
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[PHP] variables within variables - question
Im having a VERY hard time trying to explain this
Here is a bit of my code...
for ($i = 0; $i $num_fields; $i++)
{
$field = mysql_field_name($result, $i);
$field2 = .$field;
$output = str_replace($field2, $field), $output);
}
The problem I'm having here is that the variable $field contains a
string (a name of one of my fields in my table). I need for the
variable $field to actually point to the another variable, not just
contain the string.
This little loop here is supposed to go through a RTF file and replace
all the $field2 it finds with $field.
Am I making sense?
THANKS!!!
RE: [PHP] variables within variables - question
$field2 = .$$field;
-Original Message-
From: Phil Schwarzmann [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, March 27, 2002 12:33 PM
To: [EMAIL PROTECTED]
Subject: [PHP] variables within variables - question
Im having a VERY hard time trying to explain this
Here is a bit of my code...
for ($i = 0; $i $num_fields; $i++)
{
$field = mysql_field_name($result, $i);
$field2 = .$field;
$output = str_replace($field2, $field), $output);
}
The problem I'm having here is that the variable $field contains a
string (a name of one of my fields in my table). I need for the
variable $field to actually point to the another variable, not just
contain the string.
This little loop here is supposed to go through a RTF file and replace
all the $field2 it finds with $field.
Am I making sense?
THANKS!!!
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] variables within variables
On Sat, Apr 28, 2001 at 07:12:15PM +0100, Robert Morrissey wrote: I have a form on one page with a text box called 'emailtext'. This gets passed to a php script that gets info from a database (such as $name, $email, etc) and mails selected email addresses; my problem is this: Say, the form passes $emailtext as Hello $name, how're you?, and this is passed to mail() in the php form, rather than emailing Hello Bob, how're you? it prints $name literally. Seeing as $name isn't pulled from the database till the php script is it possible to get it to print the value? or do I have to pull $name from the database in the original form? (which'll is going to be a lot more complicated). Does this make sense? No, not really. What is the value of $name when the form passes $emailtext? That is the value that will be in $name. If it is actually printing $name and not an empty string, then the '$' must be getting escaped somewhere along the line, or you are using ' instead of around the variable. http://www.php.net/manual/en/function.echo.php -- Jason Stechschulte [EMAIL PROTECTED] -- I'll say it again for the logic impaired. -- Larry Wall -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP] variables within variables
Hi, I have a form on one page with a text box called 'emailtext'. This gets passed to a php script that gets info from a database (such as $name, $email, etc) and mails selected email addresses; my problem is this: Say, the form passes $emailtext as Hello $name, how're you?, and this is passed to mail() in the php form, rather than emailing Hello Bob, how're you? it prints $name literally. Seeing as $name isn't pulled from the database till the php script is it possible to get it to print the value? or do I have to pull $name from the database in the original form? (which'll is going to be a lot more complicated). Does this make sense? Thanks for any help, Robert Morrissey -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
