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Posted by Marquez Design on 10/08/24 11:32
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 "<center><p /><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 error<br>'. 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 "<center><p /><p class=\"NormalText\">$filename has been
updated.</p><p /></center>";
fclose($handle);
} else {
echo "The file $filename is not writable";
}
include "includes/footer.php";
?>
</html>
on 11/18/05 7:14 PM Lists (lists@rexruff.com) 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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