[fahuber3]

Currently I have a website that contains one php page and based on what the user selects from the pull down menu, that is what gets populated on the page from a MySQL database. The code works perfectly unless I need to

PHP Code:

<?php                            
        $value = $_GET['page'];                        
            
        $connection = mysql_connect(localhost, "ID", "PW");
        $db = mysql_select_db("DB") or die( "Couldn't select database");
        $sql = "SELECT * FROM site WHERE page_id = '$value'";
        $mysql_result=mysql_query($sql,$connection);
        $num_rows=mysql_num_rows($mysql_result);
        
                    if ($num_rows == 0){
                            echo " ";
                    } 
            
                    else {    
                        while ($row=mysql_fetch_array($mysql_result)) {
                                $pagename=$row["pagename"];
                                $content=$row["content"];

                        echo "<h1>$pagename</h1>
                                $content";
                        }
                    }
                    
                mysql_close($connection);
                ?>

I want to be able to store another php script in the $content from my database so that it is called when the user selects the page. No matter what I have tried, the php does not work and instead it prints the php code.

Is there anyone that can point me in the right direction please?

[wirehopper]

If there is PHP in the database, you'll need PHP: eval - Manual to execute it.

[fahuber3]

I tried storing just this in the content field in MySQL but when it is called, it just prints it to the screen.

mixed eval ( string $select id, teamname from team where a_flag = 'A' order by teamname )

[Jcbones]

Is your php page named with a .php ext. ?

[wirehopper]

Try something like this:

PHP Code:

<?php
echo 'Test from database';
?>

Store it in the database, then use:

PHP Code:

echo eval($content); 


to display it.

[fahuber3]

I return eval( )
using the following code in my php file

PHP Code:

echo "<h1>$pagename</h1>
        eval($content)"; 


and storing

PHP Code:

<?php
echo 'Test from database';
?>

in the content field

[wirehopper]

PHP Code:

echo "<h1>$pagename</h1>";
echo eval($content); 


[fahuber3]

I now receive the following error message on my page:


Parse error: syntax error, unexpected '<' in /home/uoszpb/public_html/nav.php(69) : eval()'d code on line 1

[job0107]

The PHP code in the database must not have opening and closing PHP tags,
just the raw PHP code.
And any HTML must be echoed.

Example:
The code in the database could look like this,
notice that there are no opening or closing PHP tags.

PHP Code:

echo "<p>Hello World.</p>"; 


Then your code will work.

PHP Code:

<?php
        $value = $_GET['page'];
        $host = "localhost";
        $user = "ID";
        $password = "PW";
        $db_name = "DB";
        $connection = mysql_connect($host, $user, $password);
        $db = mysql_select_db($db_name) or die( "Couldn't select database");
        $sql = "SELECT * FROM site WHERE page_id = '$value'";
        $mysql_result=mysql_query($sql,$connection);
        $num_rows=mysql_num_rows($mysql_result);

                    if ($num_rows == 0){
                            echo " ";
                    }

                    else {
                        while ($row=mysql_fetch_array($mysql_result)) {
                                $pagename=$row["pagename"];
                                $content=$row["content"];

                        echo "<h1>$pagename</h1>";
                        eval($content);
                        }
                    }

                mysql_close($connection);
                ?>