Need help with putting a lot of code into one variable

[Dimoop]

See I'm trying to make a variable with a lot of code in it.

<?php
include("config.inc.php");
$content = $ordertpl = file_get_contents("order1.tpl.php");
$content = . $ordertpl_sql = "SELECT * FROM template_vars WHERE template="order1.tpl.php";
$content = . $ordertpl_qry = mysql_query($ordertpl_sql);
$content = . while($ordertpl_res = mysql_fetch_array($ordertpl_qry);
$content = . {;
$content = . $ordertpl = str_replace($ordertpl_res['tvar'], $ordertpl_res['vcontent'], $ordertpl);
$content = . };
$maintpl = str_replace("{content}", $content, $maintpl);
print $maintpl;
?>

I'm not great with PHP, in fact I'm very new to it. Can anyone help me get all of that code into one variable? I just keep getting parse errors :o

MrInii

[dalecosp]

You want all this in a string variable? Then you need quotes around just about everything, I'd expect...

FWIW, you are setting the variable $content equal to 1/true on the first assignment line IF the call to "file_get_contents()" works...

My real question is, why are you doing this? If you succeed in getting all this in a string, then you're just going to print PHP code, not run it.....

[marque]

Perhaps this might help.

<?php 
include("config.inc.php"); 

$content = 
'$ordertpl = file_get_contents("order1.tpl.php"); 
$ordertpl_sql = "SELECT * FROM template_vars WHERE template="order1.tpl.php"; 
$ordertpl_qry = mysql_query($ordertpl_sql); 
while($ordertpl_res = mysql_fetch_array($ordertpl_qry); 
{; 
$ordertpl = str_replace($ordertpl_res[tvar], $ordertpl_res[vcontent], $ordertpl); 
};'
$maintpl = str_replace("{content}", $content, $maintpl); 
print $maintpl; 
?> 

My real question is, why are you doing this? If you succeed in getting all this in a string, then you're just going to print PHP code, not run it.....

I gotta know this aswell, but if you want to run the code you may like to check out the eval function
http://www.php.net/manual/en/function.eval.php

[Dimoop]

Thanks guys

With my first post just doing

print $content;

should have worked. Not with this one. So I tried eval. I have tried two different things and both give the most .... odd? parse error you've ever seen!

<?php
include("config.inc.php");
$content =
"$ordertpl = file_get_contents(\"order1.tpl.php\");
$ordertpl_sql = \"SELECT * FROM template_vars WHERE template='order1.tpl.php'\";
$ordertpl_qry = mysql_query($ordertpl_sql);
while($ordertpl_res = mysql_fetch_array($ordertpl_qry);
{
$ordertpl = str_replace($ordertpl_res['tvar'], $ordertpl_res['vcontent'], $ordertpl);
}";
$runcontent = eval($content);
$maintpl = str_replace("{content}", $runcontent, $maintpl);
print $maintpl;
?>

gives

Parse error: parse error, unexpected '=' in /home/mhbillin/wwwroot/mhbilling.co.uk/mhbilling/order.php(11) : eval()'d code on line 1

and

<?php
include("config.inc.php");
$content =
'$ordertpl = file_get_contents("order1.tpl.php");
$ordertpl_sql = "SELECT * FROM template_vars WHERE template=\'order1.tpl.php\'";
$ordertpl_qry = mysql_query($ordertpl_sql);
while($ordertpl_res = mysql_fetch_array($ordertpl_qry);
{
$ordertpl = str_replace($ordertpl_res[\'tvar\'], $ordertpl_res[\'vcontent\'], $ordertpl);
}';
$runcontent = eval($content);
$maintpl = str_replace("{content}", $runcontent, $maintpl);
print $maintpl;
?>

gives

Parse error: parse error, unexpected ';' in /home/mhbillin/wwwroot/mhbilling.co.uk/mhbilling/order.php(11) : eval()'d code on line 4

Anyone can help me with this last bit then I might tell you what it's for 🙂 but whether I do or don't you'll find out on the 1st Jan 2004...

Miniiiiiiiiiiiiii

[tsinka]

Hi,

try if this works:

<?php
include("config.inc.php");
$content = <<< EOPHP
\\$ordertpl = file_get_contents("order1.tpl.php");
\\$ordertpl_sql = "SELECT * FROM template_vars WHERE template='order1.tpl.php'";
\\$ordertpl_qry = mysql_query(\\$ordertpl_sql);
while(\\$ordertpl_res = mysql_fetch_array(\\$ordertpl_qry))
{
    \\$ordertpl =  str_replace(\\$ordertpl_res['tvar'],\\$ordertpl_res['vcontent'], \\$ordertpl);
}
EOPHP;
$runcontent = eval($content);
$maintpl = str_replace("{content}", $runcontent, $maintpl);
print $maintpl;
?>

You have to escape the $ characters in the string if you want to use the contents of $content as php code in a call to eval.

Thomas

[Dimoop]

tsinka, great, thanks all!