Variable in database field (variable in variable)

[istar9900]

I have a variable defined in the php page code, called $id.

I am also displaying a database field from a mysql database, called $text.

The $text database field needs to display and load the $id variable in it as well.

Example:

$id='1111'

$result=mysql_query("SELECT text FROM testtable WHERE id='1');
$text=mysql_result($resultlink1,$i,"text");

echo $text;

What does the $text field need to contain when I insert it into the testtable to display the $id in it as well.

Say the data in $text field is: 'here is the text and the id is $id'

What code needs to go around the $id part to display properly?

I guess it is a variable in a variable?

[BrainDonor]

echo "here is the text and the id is $id";

or

echo "here is the text and the id is " . $id;

for inserting, change echo to a variable name.

$newtext = "here is the text and the id is $id";

[DaiWelsh]

There are two basic ways to go about this - either use eval e.g.

eval('echo("'.$text.'");')

or if like me you think eval is the spawn of satan use some form of regex replacement. If you literally just want variables $xxxxx then something like

$newtext = preg_replace('/\$(\w+)/e','$$1',$text);

If you want parsing of fuller PHP code then eval may be the better (still evil) way to go.

HTH,

Dai

[istar9900]

Thank you both for your help. DailWelsh, your eval method worked perfectly. You Rock!