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up vote -1 down vote favorite

I am trying to echo a variable, from a mysql query, like this:

<?php
...//FYI: mysql connection already established
//Table: title 
//col: page | title
//row: html | "$domain_name: Welcome"

$page_id = basename(getcwd());
$domain_name = "Name of My Domain";

$sql = "SELECT title FROM mydatabase.title WHERE page = '$page_id' ";
//The query's result is 1 row
$dbq = mysql_query ($sql);
$dba = mysql_fetch_array( $dbq );
echo $dba["title"];
//it outputs: $domain_name: Welcome", instead of "Name of My Domain: Welcome"
?>

What am I doing wrong? I am trying to replace the "variable [$domain_name] in the table's content for it's php value. -I thought " (double quotes) are supposed to replace the variable with it's value.

PS. I am a beginner

EDIT 2/7/2012, 3:14pm: Forgot to mention. The query works OK. $dba['title'] has "$domain_name: Welcome" as a value. The problem is, it is not replacing $domain_name

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up vote 2 down vote accepted

I thought " (double quotes) are supposed to replace the variable with it's value.

That works only in case when you specify the string in your code. If the string comes from outside - it doesn't have such magic behaviour.

So the only solution you could go with is:

echo str_replace('$domain_name', $domain_name, $dba["title"]);

Or you could go with some sort of template engine like Twig or Smarty and treat your database value as a template, and your variables as the data.

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up vote 1 down vote

You can replace it on database level:

$sql = 'SELECT REPLACE(title, \'$domain_name\',\''.$domain_name.'\') as title FROM mydatabase.title WHERE page = '.intval($page_id);
//The query's result is 1 row
$dbq = mysql_query ($sql);
$dba = mysql_fetch_array( $dbq );
echo $dba["title"];

(Note the single quotes)

share|improve this answer
    
Your eval code won't work – zerkms Feb 7 '12 at 23:19
    
@zerkms: I see, as it's not complete code.. – konsolenfreddy Feb 7 '12 at 23:20

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