Re: Problem : replacing of variable with it's value in a string(query)

Correct me if I'm wrong, but you essentially have this scenario: A] "SELECT * FROM table WHERE column='$x'" is a string that is stored in a database. B] You use SQL to grab the string into a PHP variable. C] You then want to take that PHP variable and execute it as a SQL statement. And your trouble is when that string that you pull from the database has a PHP variable in it? If I follow you, then I believe the function that you will want to use is the eval() function. It allows a string to be evaluated as PHP. Check out the docs at @ /manual/function.eval.html . <example> Example 1. Eval() example - simple text merge 1 2 <?php 3 $string = 'cup'; 4 $name = 'coffee'; 5 $str = 'This is a $string with my $name in it.<br>'; 6 echo $str; 7 eval ("\$str = \"$str\";"); 8 echo $str; 9 ?> 10 The above example will show: 1 2 This is a $string with my $name in it. 3 This is a cup with my coffee in it. 4 </example> The general logic would be: get_sql_string_from_database(); eval_the_string(); execute_sql_string(); If you have problems doing it, you can send me a private mail and I'll see if I can help you. -- Matthew Leverton - matthew@aeroinc.net - http://www.allegro.cc -----Original Message----- From: Zak Greant <zak@nucleus.com> To: Bobby Singh <bs@corepeople.dk>; php-general@lists.php.net <php-general@lists.php.net> Date: Friday, June 16, 2000 11:05 AM Subject: Re: [PHP-GENERAL] Problem : replacing of variable with it's value in a string(query) >At 11:45 AM 6/16/00 +0200, Bobby Singh wrote: >>I have my sql-queries saved in a database. Every time I want to run a query >>I fetch the desired query from the db. It runs fine. The problem is when the >>query is dynamic. Here is one of my queries. >> >>SELECT * FROM DataTypes WHERE type = '$type' >> >>I would like the variable to be replaced with its value before the query is >>executed. How do I do that it. I have tried the query with both the single, >>double and no quotes. No luck! > >Hi Bob, > >Something like: > >$query = "SELECT * FROM DataTypes WHERE type = '$type'"; >$query_handle = mysql_query ($query); > >If it doesn't, then I usually drop a print statement in front of the $query >line. This lets me see exactly what is generated for the query. >