variable in a string

[krawat]

For a php application I want to read several sql_statements, which are stored in a database table .
May be:
"select a from b where c=$d"

Then I want make a query on the database with this sql_statement.

My probelem is:

The variable $d will not be replaced by the original value of $d, the query doesnt work.

Has anyone a idea ??

[roel_v]

Are you sure you are using " and not ' ? Variables will not be substituted in single quotes. If this still not works, do
$query = "SELECT * FROM table WHERE field=" . $var;
mysql_query($query);
or whatever.

[krawat]

Hi Roel,

it's a little arduous to cut the sql_statement in several pieces, because the statements are a little complex with several variables
may be:

$sql readden from database is
SELECT a.a1,b.b1 from table1 a,table2 b where a.c1=$d and b.e1=$f
and a.c2=b.c2

I search a syntax, where the variables are replaced online in ther own context
after reading the statement.

Roel wrote:

Are you sure you are using " and not ' ? Variables will not be substituted in single quotes. If this still not works, do
$query = "SELECT * FROM table WHERE field=" . $var;
mysql_query($query);
or whatever.

[Anon]

The variable substitution will work if you have your statement within double quotes. eg
$query = "SELECT a.a1,b.b1 from table1 a,table2 b where a.c1=$d"
is ok, $d will be replaced with the value of the variable, but
$query = 'SELECT a.a1,b.b1 from table1 a,table2 b where a.c1=$d'
is not.
Another alternative is to use sprintf(), but this also makes your code look not so nice. Anyway, just using variables within double quotes works, I do it every day.

[krawat]

I know, just using variables within doublw quotes works. How can I replace a $var in a string variable.
In your example
$query = "SELECT a.a1,b.b1 from table1 a,table2 b where a.c1=$d"
$query himself was read from a database table. How can I replace the $d with its original value?

Roel wrote:

The variable substitution will work if you have your statement within double quotes. eg
$query = "SELECT a.a1,b.b1 from table1 a,table2 b where a.c1=$d"
is ok, $d will be replaced with the value of the variable, but
$query = 'SELECT a.a1,b.b1 from table1 a,table2 b where a.c1=$d'
is not.
Another alternative is to use sprintf(), but this also makes your code look not so nice. Anyway, just using variables within double quotes works, I do it every day.

[roel_v]

I'm sorry, I still don't understand the question. You read an sql statement from a table, and now you want to replace variables in this sql statement with other values? You'd have to look at str_replace or ereg_replace for that. However, this doesn't seem good practice to me.

[krawat]

I read following sql-statement from my db-table:

select a,b,c from MyTable where d=$e

$e is set before to 2

I want to make a query like this:

select a,b,c from MyTable where d=2

but the variable $e is not resolved

I get following statement in my sql parser:

select a,b,c from MyTable where d=$e

that's not a proper sql-syntax.

I don't know how to solve this problem.

manualy replacing with str_replace make no sense, because the sql-sattements are complex and have many different variables

Roel wrote:

I'm sorry, I still don't understand the question. You read an sql statement from a table, and now you want to replace variables in this sql statement with other values? You'd have to look at str_replace or ereg_replace for that. However, this doesn't seem good practice to me.

[Anon]

You will have to use the eval() function. Please go to the following page:

http://www.php.net/manual/en/function.eval.php

You'll find good examples to solve your problem 🙂

[cfwebworks]

Very, very simple - escape php. ex.

"select a from b where c=$d"

should be

"select a from b where c=".$d.""