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I'm printer a url with php in it from a database field and all that's printing is the link and the variable code. The variable is not changing to the actual #.

 

For example this is what I'm using.

 

This in in the DB.

 

http://link.com/home.php?x=$variable

 

I use this print "$DBField"; and this prints.

 

http://link.com/home.php?x=$variable

 

The variable in not changing over to it's value when it = 1 for instance.

 

Any suggestions?

I know how variable work. What I'm asking is if I have a database field that hold a link like I posted above with a variable in it.

 

Then I print the link out in a page right now it just print the variable like $variable.

 

It's not converting the variable to it's actual value.

 

How do you get a variable to change it it's value when the variable in within a variable.

This is the field in the DB (It's a link) : http://site.com/home.php?id=$x

 

Now we're on the php page.

 

at the top of the page

 

$x = 23;

 

We pull the field into $field[dblink].

 

So $field[dblink] is the link I have above.

 

in the php page I print:

 

print "<a href=$field[dblink]>Click Here</a>";

 

instead of <a href=http://site.com/home.php?id=23>Click Here</a>

 

I get <a href=http://site.com/home.php?id=$x>Click Here</a>

 

So the $x is not changing over to 23 .....what am I doing wrong?

can you post the results of this please so i can see what your working with

 

<?php
echo "<br />";
var_dump($field);
echo "<br />";
print_r($field);
?>

 

 

as a note i tried this and it worked on my site

 

basically i createds a array (same as i would get a return from a database) and printed it

 

<?php

$field = array('dblink' => 10);
print "<a href=text2.php?x=$field[dblink]>Click Here</a>";
?>

 

string(74) "http://www.toprpgames.com/vote.php?idno=437&field1=$var[id]&field2=$var2"

http://www.toprpgames.com/vote.php?idno=437&field1=$var[id]&field2=$var2

 

when i try doing a $db['field'] i get the error about encapsulated white space

 

[Wed Feb 28 21:54:18 2007] [error] PHP Parse error:  syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING

That is different. What the OP has is a string that has the value of

'http://site.com/home.php?id=$x'

At some later point after setting the string's value, the OP sets the variable $x to 23.

 

What is needed is to use the eval() function:

<?php
$str = 'http://site.com/home.php?id=$x';
$x = 23;
echo $str;
eval("\$str2 = \"$str\";");
echo '<br>'.$str2;
?>

 

Ken

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