[SOLVED] why wont this variable declaration work?

[scarhand]

<?php
  while ($row = mysql_fetch_array($sql))
  {
    $orig_{$row["name"]} = $row["value"];
  }
  
  echo "orig_name = $orig_name";

 

any help is appreciated

[kenrbnsn]

It should be working. You're just echoing the wrong variable.

 

try:

<?php
  while ($row = mysql_fetch_array($sql))
  {
    $orig_{$row["name"]} = $row["value"];
  }
  
  echo 'orig_name = ' . $orig_{$row['name'];
?>

 

Ken

[scarhand]

well lets say that $row['name'] is = 'username'

 

im trying to get the variable to become $orig_username

[revraz]

Which variable?  Instead of echoing it, just turn it into a variable.

 

$variable = "orig_name = " . $orig_{$row['name'];

 

[teng84]

$x['test'] ='teng';
${'orig_'.$x['test']} = 'i love php';
echo $orig_teng;// echo i love php

 

that works... try 

[teng84]

 

 

IMO the reason why the above code wont work is that we cant concatenate the existing name of var

eg.. $x.'tra';  to become $xtra

 

and that should be ${'xtra'};  or ${$var.'value'}

 

[scarhand]

thanks teng