i error when try , include variable external php file: unexpected t_var on line 1
here external php code:
<?php var dbstr = 'host::,username::,password::,database'; ?>
here php file including it:
if($_post['admincreate'] == "ok") { include('db.php'); $info = explode("::,", dbstr); $con=mysqli_connect($info[0],$info[1],$info[2],$info[3]); mysqli_query($con,"insert admins (id, user, pass) values ('" . $_post['user'] . "', '" . $_post['pass'] . "',0)"); }
it works if change code to
if($_post['admincreate'] == "ok") { $dbstr = 'host::,username::,password::,database'; $info = explode("::,", $dbstr); $con=mysqli_connect($info[0],$info[1],$info[2],$info[3]); mysqli_query($con,"insert admins (id, user, pass) values ('" . $_post['user'] . "', '" . $_post['pass'] . "',0)"); }
but need included file.
if change included file to
<?php $dbstr = 'host::,username::,password::,database'; ?>
i error: parse error: syntax error, unexpected '=' in .../db.php on line 1
and find external php file has changed to
<?php = 'host::,username::,password::,database'; ?>
how include string without giving me error?
p.s. external php generated main php file use fopen , fwrite, actual values of host, username, password, , database have been censored because feel better way.
thanks in advance, -p0iz0n
var
not valid php
<?php var dbstr = 'host::,username::,password::,database'; ?>
should
<?php $dbstr = 'host::,username::,password::,database'; ?>
interestingly enough have right in second example.
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