Hello,
this works fine:
$name=fido;
$string=my dog's name is $name;
echo $string;//prints my dog's name is fido
but when I store the string my dog's name is $name in the db and pull it out:
//do the query
$row=$datab-fetch();
$name=fido;
$string=$name['db_column'];
echo $string//prints my
On 15/05/05, blackwater dev [EMAIL PROTECTED] wrote:
Hello,
this works fine:
$name=fido;
$string=my dog's name is $name;
echo $string;//prints my dog's name is fido
but when I store the string my dog's name is $name in the db and pull it
out:
//do the query
$row=$datab-fetch();
Thanks for the info but I tried it both ways and get this error:
Parse error: parse error, unexpected T_FOR, expecting ',' or ';' in
dogs.php(11) : eval()'d code on line 1
On 5/15/05, Krid [EMAIL PROTECTED] wrote:
Hi!
Try
eval(echo $string);
blackwater dev wrote:
Hello,
this works
Hi!
Try
eval(echo $string);
blackwater dev wrote:
Hello,
this works fine:
$name=fido;
$string=my dog's name is $name;
echo $string;//prints my dog's name is fido
but when I store the string my dog's name is $name in the db and pull it out:
//do the query
$row=$datab-fetch();
$name=fido;
Shouldn't that be:
eval(echo \$string\);
its pretty insecure though, be sure your users are not allowed to change
the db field, because they can do some serious damage.
grt,
Evert
Krid wrote:
Hi!
Try
eval(echo $string);
blackwater dev wrote:
Hello,
this works fine:
$name=fido;
$string=my dog's
blackwater dev wrote:
Hello,
this works fine:
$name=fido;
$string=my dog's name is $name;
echo $string;//prints my dog's name is fido
but when I store the string my dog's name is $name in the db and pull it out:
//do the query
$row=$datab-fetch();
$name=fido;
$string=$name['db_column'];
echo
On Sun, May 15, 2005 8:27 am, blackwater dev said:
Thanks for the info but I tried it both ways and get this error:
Parse error: parse error, unexpected T_FOR, expecting ',' or ';' in
dogs.php(11) : eval()'d code on line 1
On 5/15/05, Krid [EMAIL PROTECTED] wrote:
Hi!
Try
eval(echo
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