[PHP] Creating ZIP-File with password
Dear PHP-Friends,
normaly its my job to find solutions about problems around PHP. Now, i have
a problem, that seems to be unsolved in PHP.
Actually i create some zip-files and this works fine.
$zip = new ZipArchive();
$zip-open(./foo.zip, ZIPARCHIVE::CREATE);
$dir = scandir (doc/);
foreach($dir as $filename)
{
if($filename != . || $filename != ..)
{
$zip-addFile(doc/.$filename, doc/.$filename);
}
}
$zip-close();
Now, the ZIP-files must be protect. There a important files included. Each
site and forum seems to have no solution about this problem. Now, I need a
solution or a workarround. Are the some people, who have some ideas?
Regards
Sebastian
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Re: [PHP] Creating ZIP-File with password
Hmmm... its not possible to use this function, because i can't use the command line. now, i will try to crypt my data bevor i insert them into the zip file. but this needs more performance. i will wait for other comments. I will report my solution. regards Per Jessen [EMAIL PROTECTED] schrieb im Newsbeitrag news:[EMAIL PROTECTED] Sebastian Hopfe wrote: Now, the ZIP-files must be protect. There a important files included. Each site and forum seems to have no solution about this problem. Now, I need a solution or a workarround. Are the some people, who have some ideas? You could just use the zip command line utility and the '-P' option. However, the zip password mechanism does not really provide much in terms of protection - if you really need the encryption, I would look elsewhere. (PGP, X509). /Per Jessen, Zürich -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Re: back-up mysql database using PHP
Dear Vanessa You can use the SELECT ... INTO OUTFILE 'file_name' with mysql_query($vAnf, $dbconn); For the syntax you can have a look at http://dev.mysql.com/doc/refman/5.0/en/select.html Should you be allowed to send system-queries to the Server, than you should have a look to http://dev.mysql.com/doc/refman/5.0/en/backup.html There are many ways to backup the Database. I thought this will help you. Regards Sebastian Vanessa Vega [EMAIL PROTECTED] schrieb im Newsbeitrag news:[EMAIL PROTECTED] hello thereis there a way to create a back-up database through PHP?...i would like to create a file maybe an sql file that would served as back up of my database. Im using mysql database. I know i could use phpmyadmin to do this but i just like to have a function that would do this without going to phpmyadmin.any help? -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Re: enhanced_list_box, 2 tables from a database
Dear kNish,
first of all i have formated your PHP Code it looks better now.
?php
echo tr\n;
echo td height=\33\nbsp;/td\n;
echo td width=\14%\ class=\style3\Artist/td\n;
$options = mysql_query(SELECT artist_name FROM artist);
$options=mysql_fetch_array($options);
echo 'td';
enhanced_list_box(array(
'table' = 'artist',
'value_field' = 'artist_name'));
function enhanced_list_box($options)
{
$sql = select . $options['value_field'];
$sql .= from . $options['table'];
$result = mysql_query($sql)or die(error in SQL);
echo 'select name=', $options['value_field'],' size=1';
while ($row = mysql_fetch_array($result, MYSQL_NUM))
echo 'option value=' . $row[0] . '' . $row[0] . '/option';
echo '/select';
}
echo ' /td';
echo /tr\n;
echo tr\n;
echo td height=\33\nbsp;/td\n;
echo td width=\14%\ class=\style3\Project/td\n;
$options = mysql_query(SELECT project_name FROM project);
$options=mysql_fetch_array($options);
echo 'td';
enhanced_list_box(array(
'table' = 'project',
'value_field' = 'project_name'));
function enhanced_list_box($options)
{
$sql = select . $options['value_field'];
$sql .= from . $options['table'];
$result = mysql_query($sql)or die(error in SQL);
echo 'select name=', $options['value_field'],' size=1';
while ($row = mysql_fetch_array($result, MYSQL_NUM))
echo 'option value=' . $row[0] . '' . $row[0] . '/option';
echo '/select';
}
echo '/td';
?
After i had check your code, my php interpreter shows me an fatal error on
the browser.
kNish [EMAIL PROTECTED] schrieb im Newsbeitrag
news:[EMAIL PROTECTED]
Hi,
A newbie question. I have more than one table to access from a database.
When I use the code as below, it gives no response on the web page.
What may I do to run more than one table from the same database into the
script.
BRgds,
kNish
.
.
.
.
tr
td height=33nbsp;/td
td width=14% class=style3Artist/td
?php
$options = mysql_query(SELECT artist_name FROM artist);
$options=mysql_fetch_array($options);
echo 'td';
enhanced_list_box(array(
'table' = 'artist',
'value_field' = 'artist_name'));
function enhanced_list_box($options){
$sql = select . $options['value_field'];
$sql .= from . $options['table'];
$result = mysql_query($sql)or die(error in SQL);
echo 'select name=', $options['value_field'],' size=1';
while ($row = mysql_fetch_array($result, MYSQL_NUM))
echo 'option value=' . $row[0] . '' . $row[0] . '/option';
echo '/select';
}
echo '/td';
?
/tr
tr
td height=33nbsp;/td
td width=14% class=style3Project/td
?php
$options = mysql_query(SELECT project_name FROM project);
$options=mysql_fetch_array($options);
echo 'td';
enhanced_list_box(array(
'table' = 'project',
'value_field' = 'project_name'));
function enhanced_list_box($options){
$sql = select . $options['value_field'];
$sql .= from . $options['table'];
$result = mysql_query($sql)or die(error in SQL);
echo 'select name=', $options['value_field'],' size=1';
while ($row = mysql_fetch_array($result, MYSQL_NUM))
echo 'option value=' . $row[0] . '' . $row[0] . '/option';
echo '/select';
}
echo '/td';
?
/tr http://www.php.net/a
.
.
.
.
.
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[PHP] Re: Sessionvariable
Dear Ronald, I would like to ask you, want kind of session you use in you application. regards Sebastian Ronald Wiplinger [EMAIL PROTECTED] schrieb im Newsbeitrag news:[EMAIL PROTECTED] I use at the first page a session variable to set a flag. After a few pages, or if somebody wait too long, the pages have another flag. I noticed that there are often more than one session file exist. How can I avoid that? How can I make sure that for one user is only one session file? bye Ronald -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Re: Help with OOPHP
Dear Andrew,
I think normaly it isn't possible to use another class in a class, without
using extends. But you should use your array as a container. After you use
as a container, you can make new instance into a array field.
Now you can use the content of the container to administrate the instances
of the complete class. I just changed some things at your example. Please
have a look and ask if you have any questions.
?php
class fruitBasket ext
{
private $fruits = array(); //this is a class Container
public function addFruit($newFruit)
{
$this-fruits[] = new fruit($newFruit);
}
public function makeAllApples()
{
foreach($this-fruits AS $fruit)
{
$fruit-changeName(apple);
}
}
public function showAllFruits()
{
foreach($this-fruits AS $fruit)
{
echo $fruit-showFruit().br;
}
}
}
class fruit
{
private $name;
public function __construct($name)
{
$this-name = $name;
}
public function changeName($newName)
{
$this-name = $newName;
}
public function showFruit()
{
return $this-name;
}
}
$Cls = new fruitBasket();
$Cls-addFruit(test1);
$Cls-addFruit(test2);
$Cls-addFruit(test3);
$Cls-addFruit(test4);
$Cls-addFruit(test5);
$Cls-addFruit(test6);
$Cls-makeAllApples();
$Cls-showAllFruits();
?
Andrew Peterson [EMAIL PROTECTED] schrieb im Newsbeitrag
news:[EMAIL PROTECTED]
I'm hoping you guys can help me out.
I'm not sure if you can do this, but i'm trying to create a class that is
build of another class. I also want to be able to do functions on the
class1 from within class2.
example:
class fruitBasket{
private $fuit = array(); //this is a class
public function addFruit($newFruit)
{
$this-fruitBasket[] = $newFruit();
}
public makeAllApples()
{
foreach($this-fruit AS $value)
{ $value-changeName(apple);
} }
}
class fruit{
private $name;
public __construct($name)
{
$this-name = $name;
}
public changeName($newName)
{
$this-name = $newName;
}
}
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[PHP] Re: Function variables in classes
It seems to be a PHP Bug, because normaly it should very well.
but you can solve this problem, by a workarround.
using eval($a.();); instead of $a(); in the class.
Best regards
Sebastian
Paul van Haren [EMAIL PROTECTED] schrieb im Newsbeitrag
news:[EMAIL PROTECTED]
Hi there,
I'm trying to execute function variables. This works fine outside class
code, but gives a fatal error when run within a class. The demo code is
here:
?php
function bar1 () {
echo Yep, in bar1() right now\n;
}
function foo1 () {
bar1();
$a = bar1;
print_r ($a); echo \n;
$a();
}
class foobar {
function bar2 () {
echo Yep, in bar2() right now\n;
}
public function foo2 () {
foobar::bar2();
$a = foobar::bar2;
print_r ($a); echo \n;
$a();
}
}
foo1();
$fb = new foobar ();
$fb-foo2();
?
The error message reads:
Fatal error: Call to undefined function
foobar::bar2() in /home/paul/demo/demo.php on line 25
Is there anyone out there who can explain what's wrong in this code?
Thanks, Paul
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[PHP] Re: Function variables in classes
I think you should log it, because it seems to be, and you found this error. Regard Sebastian Paul van Haren [EMAIL PROTECTED] schrieb im Newsbeitrag news:[EMAIL PROTECTED] Thanks, this helps. The code now works. In case this is truely a bug in PHP, where should I log it? Anything that I should do to make sure that it gets followed up? Regards, Paul -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
