php-general Digest 15 Oct 2011 14:01:51 -0000 Issue 7521
Topics (messages 315306 through 315313):
Re: Processing newlines in a text area field
315306 by: Stephen
move_uploaded_file() does not return any value or warning
315307 by: Partha Chowdhury
315308 by: Simon J Welsh
315309 by: Partha Chowdhury
Re: Dennis Ritchie, Father of Unix and C programming language, dead at 70
315310 by: Ricardo Martinez
315311 by: Arno Kuhl
315312 by: shiplu
Extending an instantiated class
315313 by: Alain Williams
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----------------------------------------------------------------------
--- Begin Message ---
On 11-10-13 10:49 PM, Stephen wrote:
I have a web page with a form with a text area. I enter:
foo
bar
PHP processes the POST and inserts the record into MySQL.
The database field is text.
I use PDO
For testing I have removed any processing of the text area content.
Now, no matter how many blank rows I have between foo and bar, in the
database record, I always get a single newline character.
Found the problem. I was not passing the record key properly to the SQL
UPDATE transaction.
So I had WHERE id = ""
Valid SQL so no error. But no record update :(
Thanks all
Stephen
--- End Message ---
--- Begin Message ---
Hello List.
I am learning php and MySQL for my project work.
I am trying to upload files to local server.
First, a page is displayed on the browser to upload file
Relevant code:
<form method="post" action="index.php" enctype="multipart/form-data">
<input type="file" name="file">
<br>
<input type="submit" name="submit" value="upload">
</form>
After the user chooses a file and hit the submit button,the browser
displays the content from the "uploads" directory.
Php code:
<?php
if (isset ($_POST['submit']) && ($_POST['submit']=="upload")) {
$pathname = "uploads";
if (!file_exists($pathname)) {
$mkdir = mkdir($pathname);
if (!$mkdir) {
echo 'could not create directory uploads/';
exit();
}
}
$filename = $_FILES['file']['tmp_name'];
$destination = "uploads/" . time() . "_-" .
$_FILES['file']['name'];
move_uploaded_file($filename, $destination);
header("Location:uploads/");
But the uploads directory is empty and no warning or error is displayed !
Then i set a check for the return value of move-uploaded_file.
if(isset ($move_uploaded_file)):
echo ""success;
else:
echo "error in uploading";
endif;
But it does not print anything ! .According to manual,
Returns *TRUE* on success.
If /filename/ is not a valid upload file, then no action will occur,
and *move_uploaded_file()* will return *FALSE*.
If /filename/ is a valid upload file, but cannot be moved for some
reason, no action will occur, and *move_uploaded_file()* will return
*FALSE*. Additionally, a warning will be issued.
So the above code should have printed "error in uploading".
Then i tried the copy() function and it prints a warning:
*Warning*: copy() [function.copy
<http://localhost/file_uploading/function.copy>]: Filename cannot be
empty in */srv/www/htdocs/file_uploading/index.php* on line *38*
line 38 is
copy($filename, $destination);
Then i did a print_r($_FILES['file']) and found that
$_FILES['file']['error'] was 1 .The problem was the file i was trying to
upload was 6M in size,greater than upload_max_filesize in /etc/php.ini
which was 2M.I changed that and now it works.
So I want to know - does move_uploaded_file returns false on failure or
is there a bug in my code ?
I am using php 5.3.8 which is configured with
'./configure' '--prefix=/php' '--with-apxs2=/apache/bin/apxs'
'--with-config-file-path=/etc' '--with-mysql-sock=/tmp/mysql.sock'
'--with-openssl' '--with-zlib' '--enable-bcmath' '--with-bz2'
'--enable-calendar' '--with-gdbm' '--with-enchant' '--enable-exif'
'--enable-ftp' '--with-gd' '--enable-gd-native-ttf' '--with-gettext'
'--with-gmp' '--enable-mbstring' '--with-mcrypt' '--with-readline'
'--enable-soap' '--enable-sqlite-utf8' '--enable-zip'
'--with-mysql=mysqlnd' '--with-mysqli=mysqlnd'
'--with-jpeg-dir=/usr/lib' '--with-freetype-dir'
and error_reporting directive in php.ini is "error_reporting = E_ALL |
E_STRICT".
--- End Message ---
--- Begin Message ---
On 15/10/2011, at 4:01 PM, Partha Chowdhury wrote:
> Then i set a check for the return value of move-uploaded_file.
>> if(isset ($move_uploaded_file)):
>> echo ""success;
>> else:
>> echo "error in uploading";
>> endif;
> But it does not print anything !
Assuming you did $move_uploaded_file = move_uploaded_file($filename,
$destination);, then isset($move_uploaded_file) will always be true.
isset() just checks if the variable you past to it is set, not if it has a
non-false value. You could simply use if(move_uploaded_file($filename,
$destination)), or if($move_uploaded_file).
---
Simon Welsh
Admin of http://simon.geek.nz/
--- End Message ---
--- Begin Message ---
On 15/10/11 08:35 AM, Simon J Welsh wrote:
Assuming you did $move_uploaded_file = move_uploaded_file($filename,
$destination);, then isset($move_uploaded_file) will always be true.
isset() just checks if the variable you past to it is set, not if it
has a non-false value. You could simply use
if(move_uploaded_file($filename, $destination)), or
if($move_uploaded_file).
My apologies - i forgot to include the $move_uploaded_file =
move_uploaded_file($filename, $destination); line.
I saw in the manual that isset() "isset --- Determine if a variable is
set and is not NULL" and thought if move_uploaded_file() returns false,
the error checking will be triggered.But now i see the sample code in
the manual and stand corrected.
So what was happening that $move_uploaded_file was set but empty.My
understanding of isset() was wrong.
*
*
--- End Message ---
--- Begin Message ---
int main()
{
printf("R.I.P. Dennis Ritchie: 1941-2011\n");
return 0
}
sad notice... RIP
On Fri, Oct 14, 2011 at 10:01 PM, tamouse mailing lists <
tamouse.li...@gmail.com> wrote:
> On Thu, Oct 13, 2011 at 5:08 PM, Daevid Vincent <dae...@daevid.com> wrote:
> > #include <stdio.h>
> >
> > int main()
> > {
> > printf("R.I.P. Dennis Ritchie: 1941-2011\n");
> > return 0;
> > }
> >
> >
> > http://www.networkworld.com/news/2011/101311-ritchie-251936.html
> >
> >
>
> dmr--; /* :-( */
>
> --
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>
>
--
Ricardo
_______________________________________________
IT Architect
website: http://www.pulsarinara.com
--- End Message ---
--- Begin Message ---
-----Original Message-----
From: Daevid Vincent [mailto:dae...@daevid.com]
Sent: 14 October 2011 12:08 AM
To: php-gene...@lists.php.net
Subject: [PHP] Dennis Ritchie, Father of Unix and C programming language,
dead at 70
#include <stdio.h>
int main()
{
printf("R.I.P. Dennis Ritchie: 1941-2011\n");
return 0;
}
http://www.networkworld.com/news/2011/101311-ritchie-251936.html
===============================
He might have put it a bit differently:
int main()
{
printf("Goodbye world\n");
return 0;
}
--- End Message ---
--- Begin Message ---
I found this, And I liked it.
int main()
{
printf("Goodbye Daddy\n");
return 0;
}
--
Shiplu Mokadd.im
Follow me, http://twitter.com/shiplu
Innovation distinguishes between follower and leader
--- End Message ---
--- Begin Message ---
Well, that is what I think that I need. Please let me explain what I am trying
to do
and tell me how to do it or a better way of doing it.
I have an application where a Screen (web page) may contain several Forms. The
Forms
will want to access properties, etc, from their Screen. So what I want is to do
something like:
class Screen {
private $JavascriptVars = array();
function AddJsVar($name, $val) {
$this->JavascriptVars[$name] = $val;
}
}
class Form extends Screen {
public $screen; // Could set to a class Screen
function DefineForm($fName)
.....
}
$s = new Screen(....);
$s->AddJsVar('Date', '15 Oct 2011');
$f1 = new Form($s);
$f1->DefineForm('search');
$f1->AddJsVar('CompanyName', 'IBM');
$f2 = new Form($s);
$f2->DefineForm('login');
$f2->AddJsVar('Error', 'Login failed');
The trouble is that $f1->AddJsVar() will fail, class extention seems not
designed to work like that.
I could make it work by either:
1) $f1->screen->AddJsVar() --- but I would rather avoid the extra '->screen'.
2) Use of a __call() in class Form -- but that makes things slower.
Any help gratefully received.
I think that to do what I want PHP would need a syntax like:
$f2 = new $s Form();
or
$f2 = $s->new Form();
--
Alain Williams
Linux/GNU Consultant - Mail systems, Web sites, Networking, Programmer, IT
Lecturer.
+44 (0) 787 668 0256 http://www.phcomp.co.uk/
Parliament Hill Computers Ltd. Registration Information:
http://www.phcomp.co.uk/contact.php
#include <std_disclaimer.h>
--- End Message ---
