[PHP-DB] Upload_File
Hi I am trying to make a form to provide option to user to upload a file on
a server.Here is the code for my upload.php. It checks all the conditions
etc.. . This code works fine on localhost , and file is uploaded
successfully. But when I use this script online on my server, it gives me
error. couldn't figure out what is error.
What I think is error in move_upload_file, but it isn't returning any error.
File permissions on upload folder are 755.
if((!empty($_FILES['uploaded_file'])) ($_FILES['uploaded_file']['error']
== 0)){
$ok = 0;
$filename = basename($_FILES['uploaded_file']['name']);
$ext = substr($filename, strrpos($filename, '.') + 1);
if (($ext == jpg || gif || png) ($_FILES[uploaded_file][type]
== image/jpeg || image/gif || image/png)
($_FILES[uploaded_file][size] 2097152) ){
$newname = 'upload/'.$filename;
if (!file_exists($newname)) {
if
((move_uploaded_file($_FILES['uploaded_file']['tmp_name'],$newname))) {
$ok = 1;
echo It's done! The file has been saved as: .$newname;
} else {
echo Error: A problem occurred during file upload!; //This one
executes when ever i try to upload file on server.
}
} else {
echo Error: File .$_FILES[uploaded_file][name]. already
exists;
}
} else {
echo Error: Only .jpg images under 2MB are accepted for upload;
}
} else {
echo Error: No file uploaded;
}
Thanks
Bilal Farooq Ahmad
Re: [PHP-DB] Image_Problem
Hi All, Thanx Richard for your reply, but it isn't working either, please
help me out, Thanks. I can explain the problem again if some one has ne
problem in understanding of the problem. Thanks
On Thu, Jul 23, 2009 at 2:21 PM, Richard Quadling
rquadl...@googlemail.comwrote:
2009/7/22 Bilal Ahmad engg.bilalma...@googlemail.com:
Hi All, I wanna ask a question pleasenV. I am trying to create an image
on
fly, please do help me , following is the code.
*File Name : Font.php
Code: *
html
head
titleImage Creation/title
script language=javascript
var xmlhttp;
function showPic()
{
xmlhttp=GetXmlHttpObject();
if (xmlhttp==null)
{
alert (Browser does not support HTTP Request);
return;
}
var text = document.getElementById(textfield).value;
var url=image.php;
url=url+?text=text;
alert(url);
url=url+sid=+Math.random();
xmlhttp.onreadystatechange=stateChanged;
xmlhttp.open(GET,url,true);
xmlhttp.send(null);
}
function stateChanged()
{
if (xmlhttp.readyState==4)
{
if(xmlhttp.responseText == 1)
{
document.getElementById(pic).style.display=block;
}
}
}
function GetXmlHttpObject()
{
if (window.XMLHttpRequest)
{
// code for IE7+, Firefox, Chrome, Opera, Safari
return new XMLHttpRequest();
}
if (window.ActiveXObject)
{
// code for IE6, IE5
return new ActiveXObject(Microsoft.XMLHTTP);
}
return null;
}
/script
/head
body
form name=form1 method=post action= onSubmit= return false;
table width=304 border=1
tr
td colspan=2div align=centerText/div/td
/tr
tr
tdText/td
tdlabel
input type=text name=textfield id=textfield
/label/td
/tr
tr
td colspan=2label
div align=center
input type=submit name=button id=button value=Update
onClick=showPic()
/div
/label/td
/tr
tr
td colspan=2div id=pic style=display:noneimg
src=pic.jpg
//div/td
/tr
/table
/form
/body
/html
*File Name : image.php
Code:*
[php]
$name = $_GET['text'];
$pic = imagecreatetruecolor(100, 100);
$text_color = imagecolorallocate($pic, 255, 255, 255);
imagestring($pic, 10, 15, 15, $name, $text_color);
$pi = Imagejpeg($pic,pic.jpg);
echo $pi;
ImageDestroy($pic);
[/php]
*Problem: *
What this code is doing is that, it creates a new image with the text
(that
user enters) on it, but loads the image that was created previously, I
want
that it should display the text on the picture which users enter on the
fly.
( e.g. If user enter TEXT as text it should display TEXT on the picture
displayed, soon after we get response from ajax).I hope you got the
problem.
Thanks
Try changing ...
$pi = Imagejpeg($pic,pic.jpg);
echo $pi;
to ...
imagejpeg($pic);
--
-
Richard Quadling
Zend Certified Engineer : http://zend.com/zce.php?c=ZEND002498r=213474731
Standing on the shoulders of some very clever giants!
ZOPA : http://uk.zopa.com/member/RQuadling
[PHP-DB] Image_Problem
Hi All, I wanna ask a question pleasenV. I am trying to create an image on
fly, please do help me , following is the code.
*File Name : Font.php
Code: *
html
head
titleImage Creation/title
script language=javascript
var xmlhttp;
function showPic()
{
xmlhttp=GetXmlHttpObject();
if (xmlhttp==null)
{
alert (Browser does not support HTTP Request);
return;
}
var text = document.getElementById(textfield).value;
var url=image.php;
url=url+?text=text;
alert(url);
url=url+sid=+Math.random();
xmlhttp.onreadystatechange=stateChanged;
xmlhttp.open(GET,url,true);
xmlhttp.send(null);
}
function stateChanged()
{
if (xmlhttp.readyState==4)
{
if(xmlhttp.responseText == 1)
{
document.getElementById(pic).style.display=block;
}
}
}
function GetXmlHttpObject()
{
if (window.XMLHttpRequest)
{
// code for IE7+, Firefox, Chrome, Opera, Safari
return new XMLHttpRequest();
}
if (window.ActiveXObject)
{
// code for IE6, IE5
return new ActiveXObject(Microsoft.XMLHTTP);
}
return null;
}
/script
/head
body
form name=form1 method=post action= onSubmit= return false;
table width=304 border=1
tr
td colspan=2div align=centerText/div/td
/tr
tr
tdText/td
tdlabel
input type=text name=textfield id=textfield
/label/td
/tr
tr
td colspan=2label
div align=center
input type=submit name=button id=button value=Update
onClick=showPic()
/div
/label/td
/tr
tr
td colspan=2div id=pic style=display:noneimg src=pic.jpg
//div/td
/tr
/table
/form
/body
/html
*File Name : image.php
Code:*
[php]
$name = $_GET['text'];
$pic = imagecreatetruecolor(100, 100);
$text_color = imagecolorallocate($pic, 255, 255, 255);
imagestring($pic, 10, 15, 15, $name, $text_color);
$pi = Imagejpeg($pic,pic.jpg);
echo $pi;
ImageDestroy($pic);
[/php]
*Problem: *
What this code is doing is that, it creates a new image with the text (that
user enters) on it, but loads the image that was created previously, I want
that it should display the text on the picture which users enter on the fly.
( e.g. If user enter TEXT as text it should display TEXT on the picture
displayed, soon after we get response from ajax).I hope you got the problem.
Thanks
Re: [PHP-DB] Prepared Statement Insert Problem
Hi i wanna ask a question. I am trying to create an image on fly, please do
help me , following is the code.
*File Name : Font.php
Code: *
html
head
titleImage Creation/title
script language=javascript
var xmlhttp;
function showPic()
{
xmlhttp=GetXmlHttpObject();
if (xmlhttp==null)
{
alert (Browser does not support HTTP Request);
return;
}
var text = document.getElementById(textfield).value;
var url=image.php;
url=url+?text=text;
alert(url);
url=url+sid=+Math.random();
xmlhttp.onreadystatechange=stateChanged;
xmlhttp.open(GET,url,true);
xmlhttp.send(null);
}
function stateChanged()
{
if (xmlhttp.readyState==4)
{
if(xmlhttp.responseText == 1)
{
document.getElementById(pic).style.display=block;
}
}
}
function GetXmlHttpObject()
{
if (window.XMLHttpRequest)
{
// code for IE7+, Firefox, Chrome, Opera, Safari
return new XMLHttpRequest();
}
if (window.ActiveXObject)
{
// code for IE6, IE5
return new ActiveXObject(Microsoft.XMLHTTP);
}
return null;
}
/script
/head
body
form name=form1 method=post action= onSubmit= return false;
table width=304 border=1
tr
td colspan=2div align=centerText/div/td
/tr
tr
tdText/td
tdlabel
input type=text name=textfield id=textfield
/label/td
/tr
tr
td colspan=2label
div align=center
input type=submit name=button id=button value=Update
onClick=showPic()
/div
/label/td
/tr
tr
td colspan=2div id=pic style=display:noneimg src=pic.jpg
//div/td
/tr
/table
/form
/body
/html
*File Name : image.php
Code:*
[php]
$name = $_GET['text'];
$pic = imagecreatetruecolor(100, 100);
$text_color = imagecolorallocate($pic, 255, 255, 255);
imagestring($pic, 10, 15, 15, $name, $text_color);
$pi = Imagejpeg($pic,pic.jpg);
echo $pi;
ImageDestroy($pic);
[/php]
*Problem: *
What this code is doing is that, it creates a new image with the text (that
user enters) on it, but loads the image that was created previously, I want
that it should display the text on the picture which users enter on the fly.
( e.g. If user enter TEXT as text it should display TEXT on the picture
displayed).Hope you got the problem.
Thanks
For me, it shoould work like this: User enters text intext filed (in
font.html) , when pressed button, image.php should create an im
On Tue, Jul 21, 2009 at 10:46 PM, Christopher Jones
christopher.jo...@oracle.com wrote:
kesavan trichy rengarajan wrote:
could be rewritten as:
mysqli_stmt_bind_param($submitadmin, isss, $numrows, $admin,
sha1($password), $email);
Turning on E_STRICT in PHP 5.3 will show
PHP Strict Standards: Only variables should be passed by reference
This is also true in earlier versions although the warning isn't
displayed.
For compliance, try:
$s = sha1($password);
mysqli_stmt_bind_param($submitadmin, isss, $numrows, $admin, $s,
$email);
Chris
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