Comments by C0DE_007 (Top 70 by date)

i am not getting anything

yes sir ??

when i am trying to remove findViewById(R.id.checkBox1) app is crashing

But sir what should i do i can't remove findViewById(R.id.checkBox1)

actually i am making server side android app so through in android asynctask Pair pairAction = new Pair<>("action", "searchFood"); it will send action name then it will run like if ($action == 'searchFood') {

actually i am making server side android app so through in android asynctask Pair pairAction = new Pair<>("action", "searchFood"); it will send action name then it will run like if ($action == 'searchFood') {

actually i am making server side android app so through in android asynctask Pair pairAction = new Pair<>("action", "searchFood"); it will send action name then it will run like if ($action == 'searchFood') {

i don't know why it showing You are not authorized msg .. code are fine !

yes sir no password

sir can u provide me some more tutorial .. this one are not what i want..

Srry sir but still im confused . i want to make an app that can upload image in server database like i am using godaddy hosting .. so how can i upload files in godaddy db and how to fetch from there ... is there any different server for Apps ?? and WebView is only good if your website is responsive .. but what user don't have any website ? he want to make an app only just like myntra app !! PLease help me out with this confusion sir

i have make a php program image using passing trough page

like in simple term i have make 2 page in first.html i have put two button
first browse the image
<input type="file" name="file" id="file" >
2- > get the image file name and $_post to second page
<input type="submit">

and in second page sec.php

i have upload image code

where is only submit button so when i click on it .. it will get the image(file) name from first page and then start uploading
but the problem is i am success to passing the image file name to 2 page but the image is not uploading

i am using session ...

and plzz tell where i am doing wrong and what i have to do to make it work thank u

any one plzz help me i am just stuck in uploading the image

sir it required whole code without it no one will understand what i am doing and what i want to do

no bro it a link you don't need to download the code ... just open the link in your browser

noo nooo

<div id = "22" >
<form action="" enctype="multipart/form-data" method="post">
<input type="file" name="files"> "now user want to display image in div id 22 which is top right side of the page after browsing the image ... db will get the id name which user select $id = 22 now it will save in directory and insert in db like this INSERT INTO `images` (`ID`, `Path`) VALUES (22, 'path/1391936423.jpg');
<input type="submit" value="Upload"> afte this the image will display over here
</form>
query("select Path from images where images.ID in (select max(ID) from images)");

$row = $getimage -> fetch();
print '<img src="'.$row['Path'].'">';
unset($connection);
?>
</div>
id will work like rollback or commit command in each it will add the id transaction in db
now you got or not ??

commit or rollback in sql

@EZW i have make a function


<!--?php

function uploadImage($files, $post){
if(isset($post['upload'])){
$types = array('.jpg','.jpeg','.png','.gif');
$max_filesize = 10485760;
$upload_path = 'upload/';

$filename = $files['file']['name'];
$ext = substr($filename, strpos($filename,'.'), strlen($filename)-1);

if(!in_array($ext,$types))
die('The file you attempted to upload is not allowed.');

if(filesize($files['file']['tmp_name']) > $max_filesize)
die('The file you attempted to upload is too large.');

if(!is_writable($upload_path))
die('You cannot upload to the specified directory, please CHMOD it to 777.');

if(move_uploaded_file($files['file']['tmp_name'],$upload_path . $filename)) {
$query = "INSERT INTO `uploads` (name) VALUES ('".$filename."')";
$execute = mysql_query($query);
if($execute){
echo 'Your file upload was successful!';
}else{
echo 'There was an error.';
}
} else {
echo 'There was an error during the file upload. Please try again.';
}
}
}

// And call it after each form
uploadImage($_FILES, $_POST);


?>



<!--?php $connection = new PDO('mysql:host=localhost;dbname=pictures', "root", "");
??-->

showimg
<!--?php require_once("connection.php");

$getimage = $connection-?-->query("select Path from images where images.ID in (select max(ID) from images)");

$row = $getimage -> fetch();
print '<img src="'.$row['Path'].'">';
unset($connection);
?>

now how to connect all the code with each other and with my db can u plzz do this for me ...

dynamically add/generate DIVs to your page which would wrap the images (<div id="22"><img src="path/to/image.jpg" /></div>) VIA PHP by selecting image ID from the database. this part is correct like if user select (<div id="22"><form action="" enctype="multipart/form-data" method="post">
<input type="file" name="files" accept="image/*">
<input type="submit" value="Upload">'when user click on upload so it will insert value in db
INSERT INTO `images` (`ID`, `Path`) VALUES
(22, 'path/1391936423.jpg');
</form>
and then display here
query("select Path from images where images.ID in (select max(ID) from images)");

$row = $getimage -> fetch();
print '<img src="'.$row['Path'].'">';
unset($connection);
?>
</div>)

thank u bro for improving my question :) n srry next time i will remember ... n bro plzz help me out in this

like example

query("insert into images (Path) values ('".$images."');");
unset($connection);
print ("success");
}
else print "error";
}

}
?>



<div id = "image_1 ">
<form action="" enctype="multipart/form-data" method="post">
<input type="file" name="files">
<input type="submit" value="Upload">
'so here will value 1 image will display like img src = INSERT INTO `images` (`ID`, `Path`) VALUES (1, 'path/1391936423.jpg');
</form>
</div>
<div id ="image_2 ">
<form action="" enctype="multipart/form-data" method="post">
<input type="file" name="files">
<input type="submit" value="Upload">
' and here id value 2 image INSERT INTO `images` (`ID`, `Path`) (2, 'path/1391936495.jpg');
</form>
</div>

so that i can display different image in different div .... if i add more div it value increase and automatically that image which relate to those div

no answer ???

i have code which display using database

here are the codes ...




index.php

query("insert into images (Path) values ('".$filepath."');");
unset($connection);
print ("success");
header("location:index.php");
}
else print "error. First select an image";
}

?>
<form action="index.php" enctype="multipart/form-data" method="post">
<input type="file" name="files">
<input type="submit" value="Upload">
</form>

query("select Path from images where images.ID in (select max(ID) from images)");

$row = $getimage -> fetch();
print '<img src="'.$row['Path'].'">';
unset($connection);
?>

===================================================================================
connection.php


====================================================================================


pictures.sql
-- phpMyAdmin SQL Dump
-- version 4.0.4
-- http://www.phpmyadmin.net
--
-- Host: localhost
-- Generation Time: Feb 09, 2014 at 09:16 AM
-- Server version: 5.6.12-log
-- PHP Version: 5.4.16

SET SQL_MODE = "NO_AUTO_VALUE_ON_ZERO";
SET time_zone = "+00:00";


/*!40101 SET @OLD_CHARACTER_SET_CLIENT=@@CHARACTER_SET_CLIENT */;
/*!40101 SET @OLD_CHARACTER_SET_RESULTS=@@CHARACTER_SET_RESULTS */;
/*!40101 SET @OLD_COLLATION_CONNECTION=@@COLLATION_CONNECTION */;
/*!40101 SET NAMES utf8 */;

--
-- Database: `pictures`
--
CREATE DATABASE IF NOT EXISTS `pictures` DEFAULT CHARACTER SET utf8 COLLATE utf8_general_ci;
USE `pictures`;

-- --------------------------------------------------------

--
-- Table structure for table `images`
--

CREATE TABLE IF NOT EXISTS `images` (
`ID` int(11) NOT NULL AUTO_INCREMENT,
`Path` text NOT NULL,
PRIMARY KEY (`ID`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=4 ;

--
-- Dumping data for table `images`
--

INSERT INTO `images` (`ID`, `Path`) VALUES
(1, 'path/1391936423.jpg'),
(2, 'path/1391936495.jpg'),
(3, 'path/1391937295.jpg');

/*!40101 SET CHARACTER_SET_CLIENT=@OLD_CHARACTER_SET_CLIENT */;
/*!40101 SET CHARACTER_SET_RESULTS=@OLD_CHARACTER_SET_RESULTS */;
/*!40101 SET COLLATION_CONNECTION=@OLD_COLLATION_CONNECTION */;

i want to know that if i am using one form in one page to display one particular image so how to display different uploaded image using different form in same page ?? using same directory ? because declaring different variable and making new directory will become complex and so is there any easy way guyzz ??

use my code bro ..u will get the answer
.. i want to display different image in div 2

when i upload image it upload in images folder and in last folder so if i upload any other image it will remove previous img in last folder and upload new img to display in div first...
i have try to make different directory and declare new variable and then it displaying in div id 2 but its very complex method

plzz tell me okk... in pay then upload... like
<form action="process.php" method="post" enctype="multipart/form-data">
<input type="file" name="file" /><br />
<input type="submit" name="submit" />
</form>

user will select file from their computer then... user have to submit to file name ... how to add paypal button in submit so the uploading file name will also cache and it will store in session and then after payment re-direct link - process.php start !!

log in ?? bro there is not log in option there .... it just user visit the site if he/she like to upload or download he will select that file or browse file to upload ... then buynow button will be show after payment .... so plzz bro can u show me some code or demo .. plzz and thank u once again i got the answer = session .... !! how long i have been working .. hmmm 2-4 week :p

but i will post action = " upload.php " and the upload.php file put it in paypal button re direct option ... ? how to stored in the session.. any example sir plzz demo ?? so it doesn't matter there is paypal in between this.... the link whicj i provided in re-direct link will catch file name etc... ok but there is no logged user .... user do not required to log in to upload the file or download the file ... !! i think logged in user is not compulsory right ?? and if it's compulsory then plzz tell me any other way.... and thank youuuuu bro thankk you sooo much :)

@Christian Graus here is the example = > if user select xyz.mp3 i have put re direct link in paypal name { www.example.com/download_process_start.php } after payment pay pal will re direct this link .... but how the link will cache the song name ?? which item is selected by user ? is it in .mp3 or .apk or in .exe ?? how ?? and link this if user want to upload the file <form action="process.php" method="post" enctype="multipart/form-data"> <input type="file" name="file" /><br /> <input type="submit" name="submit" /> </form> so how will upload_process.php start will i was put it in paypal ? how the link will cache the file name etc ??

if any one having any better solution then plzz tell me guyzz .. thank you soo much guyz for helping me .... i really appreciate :)

Trushnak can u send me the whole working code....? i like the links... but i am llittle bit confuse in this plzz tell me ok ... example user select xyz.mp3 i have put re direct link in paypal name { www.example.com/download_process_start.php } after payment pay pal will re direct this link .... but how the link will cache the song name ?? which item is selected by user ? is it in .mp3 or .apk or in .exe ?? how ?? and link this if user want to upload the file <form action="process.php" method="post" enctype="multipart/form-data">
<input type="file" name="file" /><br />
<input type="submit" name="submit" />
</form> so how will upload_process.php start will i was put it in paypal ? how the link will cache the file name etc ??

but problem is that how to open paypal website in jquery superbox ..!!
.... this are some research
http://www.jquery4u.com/plugins/jquerypaypalbuyitnow/

superbox is like fancy box it open at same page
http://pierrebertet.net/projects/jquery_superbox/

if jquery can help in this so plzz help me it will be like step by step method by using jquery superbox effect so in jquery superbox will open paypal payment after log in and all function new jquery superbox will open in that box download button show ?? just like google wallet .. it open jquery box ask for card then user can download

Richard .. i am using buynow button but not getting it what link have to put in re direct option because after finish pay re direct and user will can start downloading that particular thing which they have selected .... so i can't use 100 different buynow button and put different different re direct link in buynow button ? i want one link which automatically catch information which item have selected by users so after payment it will re direct and start downloading ... or uploading ?

right now plzz solve pay then upload function guyzz ?

like in this upload function
<form action="process.php" method="get" enctype="multipart/form-data">
<input type="file" name="file" /><br />
<input type="submit" name="submit_file" />
</form>

form action sending information to process.php but in between first paypal payment function come then how will paypal re - direct www.example.com/process.php cache all the information ??

But with cache or session, the image will disappear after closing browser and starting again

so that we need to save settings to database and check from the database and show image.. ?? but how ??

guyz plzz edit the code and change the code as your requirement and send the code plzz

no when i will upload any other image so existing which was showing in form it will disappears

yes bro when i refresh the image disappears... !! i want when ever i will open the last uploaded image will display on form.php

if(isset($_GET["img"]))
{
$n=$_GET["img"];
echo "<img src='photos/".$n."'?-->";
}


i have try this code see above

can use my code which i have provided above and edit the code and send it with working code plzz

sorry for my bad english guyz

for ex:- /upload/users/Images/ .jpg this is path where uploaded image store in DB okk... now the the redirect page is www.example.com/upload.php .... so when payment is done i will copy on upload.php link in paypal button .. when we make paypal button it ask for after payment plz enter redirect page link... so i will put .. www.example.com/upload.php ??? is ok and bro are you asking to or or what ?? i am saying if you have better solution then tell me ?

How to add paypal payment function in between upload - display image ??

i want to display the image permanently after payment ..... and your link https://forum.jquery.com/topic/how-to-change-image-src-path not work for me ... bro ... and is there any problem in that php code ??

permanently bro ... i think i will different id for different for different img box .. as i said if you have any better way or any simple way so plzz tell me .... and what about the code it iwll not display the image permanently ??

i think i got the solution guyzz .. !! <img src="" >

$image = $_FILES["file"]["name"];
move_uploaded_file($_FILES["file"]["tmp_name"],"../upload/$image");
$query = "INSERT INTO images VALUES('','$image')";

i think it will work :p now paypal problem ? :p

right now i have make my profile picture in codeproject when there was no image the code was like this
<img id="ctl00_MC_Prof_MemberImage" class="padded-top" style="height:114px;width:85px;border-width:0px;" src=
"/script/Membership/Images/member_unknown.gif"></img>

after uploading image
<img id="ctl00_MC_Prof_MemberImage" class="padded-top" style="border-width:0px;" src="http://www.codeproject.com/script/Membership/ProfileImages/{d9491f19-c1b2-41ab-ae36-fac8b875311e}.jpg"></img>

that i want man

but why we need jquery bro ?? it can be simple done by php ?? just we need to get imagefile name which was uploaded and put the declare variable into img src tag ... but the code which i past in question bar it work like this like the image name wall.jpg so it change in directory where the uploaded image save into like this 1390298757-wall.jpg

ok i have edit code.. check it.... i have check your link and i got this
$('.changesrc').each(function(){
//Change the src of each img
$(this).attr('src', 'images/alt/imagename.jpg'); (uploaded image )
});

<img class="changesrc" src="images/es/step1green.gif">

ok now what when user upload new image in directory images/es/random.gif random.gif is image file name so this function will change the path ? like this <img class="changesrc" src="images/es/random.gif">
??
i think we have to modify this code ? for getting imagename then send to jquery function and then send to img src = " "

christain graus send me your email id i will send you code of upload image to directory via php .. right now its greatly working on local host

how ..? i want to pointing the image file name to image tag automatically .... do you have any demo of this or please provide me source code

bro 1-> step user will select image from there computer ok then image will upload or server directory 2-> then user will click on paypal buynow button after this payment user will get next button or else image will display if user not pay the amont the process will cancel and image will not display... that all if its confusing let we talk only about display image automatically .... or do you have any better way or simple way then plzz tell me it will great for me

how ?? do you have code .. plzz provide me some small demo or source code ... but bro i dont want to assume i want to attach the uploaded image automatically to img tag

i know that :p but when user pay how will upload process start do you have any code or demo so plzz provide it