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PHP - Data Only Refreshes On My Computer But Not Others! Help!
Hi all,
I've created a data-entry HTML page that posts to update.php. I am using mysql to store the data entered. When I enter data and hit "update database" button, the php file pops up and correctly shows the data entered, but the update.php file does not show that same data on another computer. I have checked the mysql database and it clearly shows that it the new data has been stored. Why isn't the data showing up on any other computer? I can provide codes as needed. Thanks. Richard Similar TutorialsNeed help with some whiles, etc., and .php page refreshes. $num; Comes from mysql_num_rows from a database table's regular 'select'. And gets the row id numbers for every row that the id numbers are listed in order. $totalrows comes from another select but is select * instead. And pulls the total rows in the whole table instead there are 49 rows in the whole table. I'm trying to get the page to show 5 table rows at a time like the following, but then when the page is still open on next page refresh to show the next 5 rows and continue that way. This following: Quote PHP Code: <?php $i=$num; if ($num == 5); {echo "The number is " . $num . "<br />";} do {$i = $i + 5; echo "The number is " . $i . "<br />"; } while ($i<$totalrows); ?> Brings up these results: The row number is 10 The row number is 15 The row number is 20 The row number is 25 The row number is 30 The row number is 35 The row number is 40 The row number is 45 The row number is 50 Shows all these on the page at the same time. I'm trying to get these to only show the 5 rows after every page refresh instead. Please let me know how to do that, and whether it can be done without any cookies. I don't know if there is a better way to show the first 5 separate like I have it. Thank you very much for your help. Hi Guy's, I've recoded allot with 3 scripts i had to my disposal. I've fixed 95% and i'm proud allready, but stuck at the following code. It does'nt execute any errors and refreshes the page without uploading the picture to the desired folder. Also folders are'nt created. I've turned on error reporting at level -1, but those errors i fixed. Can you please help me out? note: this is a partial code of a tiny usermanagement 'cms'. Code: [Select] <?php //action: add picture for user ----------------------------------------------------------------------------- if (isset($_GET['editpic']) && isset($_GET['id'])) { $id = (int) $_GET['id']; if ($id == 0) { die("Invalid ID provided."); } $sql = "SELECT username FROM `users` WHERE `id`='".$_GET['id']."'"; $res = mysql_query($sql) or die(mysql_error()); //execution when completed the add picture form and pressed submit button --------------------- if (isset($_POST['addPic'])) { $row = mysql_fetch_assoc($res); $title = protect($_POST['title']); if(!$title) { echo "<script language=\"Javascript\" type=\"text/javascript\"> alert(\"You must choose a title for your picture!\") document.location.href='profilecp.php'</script>"; } $target = $row['username']; if(!is_dir($target)) @mkdir($target); $target = $target . '/pics'; if(!is_dir($target)) @mkdir($target); $target = $target."/".basename($_FILES['pics']['name']) ; $size = $_FILES['pics']['size']; $pic = $_FILES['pics']['name']; $type = $_FILES['pics']['type']; $sql2= "INSERT INTO `user_photos` (`profile_id`,`title`,`size`,`type`,`reference`) VALUES ('".$_GET['id']."','$title','$size','$type','$pic'); "; $res2 = mysql_query($sql2) or die(mysql_error()); if(move_uploaded_file($_FILES['pics']['tmp_name'], $target)) { echo "<script language=\"Javascript\" type=\"text/javascript\"> alert(\"Your picture has been uploaded\") document.location.href='profilecp.php'</script>"; } else { echo "<script language=\"Javascript\" type=\"text/javascript\"> alert(\"There was an error, try again\") document.location.href='profilecp.php'</script>"; } $target2 = $row['username']; $target2 = $target2 . '/pics'; $target2 = $target2 . '/thumbs'; if(!is_dir($target2)) @mkdir($target2); $target2 = $target2."/".basename($_FILES['pics']['name']) ; createthumb($target,$target2,150,150); } <!-----------------------ADD PICTURES-----------------------//--> <div class="dividerp"> <form enctype="multipart/form-data" method="POST" action="administrator.php?editpic&id=<?php echo $id;?>"><br/> <strong>Upload Pictures</strong><br/><br/> <div class="formElm"> <label for="title">Title</label> <input id="title" type="text" name="title" maxlength="32"><br/> </div> <div class="formElm"> <label for="file">File</label> <input id="file" type="file" name="pics" maxlength="255"><br/> </div> <input type="submit" name="picAdd" value="Add"> </form> </div> </div> </body> </html> }?> and the mysql tables: Code: [Select] CREATE TABLE `users` ( `id` INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY , `first` VARCHAR( 32 ) NOT NULL , `last` VARCHAR( 32 ) NOT NULL , `username` VARCHAR(32) NOT NULL, `password` VARCHAR(255) NOT NULL, `email` VARCHAR(255) NOT NULL, `about` TEXT NOT NULL, `level` int(4) default '1' ) ENGINE=InnoDB AUTO_INCREMENT=27 DEFAULT CHARSET=latin1; Code: [Select] CREATE TABLE `user_photos` ( `id` INT NOT NULL AUTO_INCREMENT PRIMARY KEY , `profile_id` INT NOT NULL , `title` VARCHAR( 128 ) NOT NULL , `size` INT NOT NULL , `type` VARCHAR( 128 ) NOT NULL , `reference` VARCHAR( 255 ) NOT NULL ) ENGINE=InnoDB AUTO_INCREMENT=27 DEFAULT CHARSET=latin1; Best regards, Martijn Is there a way to get my visitors PC or Device Name? I am not looking for server name or ip. I need to get my current visitors computer name. Example: MAIN, HOME, OFFICE, TOM32154}}{|\] like when first naming a PC with a new windows install. I have used sessions before I don't know whats going on. My sessions wont die. this is what im doing: <?php session_start(); $_SESSION['user'] = $_POST['email']; $_SESSION['pass'] = $_POST['pass']; ?> ok new page <?php session_start(); echo $_SESSION['user']; echo "<br>"; echo $_SESSION['pass'] ?> OK great is show the same info but now if I close my browser and go back to the second page where it echos it still shows the user//pass I have tried restarting and it still saves the dam session! Should it kill the session if i close my browser? wtf? I am trying to pull in computer information using PHP for program compatibility and want to know if it is possible and if so, what functions to use. I am running php on one computer at work and want other users on other computers to access the index page by sharing it.after mapping the page, I see no data on the other machine. Does php, as pache need to beer installed on other computers footer it too work properly i have recently bought a new samsung netbook so i downloaded the latest version of WAMP to develop my site. However when i run my site i get the internal server error associated with the htaccess file. The thing is that this file works on my desktop pc. The version i have of WAMP on this netbook is newer than the one i have at home, socould this be the problem? i can post the contents of the htaccess file if you think it will help, its just a lot of rewrite rules. Thanks Hey, I'm trying to code a page which is supposed to bring up the time of a person's computer. Is this possible without majorly complicated coding? I'm good with GMT offsets and stuff but I don't know how to get the time of a person's computer. I don't want other stuff, just the time so I can edit the time based on where they are. I don't know if I'm making sense so here's an example... If someone does a forum post at 2pm Sydney time but it's 5pm Perth time. How can I make it so that someone in Perth sees 5pm but someone in Sydney sees 2pm...without them manually telling me their timezone? I read about using IP addresses but don't know much (Ok anything) about this , any hints or good sites to view would be appreciated? Or the answer if it's not too complex. Preferably I'd like to get the time in the 11 digit "time()" format so I can use it in tables and translate it with an existing function. IceKat I know about
<input type="file" >which allows one to upload files from one’s computer. But now I need a pop-up window that allows the user to choose first if he uploads from his computer or from the internet. How may I achieve that ? This question is perhaps better suited for the "Javascript" or even "AJAX" forum ; if someone advises me to, I will move that question and put it in the appropriate forum. Hi, i have problem with libraries, with fpdf and PHPEcel, problem is same... Can't open file C:\xampp\htdocs\skripta\.... - this is path to php script on my computer everything work ok on localhost when i test, but when i upload to web server, pdf script try to create file with old windows path: Can't open file C:\xampp\htdocs\skripta\, and not with linux path... does anyone have solution, and why is that? Hi, I have a challenging dilema. A client keeps track of his stock using a microsoft access database in store. His ecommerce website which is hosted elsewhere uses php and mysql. Is it possible to to integrate the 2 so that when someone buys a product from the website, the store MS access database is updated to reflect this and vice versa? If so how would I go about it? Thanks in advance. P.S: I'm extremely tempted to suggest that he should stop using his ms access db and instead reccommend that I set him up a suitable backend on his website for keeping stock control, then he just login to that in the store. This topic has been moved to Miscellaneous. http://www.phpfreaks.com/forums/index.php?topic=318425.0 Hello Good day,
I am have a project for company. Can you please give me advice on how will I going to determine each computer assigned to each employee.
I have my own working unit (PC), of course that i used to login on our company web site. But when someone try to log in on our web site using their own account on my working unit it will be registered as fraud.
My problem is how will i am going to check if accountTwo try to log in on accountOne pc unit. Right now we're using cookie but if they use other browser or private browser the data on the cookie was not send on the request.
Righ now I am trying to determine using Mac Address of each PC, but still can't figure out how to get the Mac Address. It gives the server address where the site was hosted.
Is there a reason why on my local machine with php, my login system works, but then when I upload my files and test it on my web host, the login system doesn't work properly. I'm using the same php version on my local machine and on my webhost, so I don't understand why it wouldn't work the same. When I try logging in on my webhost, everything processes as normal when a correct user/pass combo is found in the database, however, my session just doesn't seem to be saved, and therefore I won't be logged in. It'll end up refreshing to the home page (as I have it setup), but it won't show me as logged in. Is there something special I need to do in order for the session to be stored correctly? (I realize I haven't pasted any code, but I'm not sure exactly how much code would be needed for me to show in order to resolve the issue).
<?php
shell_exe ("172.18.9.25\photoshop\photoshop.exe");
?>
When I run this from the server computer program opening. But when i tried from a diferent computer it is only downloading the exe file to the client computer. How can I open the application from a diferent computer in the same network. heres is my code for time but it ddnt the time i get was not same in my computer Code: [Select] $date = date("y-d-m") . " at " . date("h:i:s"); $time = time(); Am new here - looks like a great foru! I would sincerely appreciate any help anyone can give me. I have been trying to solve my problem for hours and I am not having any luck, so I thought I would post and see if anyone can help. I am very stuck and am not making much progress on this project, and I am certain the answer is very simple. I am constructing a form to collect data for a specialized purpose. The form and program actually work for its intended function, but I am trying to enhance the user experience by preventing customers from having to reenter all of their data should there be a problem with any of the data submitted. I have been able to do that with the contact form portion, but what I am having trouble with is the portion which has as many as 400 possible entries. So, in a nutshell, if the customers contact data is incomplete or in error, the form will ask them to return to the page and correct things. The previous data entered has been saved in the session and the input value will equal the previous entry. i.e. <tr> <td align="right" class="infoBox"><?php echo ENTRY_EMAIL_ADDRESS; ?></td> <td align=left><?php echo "<input type=text name='cemail' value=\"$cemail\" size=35 maxlength=35>" ?></td> </tr> Works perfectly, all well and good there. On the other 400 more or less entries, I am having a difficult time tweaking the string concatenation to work to achieve similar results. There are 4 columns each with $points entries asking for a dimension in either feet or inches. The <input name=> is one of ptaf,ptai,ptbf,ptbi, appended programatically with the corresponding row number or data point. i.e. "ptaf1", "ptai1", etc... This is produced by the example below and works perfectly also. <?php { $points=100; $i=1; while ($i <= $points) {echo ' <tr><td align="center" width="6"><b> ' .$i . '</b></td> <td align="right" NOWRAP>A' .$i . ' (ft) <input type="text" name="ptaf'.$i.'" size=4 maxlength=3> </td> <td align="right" NOWRAP>A' .$i . ' (in) <input type="text" name="ptai'.$i.'" size=4 maxlength=4> </td> <td align="right" NOWRAP>B' .$i . ' (ft) <input type="text" name="ptbf'.$i.'" size=4 maxlength=3> </td> <td align="right" NOWRAP>B' .$i . ' (in) <input type="text" name="ptbi'.$i.'" size=4 maxlength=4> </td> '; $i++; } } ?> I am trying to add <input value=$ptai.$i> for each field but as I mentioned I am not having any luck. It seems as if I have tried every combination imagineable, but still no luck. My head is spinning! The closest I seem to have gotten was with this: <td align="right" NOWRAP>A' .$i . ' (ft) <input type="text" size=6 maxlength=3 name="ptaf'.$i.'" value="' . "$ptaf" . $i . '" ></td> But line 17 for example returns this: <input type="text" value="17" name="ptaf17" maxlength="3" size="6"> To recap, I am trying to have the value set to whatever the customer may have entered previously. Again, I would most appreciate any help anyone can give me. If you need clarification on anything please let me know. Thanks AJ Hello to all, I have problem figuring out how to properly display data fetched from MySQL database in a HTML table. In the below example I am using two while loops, where the second one is nested inside first one, that check two different expressions fetching data from tables found in a MySQL database. The second expression compares the two tables IDs and after their match it displays the email of the account holder in each column in the HTML table. The main problem is that the 'email' row is displayed properly while its while expression is not nested and alone(meaning the other data is omitted or commented out), but either nested or neighbored to the first while loop, it is displayed horizontally and the other data ('validity', 'valid_from', 'valid_to') is not displayed.'
Can someone help me on this, I guess the problem lies in the while loop? <thead> <tr> <th data-column-id="id" data-type="numeric">ID</th> <th data-column-id="email">Subscriber's Email</th> <th data-column-id="validity">Validity</th> <th data-column-id="valid_from">Valid From</th> <th data-column-id="valid_to">Valid To</th> </tr> </thead> Here is part of the PHP code:
<?php
while($row = $stmt->fetch(PDO::FETCH_ASSOC))
{
echo '
<tr>
<td>'.$row["id"].'</td>
';
while ($row1 = $stmt1->fetch(PDO::FETCH_ASSOC)) {
echo '
<td>'.$row1["email"].'</td>
';
}
if($row["validity"] == 1) {
echo '<td>'.$row["validity"].' month</td>';
}else{
echo '<td>'.$row["validity"].' months</td>';
}
echo ' <td>'.$row["valid_from"].'</td>
<td>'.$row["valid_to"].'</td>
</tr>';
}
?>
Thank you. I have two tables. Table Name:Users Fields: User_name user_email user_level pwd 2.Reference Fields: refid username origin destination user_name in the users table and the username field in reference fields are common fields. There is user order form.whenever an user places an order, refid field in reference table will be updated.So the user will be provided with an refid Steps: 1.User needs to log in with a valid user id and pwd 2.Once logged in, there will be search, where the user will input the refid which has been provided to him during the time of order placement. 3.Now User is able to view all the details for any refid 3.Up to this we have completed. Query: Now we need to retrieve the details based on the user logged in. For eg: user 'USER A' has been provided with the referenceid '1234' during the time of order placement user 'USER B' has been provided with the referenceid '2468' during the time of order placement When the userA login and enter the refid as '2468' he should not get any details.He should get details only for the reference ids which is assigned to him. <?php session_start(); if (!$_SESSION["user_name"]) { // User not logged in, redirect to login page Header("Location: login.php"); } $con = mysql_connect('localhost','root',''); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("login", $con); $user_name = $_POST['user_name']; $refid = $_POST['refid']; $query = "SELECT * from reference,users WHERE reference.username=users.user_name AND reference.refid='$refid' AND "; $result = mysql_query($query) or trigger_error('MySQL encountered a problem<br />Error: ' . mysql_error() . '<br />Query: ' . $query); while($row = mysql_fetch_array($result)) { echo $row['refid']; echo $row['origin']; echo $row['dest']; echo $row['date']; echo $row['exdate']; echo $row['username']; } echo "<p><a href=\"logout.php\">Click here to logout!</a></p>"; ?> <html> <form method="post" action="final.php"> Ref Id:<input type="text" name="refid"> <input type="submit" value="submit" name="submit"> </html> Here's the code that deals with the client side:
<?php
session_start();
if(!isset($_SESSION['Logged_in'])){
header("Location: /page.php?page=login");
}
?>
<!DOCTYPE Html>
<html>
<head>
<!--Connections made and head included-->
<?php require_once("../INC/head.php"); ?>
<?php require_once("../Scripts/DB/connect.php"); ?>
<!--Asynchronously Return User Names-->
<script>
$(document).ready(function(){
function search(){
var textboxvalue = $('input[name=search]').val();
$.ajax(
{
type: "GET",
url: 'search.php',
data: {Search: textboxvalue},
success: function(result)
{
$("#results").html(result);
}
});
};
</script>
</head>
<body>
<div id="header-wrapper">
<?php include_once("../INC/nav2.php"); ?>
</div>
<div id="content">
<h1 style="color: red; text-align: center;">Member Directory</h1>
<form onsubmit="search()">
<label for="search">Search for User:</label>
<input type="text" size="70px" id="search" name="search">
</form>
<a href="index.php?do=">Show All Users</a>|<a href="index.php?do=ONLINE">Show All Online Users</a>
<div id="results">
<!--Results will be returned HERE!-->
</div>
search.php
<?php //testing if data is sent ok echo "<h1>Hello</h1><br>" . $_GET['search']; ?>This is the link I get after sending foo. http://www.family-li...php?&search=foo Is that mean it was sent, but I'm not processing it correctly? I'm new to the whole AJAX thing. |
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