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PHP - Unable To Scroll My Website On Android Chrome And Safari
As mention in my title, I am using wordpress, bought a theme and the support team is not able to assist me without the FTP access.
URL: http://www.icv.solutions/swagger
Mobile: Android 4.4.2
Chrome: version 39.0.2171.93
Safari: star safari version 1.1
I am not able to scroll only on the above browser, other browser like firefox and default internet explorer on mobile have no such issue.
In fact the older version of chrome have no such issue too,
If anyone could give me a hint where might the problem lies at or any help at all, would be great.
I have been struggling with this for awhile now.
Thanks and happy holiday!
Similar TutorialsChrome is the browser I use, I have it set up so that the script detects safari and chrome as separate entities. When I browse to it, it shows: You are using Windows XP with a Chrome Web Browserwith a Safari Web Browser now I know stristr() is a case-insensitive version of strstr(), which will find the first form of a word in a string. The problem is that when I use it, the rest of the script doesn't work... it goes to the part as if nothing is there, and shows the text on the bottom. I can't figure out why. Can someone help me, is there a chrome/safari level of priority or am I just missing something? Code: [Select] <?php //Booleans to set OS and Browser to False. $os = false; $browser = false; //Booleans for Web Browser & OS Functions. $xp = xp(); $vista = vista(); $win7 = win7(); $ubuntu = ubuntu(); $chrome = chrome(); $safari = safari(); $firefox = firefox(); $ie9 = ie9(); $ie8 = ie8(); //Operating Systems function xp(){return(preg_match("/Windows NT 5.1/", $_SERVER['HTTP_USER_AGENT']));} function vista(){return(preg_match("/Windows NT 6.0/", $_SERVER['HTTP_USER_AGENT']));} function win7(){return(preg_match("/Windows NT 6.1/", $_SERVER['HTTP_USER_AGENT']));} function ubuntu(){return(preg_match("/Ubuntu/", $_SERVER['HTTP_USER_AGENT']));} //Web Browsers function chrome(){return(preg_match("/Chrome/", $_SERVER['HTTP_USER_AGENT']));} function safari(){return(preg_match("/Safari/", $_SERVER['HTTP_USER_AGENT']));} function firefox(){return(preg_match("/Firefox/", $_SERVER['HTTP_USER_AGENT']));} function ie9(){return(preg_match("/MSIE 9.0/", $_SERVER['HTTP_USER_AGENT']));} function ie8(){return(preg_match("/MSIE 8.0/", $_SERVER['HTTP_USER_AGENT']));} // If you have one of the valid Operating System. if($ubuntu){echo 'You are using Ubuntu Operating System ';} if($xp){echo 'You are using Windows XP ';} if($vista){echo 'You are using Windows Vista Operating System ';} if($win7){echo 'You are using Windows 7 Operating System ';} // If you have one of the valid Web Browser. if($chrome){echo 'with a Chrome Web Browser';} if($firefox){echo 'with an Firefox Web Browser';} if($safari){echo 'with a Safari Web Browser';} if($ie8){echo 'with an Internet Explorer 8 Web Browser';} if($ie9){echo 'with an Internet Explorer 9 Web Browser';} //If OS or Browser not found in list. if ($ubuntu || $xp || $vista || $win7) $os = true; if($firefox || $chrome || $safari || $ie9 || $ie8) $browser = true; if(!$browser || !$os){ echo'<strong>'; echo '<br />' . $_SERVER['HTTP_USER_AGENT'] . '<br /><br />Administrator someone in your work force is using an unsupported browser or OS, please email this information to the developer of the NCMR software you are using. It will allow your browser/OS combination to be used correctly. Sorry for the inconvenience.</strong> <br /><br />Please copy and paste the text above and send it to your web administrator. It will explain everything he/she needs to do.<br />';} ?> Hi.I found out that PHP developers can make apps for android too with zend which I'm really interested in.Does anyone have any resource or tutorial teaching how to do that?for now the apps I'm going to develop are just going to contain Text.not advanced apps do crazy things.
BTW I'm new to php and trying to learn fudamentals etc.If I want to go into app development wouldn't it confuse me?can I do that?thanks and sorry for my english
I want to build a online "app builder" project that users can create android / iphone applications online without any coding knowledge. These sites are very common in market. They offer onclick app builder. I have skill in development of web applications using php frameworks. My question is How we can generate android apk from our server after getting the necessary information from user? (Appname, icon, packagename etc) PHP can do this entire task?, if yes any framework for that? Can you give me some basic tips to generate apk from our server? Can we genrate both android and iphone app from one single code? What are the requirements needed for the server? Anybody having skills in these areas, please help me. I need basic tips to get started this project. Thanks Good morning! I was wondering how can i access the php data to the health data (steps, heart rate etc ..) of devices with android and ios? I don't know maybe by inserting the keys I can access the json with the related data... I tried to search a bit on the internet but I see that I can only do it if I create an app for both systems and then from there I enter the data in my mysql database.. I don't know where to start, do you know something? Thank you I have php code that uses mp3 audio files stored in a database for the user to listen to. The page loads and the audio controls work and the files play on a desktop and laptop just fine. When trying to play a file on an android or IPhone, when you press the play (or any other control) nothing happens. It acts like the page is static. There is a home button on the webpage and it works so I know the page is interactive. Below is the code. I would appreciate any help. Thank you.
<?php I inherrited a site that has PHP code that decides to either serve the dedicated mobile site or desktop site to the user. I would prefer that tablets get served the mobile site. Currently Android tablet users get served the desktop site. I'd like to figure out how I can implement User Agent to detect an android tablet (with distinction from an Android phone) This is the code I'm working with to determine which site to serve (mobile or desktop) Thanks if ((strpos($user_agent, 'Mobile') === false && // Generic mobile browser string, most browsers have it. strpos($user_agent, 'SymbianOS') === false && // Nokia device running Symbian OS. strpos($user_agent, 'Opera M') === false && // Opera Mini or Opera Mobile. strpos($user_agent, 'Android') === false && // Android devices that don't have 'Mobile' in UA string. stripos($user_agent, 'HTC_') === false && // HTC devices that don't have 'Mobile' nor 'Android' in UA string. Case insensitive. strpos($user_agent, 'Fennec/') === false && // Firefox mobile strpos($user_agent, 'Kindle') === false && // Kindle Fire tablet strpos($user_agent, 'BlackBerry') === false) || // BlackBerry strpos($user_agent, 'iPad') === false) // iPad { // Serve Desktop Site return false; } I have a little php site that looks up data in my crm database quickly and easily on my iphone. Right now it's wide open because I'm in and out of it all day and trying to input uid/pw into a secure login page is a pain in the rear. (we're not talking about sensitive data here and I'm the only user.) So, I would like to design a screen with several large graphic buttons and be able to touch those buttons in a specific sequence that would functionally allow a secure login. Similar in concept to the android screen lock. You touch a sequence of numbered buttons and voila, the screen unlocks. It does not open a text field for data entry into two separate fields. So if my access code was 1,2,3,4 and the screen was divided into quarters with a large number on each one, I would touch the buttons in that order and it would open the main page. How would this best be accomplished? Is there a php function to essentially build a variable through successive key presses? then with a final submit it would just check that against a "secret code" and allow access. Thanks for any input. Hello, I've never really used the update command before for mysql and I'm attempting to use it and struggling a little bit. I'm trying to use mysqli prepared statements.. here's the code that I have thus far: if($query = $database->connection->prepare("UPDATE videos SET comments=?, views=?, uploader=? WHERE title = ?")) { $query->bind_param('iiss', $comments, $views, $uploader, $title); $query->execute(); $result = $query->affected_rows; $query->close(); } For some reason I cannot get this working. I have created a modification page for the administrators to be able change any of the values and wanting to update the database to reflect the changes. When using the MySQL UPDATE command do all of the values have to get changed or modified, or am I able to pass back some of the same values? Like with the above code.. if I only wanted to update the views, would I still be able to just pass in the same values for comments and uploader and it would just replace the values? Hi all,
I am not sure why my header is not displaying the header image after using the CSS
I have a png file that repeats horizotally.
Please help
<!DOCTYPE html> <html> <head> <meta charset="UTF-8"> <title>Wikigets The online store</title> <style type="text/css"> body { margin:0 px;} #pageTop { background: url(style/headerline1.png); height:110 px; } </style> </head> <body> <div id="pageTop"> </div> <div id="pageMiddle"></div> <div id="pageBottom"></div> </body> </html> headerline1.png 2.82KB 0 downloads headerline1.png 2.82KB 0 downloads I get the unable to jump to row zero mysql error. Code: [Select] function is_admin($uid, $cid) { $uid = (int)$uid; $cid = (int)$cid; $sql = "SELECT `users`.`id` AS `uid`, `companies`.`companyid` AS `cid`, `companies`.`adminid` AS `aid` FROM `companies` LEFT JOIN `users` ON `users`.`id` = companies.adminid WHERE `users`.`id` = {$uid} AND `companies`.`companyid` = {$cid}"; $user = mysql_query($sql); return (mysql_result($user, 0) == '1') ? true : false; } Code: [Select] <?php function checking_out() { $conn = db_connect(); $nickname=$_SESSION['valid_user']; $query="select sum(price) from preorders where name='".$nickname."'"; $result = $conn->query($query); if ($result) { echo '<h1>'.$result.'</h1>'; } } ?> This is not working, there is no result in the browser, any idea ? I'm trying to make a League of Legends (a video game) community website, both as a personal project and for practice. Now the game has a lot of champions, each of whom have 5 unique abilities. Now, I thought about manually inputting all the details about each champion into a MySQL database, but that would long and tedious, and I don't really have the time for it now. Also, the game patches very oftern (like, once every 2 weeks) which changes many of the stats, etc. of the champion, and it is not possible for me to keep manually updating these every time there is a patch. Fortunately, there is a League of Legends Wiki which has all the data I need in their specific champion pages, which they keep updated per patch. So I was wondering if there was any way to get the data from the divs in the wiki, and have it display on my site. What I want to do in my website is that whenever someone types a champion's name (in a post or whatever), I want it to display a hover-over dialog with some of the champions details. And a lot of other features such as that. In plain English I need a way to : > Tell PHP to go to the wiki's source code on a specific page > Find a specific div container > Get X data from there > Pass X data into a function to display the hover-over I think this way, I would not have to maintain a database as I can leech off the wiki's data. I have not coded anything like this before, so I would like a few pointers as to how to achieve this. Any help will be appreciated! my firefox won't open up .tpl format im running apache...how can i view that format on my apache please? Hi All,
I have this issue when upload a file using an uploader i made. For the life of me I cant figure out why it won't write the file.
Error:
[Sat Jun 21 20:45:40 2014] [error] [client xxxxxxx] AH01215: PHP Warning: move_uploaded_file() [<a href='function.move-uploaded-file'>function.move-uploaded-file</a>]: Unable to move '/tmp/phpUsCyXG' to '/home/sites/xxxxxx.co.uk/public_html/yourphotos/uploads/' in /home/sites/xxxxxxx.co.uk/public_html/yourphotos/index.php on line 31My PHP
<?php
//if they DID upload a file...
$message = '';
if($_FILES['photo']['name'])
{
$valid_file = true;
//if no errors...
if(!$_FILES['photo']['error'])
{
//now is the time to modify the future file name and validate the file
$new_file_name = strtolower($_FILES['photo']['tmp_name']); //rename file
if($_FILES['photo']['size'] > (26214400)) //can't be larger than 25MB
{
$valid_file = false;
$message = 'Oops! Your file\'s size is to large.';
echo $_FILES['photo']['size'];
}
if($_FILES['photo']['size'] < (1572864)) //can't be smaller than 1.5MB
{
$valid_file = false;
$message = 'Oops! Your file\'s size is to small.';
echo $_FILES['photo']['size'];
}
//if the file has passed the test
if($valid_file)
{
//move it to where we want it to be
move_uploaded_file($_FILES['photo']['tmp_name'], '/home/sites/xxxxxxxx.co.uk/public_html/yourphotos/uploads/');
$message = 'Congratulations! Your file was accepted.';
}
}
//if there is an error...
else
{
//set that to be the returned message
$message = 'Ooops! Your upload triggered the following error: '.$_FILES['photo']['error'];
}
}
?>
<html>
<body>
<form action="index.php" method="post" enctype="multipart/form-data">
Your Photo: <input type="file" name="photo" size="25" />
<input type="submit" name="submit" value="Submit" />
<?PHP echo $message; ?>
</form>
</body>
</html>
I have set the uploads file to have write permissions as well.
Sam
Edited by samtwilliams, 21 June 2014 - 02:51 PM. I wrote a very simple web form that allows my user to view text files from within their Internet browser. Occasionally, the search criteria entered returns more than one file. So I want to implement a feature whereby the text files returned by the search are compressed into a ZIP. I got a prototype working but it only compresses the first file. The second or third files are ignored. Here's my code Code: [Select] <HTML><body><form name="myform" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> <fieldset><label for="DBRIDs">RIDs</label><input type="text" id="DBRIDs" name="DBRIDs" > </fieldset></form></body></HTML> <?php function check_search() { if (isset($_POST['submit'])) {if (!empty($_POST['DBRIDs'])) { $results = getFiles(); } } else $errors = "Please enter something before you hit SUBMIT."; return Array($results, $errors); } function getFiles() { $result = null; $ZIPresult = null; if (empty($_POST['DBRIDs'])) { return null; } $mydir = MYDIR; $dir = opendir($mydir); $DBRIDs = $_POST['DBRIDs']; $getfilename = mysql_query("select filename from search_table where rid in (" . $DBRIDs . ")") or die(mysql_error()); while ($row = mysql_fetch_array($getfilename)) { $filename = $row['filename']; $result .= '<tr><td><a href="' . basename($mydir) . '/' . $filename . '" target="_blank">' . $filename . '</a></td></tr>'; $ZIPresult .= basename($mydir) . '/' . $filename; } if ($result) { $result = "<table><tbody><tr><td>Search Results.</td></tr> $result</table>"; shell_exec("zip -9 SearchResult.zip ". $ZIPresult ." > /dev/null "); } return $result; } The hyperlinks pointing to the file(s) are generated just fine. The ZIP file however only contains the first file listed in the results. How can I get the ZIP file to capture ALL the files returned?? Thanks for your input. **PS: The new ZipArchive() library/class is not available on our production environment so I chose to use the Unix utility ZIP instead.** Hi all, Newbie here, i am having a problem to get my images to show which are stored in mysql database as a mediumblob. I get id number to print in table ut am just getting empty square with red cross in where my image should be. Is my code incorrect or is it something else? Appreciate your help with this. I have included both of the pages codes i am using. Thanks Tony image2.php <?php include("common.php"); error_reporting(E_ALL); $link = mysql_connect(host,username,password) or die("Could not connect: " . mysql_error()); mysql_select_db(db) or die(mysql_error()); $sql = "SELECT id FROM photos"; $result = mysql_query("$sql") or die("Invalid query: " . mysql_error()); ?> <table border="1"><tr><td>id</td><td>image</td></tr> <?php while($row=mysql_fetch_assoc($result)){ print '<tr><td>'.$row['id'].'</td><td>'; print '<img src="image1.php?id='.$row['id'].'height="75" width="100"">'; } echo '</td></tr></table>' ?> image1.php <?php ob_start(); include("common.php"); mysql_connect(host,username,password) or die(mysql_error()); mysql_select_db(db) or die(mysql_error()); $query = mysql_query("SELECT imgage FROM photos WHERE id={$_GET['image_id']}"; $row = mysql_fetch_array($query); $content = $row['image']; header('Content-type: image/jpg'); echo $content; } ob_end_flush(); ?> Hi I have got a database table for logging in. One user will not log in even though the data is valid Code: [Select] $user = ValidateKey($_SESSION["Name"] , $_SESSION["PWD1"] , $_SESSION["PWD2"]); // User if($Error == 1 || $user == 0 || $user == -1) { // header('Location: index.php'); echo $Error . " , " . $user; // result is 0, caf9eba77c55ab5ae81a01c25d1987d3 exit; } All other user are OK! Strange Desmond. Unable to execute query (SELECT * FROM lb-players) in the database : You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '-players' at line 1 I don't know why i am getting this error. All i can think of that is taking away lb from "lb-players" when you add a " - "? Code: [Select] [m]<?php $conn = mysql_connect("23.23.23.23", "unknown", "itsAsecertxD");//ignore this xD if (!$conn) { echo "Unable to connect to the database : " . mysql_error(); exit; } if (!mysql_select_db("minecraft-rc")) { echo "Unable to select database mydbname : " . mysql_error(); exit; } $sql = 'SELECT * FROM lb-players'; $result = mysql_query($sql); if (!$result) { echo "Unable to execute query ($sql) in the database : " . mysql_error(); exit; } if (mysql_num_rows($result) == 0) { echo "No rows found, nothing to display."; exit; } while ($row = mysql_fetch_assoc($result)) { echo $row["playername"]; } mysql_free_result($result); ?>[/m] tell me if i posted this in the wrong place its been forever since i last been on this site :$ In the given code below I am getting this error Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\samples\getcustomers.php on line 20 WHY? Code: [Select] <?php $value = $_GET["q"]; $con = mysql_connect('localhost', 'root', ''); //db_connect(); if (!$con) { die('Could not connect: ' . mysql_error()); } $db_selected = mysql_select_db("ajaxDb",$con); $sql = "SELECT * FROM 'cutomers' WHERE Name = '".$value."'"; // echo $sql; $result = mysql_query($sql); echo "<table border='1'> <tr> <th>Name</th> <th>Address</th> <th>Country</th> </tr>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['Name'] . "</td>"; echo "<td>" . $row['Address'] . "</td>"; echo "<td>" . $row['Country'] . "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> |
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