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PHP - Moved: How Can I Make This Clock Ticking On My Website??(without Refreshing The Page)
This topic has been moved to JavaScript Help.
http://www.phpfreaks.com/forums/index.php?topic=349726.0 Similar TutorialsThis topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=330523.0 This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=347117.0 This topic has been Ctrl+X/Ctrl+V'd to Miscellaneous. http://www.phpfreaks.com/forums/index.php?topic=347400.0 This topic has been moved to Third Party PHP Scripts. http://www.phpfreaks.com/forums/index.php?topic=354321.0 Stop posting non-PHP questions in the PHP Coding Help section. We have many sub-forums, including both a JavaScript forum and one dedicated entirely to Ajax. This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=327454.0 This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=317523.0 It seems to only wanting to tick the enabled radio button even if the row value is 1 for disabled in the database. <?php while ( $row = mysqli_fetch_array ( $result, MYSQL_ASSOC ) ) { echo ' <tr> <td>' . $row['fullname'] . '</td> <td style="text-align:center;"><input type="radio" value="0" name="' . $row['fullname'] . '" class="status"'; if($enabled == 0) echo ' checked="checked"'; echo '/>Enabled'; ?> <?php echo '<input type="radio" value="1" name="' . $row['fullname'] . '" class="status"'; if($enabled == 1) echo ' checked="checked"'; echo ' />Disabled</td> </tr>'; } ?> This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=323252.0 Hi, In the following code I'm using a combo box to select a relay and activate it. The code works but if I refresh the page, the last selected relay is activated again. How can I reset the page to default to value '0' which has no relay control? TIA
<!DOCTYPE html>
<html>
<head>
<title>Relay control</title>
<meta content="text/html; charset=utf-8" http-equiv="Content-Type" />
<meta name="apple-mobile-web-app-capable" content="yes" />
<meta name="apple-mobile-web-app-status-bar-style" content="black-translucent" />
<link rel="stylesheet" type="text/css" href="style.css" />
</head>
<body>
<h2><center>Status y control de relés</center></h2>
<?php
$arrleds = array();
if ($_SERVER["REQUEST_METHOD"] == "POST") {
$selected_val = $_POST['sel'];
if ($selected_val == '1'){
$cmd = exec("sudo ./trigger.py b 0");
}
elseif ($selected_val == '2'){
$cmd = exec("sudo ./trigger.py b 1");
}
elseif ($selected_val == '3'){
$cmd = exec("sudo ./trigger.py b 2");
}
....
else {}
}
?>
<table class="center">
<tr>
<td></td>
<td><h1>Activación de relé</h1>
<form action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]); ?>" method="post">
<select name='sel' onchange='submit();'>
<option value='0'>Sin activar</option> <!-- no action -->
<option value='1'>Secadora 1</option>
<option value='2'>Secadora 2</option>
<option value='3'>Lavadora S</option>
....
</select>
</form>
</td>
<td></td>
</tr>
<tr>
<td></td>
<td><a href="index.php">Home</a>
<td></td>
</tr>
</table>
</body>
</html>
i have a form with two buttons, one is for submitting the form while button is for generating a random a code and placing it in a textfield. this works fine but the problem is anytime i click on the generate button, it refreshes the page thereby validating the form , which i don't want. How do i make it perform the function without refreshing the page.... Here is my code if(isset($_POST['pass_btn'])) { $numchars = 8; $chars = explode(',','2,3,4,5,6,7,8,9,a,b,c,d,e,f,g,h,i,j,k,m,n,p,q,r,s,t,u,v,w,x,y,z,2,3,4,5,6,7,8,9,A,B,C,D,E,F,G,H,J,K,L,M,N,P,Q,R,S,T,U,V,W,X,Y,Z,2,3,4,5,6,7,8,9'); $password=''; for($i=0; $i<$numchars;$i++) { $password.=$chars[rand(0,count($chars)-1)]; } } This is set up so when a person pressing the pass_btn, a password is provided. This works just fine. However, it refreshes the page and clears all other data that user may have entered. Suggestions? Hi, I have a "user profile" page and it has a profile image upload form. I have it so that the old profile image is deleted, and the new image is uploaded and the new image is echoed out using the name of the new file from the database. The problem is, when I click submit to add the new image, the old image still stays there and the new image does not show up. Only when I manually click refresh on my browser does the new image show up. That is going to be bad for my users. Please if anyone can help, I'd greatly appreciate it. Below is the code I am using to delete the old image, and code to upload the image as well, when the photo upload form is used. Code: [Select] if(isset($_POST['photoUpload'])) { if(file_exists("images/".$currentUser.".jpg")){ unlink("images/".$currentUser.".jpg"); clearstatcache(); } //reads the name of the file the user submitted for uploading $image=$_FILES['image']['name']; //if it is not empty if ($image) { //get the original name of the file from the clients machine $filename = stripslashes($_FILES['image']['name']); //get the extension of the file in a lower case format $extension = getExtension($filename); $extension = strtolower($extension); //if it is not a known extension, we will suppose it is an error and will not upload the file, //otherwise we will do more tests if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif")) { //print error message echo '<h1>Unknown extension!</h1>'; $errors=1; } } //end if here //was else here //get the size of the image in bytes //$_FILES['image']['tmp_name'] is the temporary filename of the file //in which the uploaded file was stored on the server $size=filesize($_FILES['image']['tmp_name']); //compare the size with the maxim size we defined and print error if bigger if ($size > MAX_SIZE*1024) { echo '<h1>You have exceeded the size limit!</h1>'; $errors=1; } //we will give an unique name, for example the time in unix time format $image_name= $currentUser .'.'.$extension; //$image_name=time().'.'.$extension; //the new name will be containing the full path where will be stored (images folder) $newname=$image_name; //"images/". // Connects to your Database mysql_connect("host", "user", "pass") or die(mysql_error()); mysql_select_db("database") or die(mysql_error()) ; //we verify if the image has been uploaded, and print error instead $copied = copy($_FILES['image']['tmp_name'], "images/".$newname); // update the photo name in the database $result = mysql_query("UPDATE users SET profilePhoto='$newname' WHERE username='$currentUser'") or die(mysql_error()); if (!$copied) { echo '<h1>Copy unsuccessful!</h1>'; $errors=1; } else{ $dir="images/"; echo "<p>Profile Picture Change Successful!</p>"; echo "<img src='{$dir}{$newname}' alt='{$newname}' height='200' width='200' />"; echo "<p></p>"; } } Here is my add driver button I'm using paypal PDT, so when the user has made payment he gets directed back to my site, on that page it executes a special sql query. I noticed I can refresh the page to make duplicate sql queries. How can I avoid this?? <!DOCTYPE html> <html> <head> <script> $(document).ready(function() { if($("#getPIN").click(function(){ <?php $char = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVW XYZ1234567890'; $generated_password = substr(str_shuffle($char), 0, 10); ?> var genNumber; $("#txtpin").val() = genNumber; return false; })); }); </script> </head> <body> <input type = "text" name = "txtpin" id = "txtpin" value = ""> <input type = "button" name = "getPIN" id = "getPIN" value = "getPIN"> </body> </html> Hey you guys need your help... Basically what im trying to do is when the user clicks the getPIN button, the value generated using php is assigned to the txtpin textfield without refreshing the page.... thanks alot guys.... Hello everyone. This is my first post, so be nice! I am building a website that will have a lot of content, similar to a newspaper. I have some pretty good HTML/CSS pages written, but the problem is that I need a way to make things more dynamic. One of my templates has a Header, Left Column, Middle Column, Right Column, and Footer. Everything stays the same from page to page except for the Middle Column (which holds each article). As it stands now, if I had 12 articles, I would have to have 12 nearly duplicate HTML pages which isn't good! I started studying PHP a while ago, but put that on hold to learn HTML/CSS, so I've kinda forgotten how PHP can help me out! Can someone help me figure out how to use PHP to my benefit? Thanks, Debbie How can i make/add facebook like on my website or blog? Can anyone post sample example code? Hey guys I had created a while ago a script for my friend where you can buy points and then redeem stuff with those points, i'm looking for ways to keep my site secu currently what i have done- - protected all mysql queries with mysql_real_escape_string, strip_tags, and addslashes - have a valid SSL certificate on my website - checked if emails are valid for account creation what else can I do? Thank you. Hello !
How I can make a script of my PHP code to work even if my website is not running
Can I make this with php ?
This topic has been moved to HTML Help. http://www.phpfreaks.com/forums/index.php?topic=307669.0 |
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