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Found 10 results

  1. Using Joomla 2.5 and Virtuemart 3. I've had a client tell us that they're getting e-mails (admin order confirmation email) that are different orders from the same person at the same time. After some test orders I've seen that what happens is when PayPal is used, and the customer is being redirected or has been, they then press the back button in the browser, amend the order and send it again. This sends 2 e-mails to the shop and causes confusion. Is there a way to stop this? (Other than a big bold message warning not to do it)
  2. Say I have button 1 and I want to replace it with button2 on click. How would I do that? <button id="button1">butto 1</button> <button id="button2">butto 2</button>
  3. This twitter style button I found. http://jsfiddle.net/jakie8/ZECEN/ I have managed to modify it to my liking. And yes I know the ".live" is depreciated and instead use "on". Anyways, the only thing I noticed is that if I click "Follow" button and it turns green; once I refresh the page, it goes back to default "follow" mode. I was wondering how can I keep it green even if I refresh the page?
  4. The like/dislike button script found here: http://wcetdesigns.com/tutorials/demos/like-dislike-button.html works perfectly for what I am trying to do on my webpage. The issue I am having is I can connect to one table perfect but the script breaks if I try to add another set of like/dislike buttons that connects to another table. How can I add 2 additional like/dislike button sets that can connect to 2 addtional tables in my database. For example Topic name (like/dislike button set that connects to table 1) Messages (like/dislike button set that connect to table 2) Message Replies (like/dislike button set that connects to table 3) any help would be appreciated. Thanks
  5. $liked = explode(":", $_POST[submit]); if($liked[0] == "like"){ //Select from database where id is $liked[1] } <input type="image" src="images/arrowup.gif" name="submit" value="like:<? echo $right->id; ?>"/> Ok, so i have made a forum and i have a like function running on the comments of a post. I needed to find a way when the image was clicked it updated the database. So i experimented and came up with the solution above having it explode the value of the button to read if it was press and get the id. It works perfectly but i just want to be convinced it is safe.
  6. hi ive been trying to sort this out all day, im very inexperianced in php but really want to do well in it and at the moment ive got a table working linked to a database and which pulls records but now i want to get an update and delete button but i cant seem to hack it heres the code. <?php $con = mysqli_connect("localhost","root","","lcm"); // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } //updates table. if (isset($_POST['update'])){; $UpdateQuery = "UPDATE tbl_contactinfo SET Name='$_POST[Name]', Address='$_POST[Address]' WHERE Name='$_POST[hidden]'"; mysql_query($UpdateQuery, $con); } if (isset($_POST['delete'])){; $DeleteQuery = "DELETE FROM tbl_contactinfo WHERE Name='$_POST[hidden]'"; mysql_query($DeleteQuery, $con); } //selects the table. $sql = "Select * from tbl_contactinfo"; if ($result = mysqli_query($con, $sql)) { echo "<table border='1'> <tr> <th>Name</th> <th>Address</th> </tr>"; while($row = mysqli_fetch_array($result)) { echo '<form action="index1.php" method="post">'; echo '<tr>'; echo '<td><input type="text" name="name" value="' . $row['Name'] . '" /></td>'; echo '<td><input type="text" name="address" value="' . $row['Address'] . '" /></td>'; echo '<td><input type="hidden" name="hidden" value="' . $row['ID'] . '" /></td>'; echo '<td><input type="submit" name="update" value="update" /></td>'; echo '<td><input type="submit" name="delete" value="delete" /></td>'; echo '</tr>'; echo '</form>'; } echo "</table>"; } ?> <html> <body> <form action="insert.php" method="post"> Name: <input type="text" name="Name"> Address: <input type="text" name="Address"> <input type="submit"> </form> </body> </html> and error messages ive been getting. ( ! ) Notice: Undefined index: Name in C:\wamp\www\Index1.php on line 12 Call Stack # Time Memory Function Location 1 0.0000 147208 {main}( ) ..\Index1.php:0 ( ! ) Notice: Undefined index: Address in C:\wamp\www\Index1.php on line 12 Call Stack # Time Memory Function Location 1 0.0000 147208 {main}( ) ..\Index1.php:0 ( ! ) Warning: mysql_query() expects parameter 2 to be resource, object given in C:\wamp\www\Index1.php on line 13 Call Stack # Time Memory Function Location 1 0.0000 147208 {main}( ) ..\Index1.php:0 2 0.1562 154608 mysql_query ( ) ..\Index1.php:13 these are for update and for delete ( ! ) Warning: mysql_query() expects parameter 2 to be resource, object given in C:\wamp\www\Index1.php on line 18 Call Stack # Time Memory Function Location 1 0.0000 147208 {main}( ) ..\Index1.php:0 2 0.1406 154456 mysql_query ( ) ..\Index1.php:18 ive stoped the error messages before dont no how i did it but they wernt showing, only problem is when i pressed them nothing happened any help will be great.
  7. hi ive been trying to sort this update button out but i keep getting errors and got no clue how to fix it. heres the code. <?php $con=mysqli_connect("localhost","","","lcm"); // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } $sql = "Select * from tbl_contactinfo"; if (isset($_POST['update'])){ $UpdateQuery = "UPDATE tbl_contactinfo SET Name='$_POST[Name]', Address='$_POST[Address]' WHERE Name = '$_post[hidden]'"; mysql_query($UpdateQuery,$con); }; if ($result = mysqli_query($con, $sql)){ echo "<table border='1'> <tr> <th>Name</th> <th>Address</th> </tr>"; while($row = mysqli_fetch_array($result)) { echo "<form action = index1.php method =post>"; echo "<tr>"; echo "<td>" . "<input type= text name =Name value =" . $record['Name'] . "</td>"; echo "<td>" . "<input type= text name =address value =" . $record['Address'] . "</td>"; echo "<td>" . "<input type= hidden hidden =Name value =" . $record['hidden'] . "</td>"; echo "<td>" . "<input type= submit name = update value=update". " </td>"; echo "</form>"; } echo "</table>"; mysql_close($con); ?> if someone could help id really apreciate it.
  8. I have created a PHP web page that allows me to control a serial device (a speaker selector matrix), and an Arduino (to switch an amplifier off and on) over my network. I am trying to make it so the text color changes on what ever button is pressed. The button presses are stored as a variable and then POSTed. I would like the text on the current button selected to turn red. There are 5 separate sets of buttons, so each set could have a button with red text at any time. Eventually, I would like to have a graphic next to the button which would change based on a button press. (baby steps!) Attached is the web page. Please keep any answers in laymans terms, since I am a real PHP noob! This webpage is the result of A LOT of trial and error, but it works! Any suggestions are welcome! audio.php
  9. I have this array that has n elements ---> list_array = new Array(); it contains ['Product 1','Product 2','Product 3','Product 4','Product 5','Product 6','Product 7','Product 8'] what I am trying to do is that starting a loop from j=0 to j<n, I am adding button on html page, now when I click on button, it must remove itself from the page. I want to do this with the help of javascript...Is there anyone who can help ??? If I click on "product 8", the page must display other 7 buttons and it must delete the button itself on the page..... function create() { for(j=0;j<i;j++) { "<button onclick='remove_me(this)' id="+list_array[j]+">Remove</button><br>"; } } function remove_me(obj) { _button = obj.id; remove(_button); } I am trying to perform this operation, however, it only receive "Product" in obj.id.............It skips the text after the first whitespace..... I hope that you got my concern, looking for quick reply.......
  10. I have integrated a facebook share button with the script from developers and now when I use the share button the data fetched is the old data which is not even in my database what can be the problem. I cleared all the cache data from browser still it takes the data that has already been deleted from the website. Please help me,I am not getting the problem.
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