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PHP - Load Text Dynamically Depending On Button Press?
I have a site that I have an about page with pictures of people on it. I would like to create something that would do the following:
* When you click on one of the images, i need text for that image to load in a text area. I would like to do this rather than have to make a different HTML page for each person, but I am not sure how to go about doing it. I do not have a database for this site, so I will need to pull the text from a text file or something like that. Any ideas on how to get started would be helpful. Thanks in advance! Similar TutorialsI am making a pos(point of sale) application using object oriented php. I have found thermal receipt printers support ESC/POS . i want that when i will press button to print directly , it will do automatically. anyone can help me please. Thanks in advance. I am trying to find a good IDE for PHP and JavaScript. I fiddled around with Visual Code and it was very buggy. The GitHub page said over 800 open errors?!?!? Now I'm looking at PHPStorm and it reads that only PHP in a PHP page or JavaScript in a JavaScript page may be debugged. I am wondering if this is always possible to separate the code like this.
Also, in the following code, what do you all think about using a XMLHttpRequest() to call this right in the beginning of the page with no button or anything. Is this bad form? It creates a dropdown list from the database. Right now it just sits there. This code isn't ready for prepare yet.
<?php
$host = 'localhost';
$user = 'root';
$pass = '';
$database = 'ecommerce';
$options = array(
PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION,
PDO::ATTR_EMULATE_PREPARES => false
);
$dbo = new PDO("mysql:host=$host;dbname=$database", $user, $pass, $options);
$q4 = "SELECT categories.title FROM categories INNER JOIN customers on customers.CustomerID = categories.CustomerID and customers.CustomerID = 1";
echo '<select id = "dropDown1" >';
foreach ($dbo->query($q4) as $row) {
echo "<br>";
echo '<option value = " ';
echo $row['title'];
echo '">';
echo $row['title'];
echo '</option>';
}
echo "<br>";
echo '</select>';
?>
I'm using AJAX.
Edited January 2 by JoshEir I have a template written for me on another project that searches a MySQL database. I'm moving this template to work with a database that only have 10 records in it and instead of having a search option I would just like it to automatically display the database when the page opens. Is this possible? The php page in close to 200 lines. This is the line of the code that has the search option. If you need more I can post it. Thanks Bob Code: [Select] <div id="search"> <form action="parts.php" method="GET"> Search: <input type="text" value="'.htmlentities($_GET['s']).'" name="s"/> <input type="submit" value="Search!"/> </form> </div> I have 10 tables in my database. I want load these 10 table names into a drop-down list. If a user selects a value form the drop-down list, a form should be displayed with related to the table. (form fields should be appropriate table's column headers) When user fill the form and submitted, data should be saved in the selected table. Give me a code example to do this. I have to radio buttons at the login page. Both of them are in group1. The value of the first is 'fancy', the value of the second is 'fast'. Also note that the buttons are on the login page, and i want to change the bg on the main page. Hi, I have a table row that has a dropdown and two textboxes in it. I would like to use a button that allows me to add another row beneath the existing row. It cannot be added to the bottom of the table as there is further content beneath. The content of the dropdown comes from a database query. The dropdown on the new row should have the selection in the first row greyed out. I expect this will need to be a js function but i dont really know where to start with it being dynamic content. This is the table as it stands if this provides clarity of my intentions - picture below
echo "<form method='post' id='staffOrderForm'>";
$stmt->close();
echo "<table id='staffOrderTable' class='table table-striped table-bordered mt-3 text-center'>";
foreach($daterange as $date){
echo"
<tr>
<th class='table-dark' colspan='3'>".$date->format("l - jS F Y")."</th>
</tr>
<tr>
<th class='' colspan='3'>Management </th>
</tr>
<tr>
<th class='col-4'>Name</th>
<th class='col-4'>Start Time</th>
<th class='col-4'>End Time</th>
</tr>
<tr>
<td>
<select class='custom-select managerSelect'>";
$managerRoleId = 1;
$stmt = $conn->prepare("
SELECT user_firstname, user_lastname, user_id
FROM ssm_user
WHERE user_role_id = ? ");
$stmt->bind_param("i", $managerRoleId);
$stmt->execute();
$stmt->store_result();
$stmt->bind_result($ufn, $uln, $uid);
while($stmt->fetch()){echo "<option>".$ufn." ".$uln."</option>";};
echo"</select>
</td>
<td><input class='form-control' type='' name=''></td>
<td><input class='form-control' type='' name=''></td>
</tr>
<tr>
<th colspan='3'>Chefs</th>
<tr/>
<tr>
<th class='col-4'>Name</th>
<th class='col-4'>Start Time</th>
<th class='col-4'>End Time</th>
</tr>
<tr>
<td>
<select class='custom-select chefSelect'>";
$chefRoleId = 2;
$stmt = $conn->prepare("
SELECT user_firstname, user_lastname, user_id
FROM ssm_user
WHERE user_role_id = ? ");
$stmt->bind_param("i", $chefRoleId);
$stmt->execute();
$stmt->store_result();
$stmt->bind_result($ufn, $uln, $uid);
while($stmt->fetch()){echo "<option>".$ufn." ".$uln."</option>";};
echo"</select>
</td>
<td><input class='form-control' type='' name=''></td>
<td><input class='form-control' type='' name=''></td>
</tr>";
}
echo "</table></form>";
I appreciate that i will likely have to add a button with an onclick but from there i am pretty lost. As always i appreciate the help provided. Right, Hopefully I can make you understand what I am trying to do. I am currently creating a PHP chat. Where I use php and mysql. Right now I have only have one chat board and everything works, but I want to add a button that will stay on the same page but load another pvp script instad of the one I an currently running and that will load another sub chat board such as game discussion.Like a mini-irc. Now my question is, is there a php way to this or will I have to script it and if there is ways of this out there could you point me in the right direction. Thanks, ScoreKeeper I am building an Event Registration system, and am thinking that it should put the Attendee's name on the Ticket (and in the database) as a backup. Is there a way to easily (and dynamically) display the same number of Text-Boxes as the # of Attendees entered? For example, let's say the Customer (payor) wants to buy Tickets for herself and 5 friends to a Banjo Festival. Right now, I just have a drop-down box to capture the head-count, but it would be nice to have 5 Text-Boxes (or something appear) so that the Customer can easily type in the Names of the 5 Attendees *without* having to suffer numerous other clicks and/or screens. Follow me? Is there a way to do that? Thanks, Debbie Hi guys i have to create text field & enter data and store in the data base. here im able to create text field but couldn't insert the data. so could anyone please check this code for me. <body> <form method="POST" action="cell.php"> Enter the number of question <input type="text" name="question"> <input type="submit"> <?php $value=$_POST['question']; for($i=0;$i<$value;$i++) { echo "Question NO:"; echo '<input type="text" name="qno">'."\t"; echo "Enter Marks:"; echo '<input type="text" name="marks">'."\t"; echo "<br>"; } ?> </form> <form name="form1" method="post" action="cellresult.php"> <label> <input type="submit" name="Submit" value="Submit"> </label> </form> </body> cellresult.php <body> <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("test", $con); $sql="INSERT INTO cell (QNO,MARKS) VALUES ('$_POST[qno]','$_POST[marks]')"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } echo "1 record added"; mysql_close($con) ?> </body> my name is fairooj and ama new to php and jquery. i have a proplam. i want your help.
<script type="text/javascript">
var count = 0;
$(function(){
$('p#add_field').click(function(){
count += 1;
$('#container').append(
'<strong>Link #' + count + '</strong><br />'
+ '<input id="field_' + count + '" name="fields[]' + '" type="text" />'
+ '<input id="code_' + count + '" name="code[]' + '" type="text" /><br />' );
});
});
</script>
Hope I'm not in the wrong forum here as javascript is required, BUT my page is written in PHP and will need to work from any number of records from the database. Basically, I have a drop-down box that contains product sizes, for example:- Small Medium Large Extra Large If the user selects Large for example, I want to change 'Was Price', 'Now Price' and 'Product Code' displayed on the page. If they select another size from drop-down the prices and code need to change again. Posted in the PHP section as PHP code is needed as well. Any help much appreciated. Hello all , here is another problem of my project. I need to create a textarea , drop down list and submit button . At first , I can type whatever I want in the textarea , but for certain part I can just choose the word I want from drop down list and click submit , then the word will appear in the textarea as my next word . But I have no idea how to make this works , is there any simple example for this function ? Thanks for any help provided . I’m trying to construct a button that simply writes an "aleph" character into a text area, see below.
My code does not work, can anyone tell me why ? How should I fix it ?
<!DOCTYPE html>
<html>
<meta charset="UTF-8">
<head>
<title>Example</title>
<script type="text/javascript">
//JavaScript code goes here
function insertAtEnd(text)
{
var theArea = document.getElementById("thisArea");
theArea.value += '' + text + '';;
}
</script>
</head>
<body>
<input type="button" id="aleph" name="aleph" value="Write an aleph"
onClick="javascript:insertAtEnd(\'<span>א</span>\');return(false)" />
<textarea id="thisArea"> </textarea>
</body>
</html>
Hi, I am wanting to create a random fact script for a small website project. The idea is when the user loads a page a random fact will appear at the top ( I guess from an array of strings I will make?) Can anyone recommend the most efficient and easiest way to go about this please? Hello, what's the correct use of GD library to compress JPG images please? I am using this upload script with TinyMCE. See below : I have this snippet but it doesn't compress the image:
// Accept upload if there was no origin, or if it is an accepted origin
// If image is JPEG compress with GD library
$filetowrite = $imageFolder . $temp['name'];
if (in_array(strtolower(pathinfo($temp['name'], PATHINFO_EXTENSION)), array("jpg", "jpeg"))) {
imagejpeg($temp['tmp_name'], $filetowrite, 75);
} else {
move_uploaded_file($temp['tmp_name'], $filetowrite);
}
<?php
$accepted_origins = array("http://localhost", "http://127.0.0.1", "http://192.168.1.1", "http://example.com");
$month = strtolower(date('M'));
$year = date('Y');
$newspath = "img/news/";
if (!is_dir("$newspath/$year/$month")) {
mkdir("$newspath/$year/$month", 0777, true);
}
$imageFolder = "$newspath/$year/$month/";
reset ($_FILES);
$temp = current($_FILES);
if (is_uploaded_file($temp['tmp_name'])){
if (isset($_SERVER['HTTP_ORIGIN'])) {
// same-origin requests won't set an origin. If the origin is set, it must be valid.
if (in_array($_SERVER['HTTP_ORIGIN'], $accepted_origins)) {
header('Access-Control-Allow-Origin: ' . $_SERVER['HTTP_ORIGIN']);
} else {
header("HTTP/1.1 403 Origin Denied");
return;
}
}
/*
If your script needs to receive cookies, set images_upload_credentials : true in
the configuration and enable the following two headers.
*/
// header('Access-Control-Allow-Credentials: true');
// header('P3P: CP="There is no P3P policy."');
// Sanitize input
if (preg_match("/([^\w\s\d\-_~,;:\[\]\(\).])|([\.]{2,})/", $temp['name'])) {
header("HTTP/1.1 400 Invalid file name.");
return;
}
// Verify extension
if (!in_array(strtolower(pathinfo($temp['name'], PATHINFO_EXTENSION)), array("gif", "jpg", "jpeg", "png"))) {
header("HTTP/1.1 400 Invalid extension.");
return;
}
// Accept upload if there was no origin, or if it is an accepted origin
// If image is JPEG compress with GD library
$filetowrite = $imageFolder . $temp['name'];
if (in_array(strtolower(pathinfo($temp['name'], PATHINFO_EXTENSION)), array("jpg", "jpeg"))) {
imagejpeg($temp['tmp_name'], $filetowrite, 75);
} else {
move_uploaded_file($temp['tmp_name'], $filetowrite);
}
// Respond to the successful upload with JSON.
// Use a location key to specify the path to the saved image resource.
// { location : '/your/uploaded/image/file'}
// echo json_encode(array('location' => $filetowrite));
$passpath = $year.'/'.$month.'/'.$temp['name'];
echo json_encode(array('location' => $passpath));
} else {
// Notify editor that the upload failed
header("HTTP/1.1 500 Server Error");
}
?>
This topic has been moved to PHP Applications. http://www.phpfreaks.com/forums/index.php?topic=326314.0 This topic has been moved to PHP Applications. http://www.phpfreaks.com/forums/index.php?topic=330095.0 Hi guys,
Having some trouble entering some php into my wordpress site that runs on Optimize Press.
I'm trying to insert this URL/ tracking code into my page as a hyperlink - 'Click Here Now'.
http://www.mysite.co...1_50onRed&sub1=<?php echo $sub1; ?>&sub2=<?php echo $sub2; ?>&sub3=<?php echo $sub3; ?>&sub4=<?php echo $sub4; ?>&sub5=<?php echo $sub5; ?>&lpid=<?php echo $lpid; ?>
This is what I have done so far:
<a href="http://www.mysite.co...;sub1=<?php echo $sub1; ?>&sub2=<?php echo $sub2; ?>&sub3=<?php echo $sub3; ?>&sub4=<?php echo $sub4; ?>&sub5=<?php echo $sub5; ?>&lpid=<?php echo $lpid; ?>">Click Here Now</a>
This does generate a hyperlink, however when I click it, I get: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '=' at line 2
I understand for this code to work the page has to be saved in php. I thought wordpress did this automatically?
Can anyone help please? Also, please bear in mind I am using the Optimize Press plugin.
Many thanks in advance!
Tom
I'm having a problem and need an answer to why its happening and how to prevent it. Scenario: I begin load my home page which starts with a session_start(); .... Before it FULLY completes loading I try to navigate to another page and BOOM, that page will not load and any other page that begins with session_start(); will not load unless I close and restart the entire browser or wait about 10 minutes.... I will note my website makes ajax calls every 5 seconds or so, but I use setTimeout for them. Any help??? Thanks ahead! I have a comment section. It is comprised of 3 major files. One file creates the form where you fill in your name and comment. Another file sends your comment to the DB. The third file displays the already existing comments on the page. You have the option to reply to the existing comments on the page. When you click the reply button I would like the comment text box to light up so the user knows its working but I have hit a brick wall.
This is the comment form the user actually types into:
<div id="comment-container" > <form id="comment_form" action="/comm_1/post_comment.php" method='post' onsubmit=" return validateForm()"> <table> <tr> <td id="error"></td> </tr> <tr> <td><textarea name="comment_body" id='comment_body' placeholder="Comment"></textarea></td> </tr> <?php if(!$username && !$userid): ?> <tr> <td><input type="text" name="name" class="input_style" placeholder="Name"/> <input type="submit" id="loginbtn" value="Or Login" onclick="window.location='/login_scripts/login.php'" /></td> </tr> <tr> <td><input type="email" name="email" class="input_style" placeholder="Email"/></td> </tr> <?php endif; ?> <?php if($username && $userid): ?> <input type="hidden" name="name" value="<?php echo htmlspecialchars($username) ?>"/> <input type="hidden" name="email" value="<?php echo htmlspecialchars($email) ?>"/> <?php endif; ?> <tr> <input type='hidden' name='parent_id' id='parent_id' value='0'/> <td><input type="submit" name="submitbtn" id="submitbtn" value="Add comment"/></td> </tr> </table> </form>This how the existing comments are displayed on the page: You can see on line 7 is the button to the reply to a comment. When this is clicked I need the "comment_body" from the above code to highlight.
<?php
function getComments($row) {
echo "<li class='comment'>";
echo "<div class='aut'>".$row['author']."</div>";
echo "<div class='timestamp'>".$row['created_at']."</div>";
echo "<div class='comment-body'>".$row['comment']."</div>";
echo "<a href='#comment_form' name='replybtn' class='reply' id='".$row['id']."'>Reply</a></script>";
$q = "SELECT * FROM threaded_comments WHERE parent_id = ".$row['id']."";
$r = mysql_query($q);
echo "</li>";
if(mysql_num_rows($r)>0)
{
echo "<ul>";
while($row = mysql_fetch_assoc($r)) {
getComments($row);
}
echo "</ul>";
}
}
?>
I tried this with JS, but it does not work:
$(function(){
$("a.reply").click(function() {
var id = $(this).attr("id");
$("#parent_id").attr("value", id);
$("#comment_body").focus();
});
});
Edited by ryanmetzler3, 26 January 2015 - 09:48 PM. |
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