I'm new in php how can I make this work
delete.php
<?php
include_once 'dbconfig.php';
if($_POST['del_id'])
{
$id = $_POST['del_id'];
$stmt=$db_con->prepare("UPDATE tbluser set status=1 WHERE id=:id");
$stmt->execute(array(':id'=>$id));
}
?>
I want to add an active/inactive dropdown choice in my edit form page but I dont know how to make it work I dont know how to call the delete.php so that when i choose inactive and submit the form it will not show on my datatable
edit_form.php
<?php
include_once 'dbconfig.php';
if($_GET['edit_id'])
{
$id = $_GET['edit_id'];
$stmt=$db_con->prepare("SELECT * FROM tbluser WHERE id=:id");
$stmt->execute(array(':id'=>$id));
$row=$stmt->fetch(PDO::FETCH_ASSOC);
}
?>
<div id="dis">
</div>
<form method='post' id='emp-UpdateForm' action='#'>
<table class='table table-bordered'>
<input type='hidden' name='id' value='<?php echo $row['id']; ?>' />
<tr>
<td>Status</td>
<td><select name=stats>
<option value="1">Active</option>
<option value="0">Inactive</option>
</select>
</td>
</tr>
<tr>
<td colspan="2">
<button type="submit" class="btn btn-primary" name="btn-update" id="btn-update">
<span class="glyphicon glyphicon-plus"></span> Save Updates
</button>
</td>
</tr>
</table>
everything is working it's just an additional functionality in the edit form page
Get the stats like $status= $_POST['stats']; in delete.php
change you $_GET['edit_id'] to $_GET['id']
modify you updated to set the status
$id = $_POST['id'];
$status = $_POST['stats']
$stmt=$db_con->prepare("UPDATE tbluser set status=:status WHERE id=:id");
$stmt->execute(array(':status'=>$status,':id'=>$id));
If you want to call the delete.php, you can use ajax.
<script>
$.ajax({
url:'delete.php',
type:'post',
data: $('form').serialize()
});
</script>
You can get values on the delete.php, with the help of $_POST or $_REQUEST.
And if the delete.php code is right, then on submit of the edit_page.php your datatable can get affected.
Related
I have 2 dropdown menus. I want to dynamically populate second dropdown based on the first one. When changing the value in first dropdown, it has to pass the selected value to an invisible php array to update it. So here is the problem: passing the js variable(strUser) to php {{ the php var }} (used {{}} only to make it more simple to see).
<form class="formatted-form" action="" method="POST">
<div class="message">
<?php ThemexInterface::renderMessages(isset($_POST['success'])?$_POST['success']:false); ?>
</div>
<table class="profile-fields">
<tbody>
<?php if(!ThemexCore::checkOption('user_location')) { ?>
<tr style="display: none">
<th><?php _e('Hidden Country', 'lovestory'); ?></th>
<td>
<div class="select-field">
<span></span>
<?php
echo ThemexInterface::renderOption(array(
'id' => 'country',
'type' => 'select',
'options' => array_merge(array('-' => '–'), ThemexCore::$components['countries']),
'value' => !empty(ThemexUser::$data['user']['profile']['country'])?{{ThemexUser::$data['user']['profile']['country']}}-this should be replaced with the strUser variable we get in js :null,
'wrap' => false,
));
?>
</div>
</td>
</tr>
<tr>
<th><?php _e('Visible_Country', 'lovestory'); ?></th>
<td>
<span>
<?php
$post_ids2 = $wpdb->get_col( "SELECT DISTINCT country FROM countries" );
echo "<SELECT id=tara_1 onChange='reload()' >";
foreach ($post_ids2 as $value){
echo "<option value=$value </option>";
echo $value;
}echo "</select>";
?>
</span>
</div>
</td>
</tr>
<script>
function reload(){
var e = document.getElementById("tara_1");
var strUser = e.options[e.selectedIndex].text;
}</script>
<tr>
<th><?php _e('City', 'lovestory'); ?></th>
<td>
<span>
<?php
global $wpdb;
$post_ids = $wpdb->get_col( "SELECT city FROM countries WHERE tara='".ThemexUser::$data['user']['profile']['country']."'" );
echo "<SELECT id=tara_1 name=cities>";
foreach ($post_ids as $value){
echo "<option value=$value </option>";
echo $value;
}echo "</select>";
?>
</span>
</div>
</td>
</tr>
<?php } ?>
<?php
ThemexForm::renderData('profile', array(
'edit' => true,
'before_title' => '<tr><th>',
'after_title' => '</th>',
'before_content' => '<td>',
'after_content' => '</td></tr>',
), ThemexUser::$data['user']['profile']);
?>
</tbody>
</table>
<?php _e('Save Changes', 'lovestory'); ?>
<input type="hidden" name="update" value="1" />
<input type="hidden" name="user_action" value="update_profile" />
<input type="hidden" name="nonce" value="<?php echo wp_create_nonce(THEMEX_PREFIX.'nonce'); ?>" />
</form>
I know php is server-side and js works on client-side, that's why i am confused on how could i dynamically refresh the second dropdown AND pass the value selected by user to PHP POST. I am new to PHP and JS so please be kind, i would appreciate any hint on how I should approch the problem. In the end i want it to behave like this:
Dropdown1 selected>Italy, Dropdown2 > {gets populated with italian cities from database}
Dropdown1 selected>France, Dropdown2 > {gets populated with frech cities from database}
Well, the real problem is passing JS to PHP- which i think is the wrong approach.
Thanks in advance.
how do I display data in the textbox automatically when I click on the edit button, I want to display data from the database using jquery chosen like this:
enter image description here
, sorry if my question is unclear, if anyone is unable to ask me, please help
This is my code
<?php
$main_controll = new App_UserManagement_Control_UserManagement();
$data_from_ctr = $main_controll->edit();
echo json_encode( $data_from_ctr);
?>
<tr class="odd">
<td width="150"><strong>Account Access</strong></td>
<td width="10">:</td>
<td colspan="2">
<select id="acctAccessed" name="acctAccessed[]" multiple="multiple" class="chosen-select" style="width:350px;" data-placeholder="Select account">
<?php foreach ($data_from_ctr['rek'] as $data) :?>
<option <?php if ($data['selected'] == "1"){ ?> selected= "" <? } ?> value="<?php echo $data['idRek'];?>"><?php echo $data['idRek'];?></option>
<?php endforeach; ?>
</select>
<input id="chkall" type="checkbox" >Select All</input>
</td>
<script type="text/javascript">
$("#acctAccessed").chosen();
</script>
but my error like this, I don't get data for account access
enter image description here
Can you post $data_from_ctr as , the above mentioned code by you is working.
<?php session_start() ;
include 'databaseConnection.php' ;
$sqlCommand3 = "SELECT * FROM posts ;" ;
$result3 = mysqli_query($conn , $sqlCommand3);
$resultCheck = mysqli_num_rows($result3);
while ($row = mysqli_fetch_array($result3)) { ?>
<tr>
<td style="padding-left: 10px;">
<?php echo $row['poster']; ?>
</td>
<td style="padding-left: 45px;">
<?php echo $row['post_destination']; ?>
</td>
<td style="padding-left: 30px;">
<?php echo $row['post_readyTime']; ?>
</td>
<td style="padding-left: 45px;">
<?php echo $row['post_numberOfPassenger'];?>
</td>
<td style="padding-left: 32px;">
<?php echo $row['post_submitTime']; ?>
</td>
<td>
<form action="riderJoin.php" method="post">
<input type ="hidden" name ="id" value="<?php echo $row['post_id']; ?>" >
<input type =submit class="edit_btn" name = submit value = "Join">
</form>
</td>
<!--<td>
Delete
</td>-->
</tr>
<?php } ?>
</table>
</div>
</div>
<script type="text/javascript">
$("td:nth-child(4):contains('5')").addClass('redWord');
var x = document.getElementsByClassName('edit_btn');
<?php if (isset($_POST['submit'])) { ?>
var i = <?php echo $_SESSION['postId'] ; ?> - 1 ;
$(x[i]).val('Delete').css('background-color','red').css('color','white');
<?php }?>
</script>
How to add a new input (type = "submit" value="delete") with a new action (before the "join" button) whenever the $_POST['submit'] is set? I don't want to add a new input before each "join" button. Instead, I just want to add after a specific "join" button inside a cell when that specific "join" button is clicked. Also, if I click another 'join' button in other cell, a similar effect would happen too. So in the end, after clicking two "join" button, there should be only two delete button each added before the specific clicked "join" buttons. How to solve it ?
Thanks a lot :)
I want to fetch data from a database related to a selected option when it is clicked.
When I select another option then fetch the data for that selection.
When I fetch the new data option the old data is replaced.
I want to display each set of data in a new row so I do not want the new data to replace the old data.
<form action="" method="post">
<div class="form-group">
<label for="sel1">Select list:</label>
<select multiple name="names" class="form-control" id="sel1">
<?php
$result = "SELECT Name FROM coder";
$sql = $conn->query($result);
if ($sql->num_rows > 0)
while($row = $sql->fetch_assoc()) {?>
<option value='<?php echo $row['Name'] ?>'><?php echo $row['Name'] ?></option>
<?php }
?>
</select>
</div>
<input type="submit" name="submit" class="btn btn-info">
</form>
<table>
<thead>
</thead>
<tbody>
<?php
$result = "SELECT * FROM coder WHERE Name='$name'";
$sql = $conn->query($result);
if ($sql->num_rows > 0)
while($row = $sql->fetch_assoc()) {{?>
<tr>
<td><?php echo $row['Name']?></td>
</tr>
<?php
}
}
?>
</tbody>
</table>
I suggest to use jquery for binding the div once the result is placed inside it.
Also Instead of using direcy Query, you also can use onChange event of jQuery for fetching the data according to your need. It will accelerate the speed of fetching of data.
Divide the div into samall child divs and store the data into them.
Your problem will be resolved in this way for sure
Im using ajax to load the testNew.php file into the Cart.html file. It loads the testNew.php file but when i click on the button add which is in the testNew.php, 0 is being entered in the database and as soon as i click on the add button the page refresh by itself. My problem is that i dont want the page to refresh and want the add button to do the same action as in the testNew.php file(which works correctly).
<script type='text/javascript' src='http://code.jquery.com/jquery-1.6.2.js'>
</script>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js">
</script>
<script>
$(document).ready(function() {
$("#product").click(function() {
$.ajax({ //create an ajax request to load_page.php
type: "GET",
url: "testNew.php",
dataType: "html", //expect html to be returned
success: function(response){
$("#responsecontainer").html(response);
//alert(response);
}
});
});
});
</script>
</head>
<body>
<table border="1">
<tr>
<td>
<input type="button" id="product" name="product"value="View all products"/>
</td>
</tr>
</table>
<div id="responsecontainer"></div>
Here is the testNew.php which works correctly.
<?php
include'connect.php';
$image = isset($_REQUEST['image']) ? $_REQUEST['image'] : "";
$id = isset($_REQUEST['id']) ? $_REQUEST['id'] : "";
$name = isset($_REQUEST['name']) ? $_REQUEST['name'] : "";
$price= isset($_REQUEST['price']) ? $_REQUEST['price'] : "";
$sql="SELECT * FROM product";
$result = mysql_query($sql);
if($result>0){
?>
<table border='1'>
<tr>
<th>Id</th>
<th>Image</th>
<th>Name</th>
<th>Price MUR</th>
</tr>
<?php
while ($row = mysql_fetch_array($result)){
?>
<tr>
<td><?php echo ($row['id']); ?></td>
<td><img src=<?php echo $row['image'] ?> width='120'
height='100'/></td>
<td><?php echo htmlspecialchars($row['name']); ?></td>
<td><?php echo htmlspecialchars($row['price']); ?></td>
<td>
<form method="POST" action="" >
<input type="hidden" name="id" value="<?php echo $row['id']; ?>" />
<input type="hidden" name="name" value="<?php echo $row['name']; ?>" />
<input type="hidden" name="image" value="<?php echo $row['image']; ?>" />
<input type="hidden" name="price" value="<?php echo $row['price']; ?>" />
<input id="submit" type="submit" name="submit" value='Add to cart'
onclick="add()"/>
</form>
</td>
</tr>
<?php
}
?>
</table>
<?php
}
$insert = "INSERT INTO product_add(id, name, price) VALUES ('$id', '$name','$price')";
$insertQuery=mysql_query($insert);
?>
Your error in thought is actually that you are including a form through AJAX of which the HTML is then loaded into your page after which the action attribute on the form refers to the page itself (loading HTML into your page with AJAX does not work the same as an iframe) which does not have the relevant code to actually parse and insert the database.
You need to make a separate page that only accepts a few parameters and inserts those into the database. However, before you do that you need to read this:
I'm going to go off on the safety of your code for a tad.
$_REQUEST refers to both $_GET and $_POST, you really want only $_POST as you just want to deal with what gets submitted through a form.
You never sanitize your input, this way a person could craft an URL with say phpNew.php?id='; DROP TABLE wooptiedoo. This is a simplified example but nevertheless you should take a closer look at the mysql_real_escape_string documentation and possibly some guides on "sql injection".
After that those same variables can be used for XSS which means someone can use your page to serve random HTML to people visiting that site and trusting your domain. I suggest you look up the htmlentities function and/or read up on "cross-site scripting" attacks.
And to end it all, the mysql_* functions are deprecated and you should probably be using mysqli_* functions.