On a page I insert a button that will direct the user to another view (newview.ctp)
in the controller I create a function
function newview()
{
if($this->Session->read($this->_userName))
{
$this->loadSkinForAction();
$user = $this->user->findByUsername($this->Session->read('User'));
$this->set('item',$user);
}
}
In the page as I mention I simply insert a button this way
<input type="button" value="Change Your self" onclick="window.location.assign('../users/newview')" />
The view newview.ctp looks something like this
<?php
if(isset($item))
{
echo $form->create(null,array('url' => '/users/newview/'.$item['User']['id']));
}
?>
<h1>Change your password</h1>
<table>
<tr>
<td>New password</td>
<td><?php $form->password('password',$newPassword);?></td>
</tr>
<tr>
<td>Confirm password</td>
<td><?php echo $form->password('password_',$newPasswordConfirm);?></td>
</tr>
</table>
<?php
echo "<br/>";
echo $form->end('Save');
}?>
When i click the button in the page, it doesn't show the new page but a blank page instead,
Does the page which links to the "newview()" page need to be a HTML button? Or simply a link?
If it is a link, use CakePHP's Html helper functions. Link
<?php echo $this->Html->link('Change Password', '/users/newview'); ?>
Related
I have an php form utilizing several inputs that is driving a kiosk page. If a text input is blank I want this update a separate input with the word "hidden" If there is text I would like the word "visible" to show. Currently my code works if you click submit twice but will not work on the first submit. Here is my current code:
The if function that is current working on second submit:
if (strlen($something)>0) {
$_POST['someone'] = "visible";
} else {
$_POST['someone'] = "hidden";
}
input form:
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<table>
<tr>
<td>something : </td>
<td><input type="text" id="something" name="something" value="<?php echo htmlspecialchars($something); ?>"/></td>
<td></td>
</tr>
<tr>
<td></td>
<td><?php echo $_SERVER['PHP_SELF']; ?></td>
<td></td>
</tr>
<tr>
<td>someone:</td>
<td><input type="text" id="someone" name="someone" value="<?php echo htmlspecialchars($someone); ?>"/></td>
<td></td>
</tr>
<tr>
<td></td>
<td><input type="submit" name='submit' value="Submit"/></td>
<td></td>
</tr>
</table>
</form>
Here is the update code:
$usql = "UPDATE test SET something= '".$_POST['something']."', someone= '". $someone ."' WHERE ID='a';";
Currently the "someone" input has a display of none so it cannot be seen by the user. This is not necessary but if someone could tell me how to bypass adding an input altogether and tweak the update statement itself to update something that would be great as well! Thanks!
Any help would be appreciated!
Using session variables can save the page state, even after reload.
First page save:
session_start();
$_SESSION['someone'] = $_POST['someone'];
Then:
if(isset($_SESSION['someone']))
{
if (strlen($_SESSION['someone'])>0) {
$_POST['someone'] = "visible";
} else {
$_POST['someone'] = "hidden";
}
}
I'm having some difficulty with php forms.
I have created a page called 'post_details.php' (this simply displays a photo of a product & description). Each product has a unique id
Within posts_details.php, I have used to include command to include a form. This form allows users to send me feedback regarding the product.
For some reason the form is not workin. Everytime the submit button is clicked, the alert box warns me that I need to complete the form (even if the form is complete)
The last part of line one doesn't seem to work. It's not picking up the post_id
Can anyone please help ??
post a comment
<form method="post" action="post_details.php?post= <?php echo $post_id; ?>">
<table width "600">
<tr>
<td>Your email:</td>
<td><input type="text" name="comment_email"/></td>
</tr>
<tr>
<td>Your Comment:</td>
<td><textarea name="comment" cols="35" rows="16"/></textarea></td>
</tr>
<tr>
<td><input type="submit" name="submit" value="postcom"/></td>
</tr>
</table>
</form>
<?php
if(isset($_POST['comment'] )) {
$comment_email = $POST['comment_email'];
$comment = $POST['comment'];
if( $comment_email=='' OR $comment=='') {
echo "<script>alert('Please complete form')</script>";
echo "<script>window.open('post_details.php?post=post_id')</script>";
exit();
}
else {
echo "complete";
}
}
?>
You have error here
if(isset($_POST['comment'] )) {
$comment_email = $POST['comment_email'];
^
$comment = $POST['comment'];
^
....
Instead of $POST it must be $_POST['comment_email'] and $_POST['comment']
I'm attempting to make a master-detail search page for my intranet that will allow me to search both staff and hardware from the same page. I've gotten part of the way there but I've gotten myself stuck and need some help.
I have the inventory page (index.php) setup and I've included a form with two button - "Staff" and "Hardware" - to include the master-details .php pages I've designed.
The code for this is as follows:
<div class="content">
<h1>Inventory</h1>
<p></p>
<form action="" method="post">
<input type="submit" name="Staff" value="Staff" />
<input type="submit" name="Hardware" value="Hardware" />
</form>
<p></p>
<p></p>
<?php
if (isset($_POST['Staff']))
{
include 'details.php';
include 'lib/staffsearch.php';
};
if (isset($_POST['Hardware']))
{
include 'detailhw.php';
include 'lib/hwsearch';
};
?>
</div>
Again, this works fine - I click "staff" and it loads both my "staffsearch.php" and "details.php" pages. This is where I run into trouble however. My staffsearch.php includes a form that queries my staff database and returns the matching results in a table. I should be able to click on one of the results and the detailed information is displayed in the details box above. What happens though is that I click on the "Staff" button, my includes are displayed, I enter the search criteria, click "search" and get a blank page as a result.
If I click on the "staff" button again though my includes are reloaded along with the correct search result. Accessing the details however results in the same problem - once my results are displayed and I click on one instead of the details I receive a blank page until I click the "staff" button again - then my includes are reloaded and my results are displayed.
Is there anyway that when I click the search button on my staffsearch form that I can force the index page to reload the includes without having to click the "staff" button again?
Thanks, ahead of time for any advice as I'll throw in the mandatory. "I'm new to php" and my Googling hasn't really netted me any good results. If you need anymore information just ask!
EDIT 1
The code for my searchstaff.php page is as follows:
<table width="598px" border="1" align="center">
<tr valign="middle">
<td alight="left" colspan="3"> <form id="form1" name="form1" method="get" action="">
<label for="name">Search:</label>
<input type="text" name="name" id="name" /><input type="submit" name="search" id="search" value="Submit" style="margin-left:10px";/>
</form>
</td>
<tr>
<td align="center" width="40px">No.</td>
<td width="300px">Name</td>
<td width="158px">IP Address</td>
</tr>
<?php do { ?>
<tr>
<td align="center"><?php echo $row_inventorystaff['staff_id']; ?><?php echo $row_DetailRS1['staff_id']; ?></td>
<td> <?php echo $row_inventorystaff['f_name']; ?> <?php echo $row_inventorystaff['l_name']; ?><?php echo $row_DetailRS1['f_name']; ?> <?php echo $row_DetailRS1['l_name']; ?></td>
<td><?php echo $row_inventorystaff['ip_address']; ?><?php echo $row_DetailRS1['ip_address']; ?></td>
</tr>
<?php } while ($row_inventorystaff = mysql_fetch_assoc($inventorystaff)); ?>
</table>
<br />
<table border="0" style="margin-left:15px;" >
<tr>
<td><?php if ($pageNum_inventorystaff > 0) { // Show if not first page ?>
First
<?php } // Show if not first page ?></td>
<td><?php if ($pageNum_inventorystaff > 0) { // Show if not first page ?>
Previous
<?php } // Show if not first page ?></td>
<td><?php if ($pageNum_inventorystaff < $totalPages_inventorystaff) { // Show if not last page ?>
Next
<?php } // Show if not last page ?></td>
<td><?php if ($pageNum_inventorystaff < $totalPages_inventorystaff) { // Show if not last page ?>
Last
<?php } // Show if not last page ?></td>
</tr>
</table>
<p style="margin_left:15px">Records <?php echo ($startRow_inventorystaff + 1) ?> to <?php echo min($startRow_inventorystaff + $maxRows_inventorystaff, $totalRows_inventorystaff) ?> of <?php echo $totalRows_inventorystaff ?>
</p>
I have screen shots of the process of clicking through the index and performing a search that I can upload if it helps the explanation.
I would use a radio button to make a decision which will be easier to parse. Double submit buttons are a funny thing and some browsers dont like that.
<input name="dept" value="staff" type="radio"> Staff
<input name="dept" value="hardware" type="radio"> Hardware
<?php
if( isset($_POST['dept'] && $_POST['dept'] == 'staff' ) {
include 'staffForm.php';
} else {
include 'hardwareForm.php';
}
?>
I have a parent page which has a drop down list in it. using the onchange event, data is posted to a second page using $.post(). This page uses the posted variables to output mysql data with a checkbox for each line. This page output is then inserted into the parent page using jquery $('#DIV name').html(output).show();
The user can then see the mysql table rows with corresponding checkboxes. They then select the checkboxes they want and say delete. Delete is a form submit button.
My question is, when they click delete how do I take the form data from the second page and post it with that of the parent page so that I can then use $_POST[] to get the checkbox info and delete the selected table rows?
example of the parent page code is:
javascript/jquery
<script type="text/javascript">
function get(row){ //row being processed, defined in onchange="get(x)"
('getpeopleinjobs.php',{ //data posted to external page
postvarposition: form["position"+row].value, //variable equal to input box, sent to external page
postvarjob: form["job"+row].value, //variable equal to input box, sent to external page
postvarperson: form["person"+row].value, //variable equal to drop down list, sent to external page
postrow: row}, //variable equal row being processed, sent to external page
function(output){
$('#training'+row).html(output).show(); //display external results in row div
//popupWindow = window.open('t4.php?variable1Name=Vicki&variable2Name=Maulline','popUpWindow','height=400,width=1000,left=10,top=10,resizable=yes,scrollbars=yes,toolbar=no,menubar=no,location=no,directories=no,status=yes')
});
}
</script>
Form data is
<tr>
<td>
<?PHP echo $i; ?>
</td>
<td>
<input type=text NAME="position<?PHP echo $i; ?>" id="position<?PHP echo $i; ?>" style="border: 1px solid #2608c3;color:red; width=200px" value="<? echo mysql_result($resultpositionjob,$r,0);?>">
</td>
<td>
<input type=text NAME="job<?PHP echo $i; ?>" id="job<?PHP echo $i; ?>" style="border: 1px solid #2608c3;color:red; width=200px" value="<? echo mysql_result($resultpositionjob,$r,1);?>">
</td>
<td>
<SELECT NAME="person<?PHP echo $i; ?>" id="person<?PHP echo $i; ?>" style="border: 1px solid #2608c3;color:red; width=200px" onchange="get(<? echo $i; ?>);">
<OPTION VALUE=0 >
<?=$optionpeople?>
</SELECT>
</td>
<td onclick="train(<? echo $i; ?>);" style="color:grey; cursor: pointer;">
<div id="training<?PHP echo $i; ?>"><font color=grey size=2></div>
</td>
<td>
</td>
</tr>
<?PHP
$i++;
$r++;
}
?>
The second page or page called by jquery, the output is:
echo
"
<table border=0 width=400>
<tr>
<td width=20>
</td>
<td width=150>
<b>Position<b>
</td>
<td>
<b>Job<b>
</td>
</tr>
";
while($line = mysql_fetch_array($result))
{
echo "
<tr>
<td>
";
?>
<input type=checkbox name="delete[]" id="delete[]" value="<?php echo $rows['id']; ?>">
<?PHP
echo "
</td>
<td>
";
echo $line['position'];
echo "
</td>
<td>
";
echo $line['job'];
echo "
</td>
</tr>
";
}
?>
<tr>
<td>
<input type=submit name="update" id="update" value="Update">
</td>
</tr>
</table>
<?PHP
}
To repeat I want the table checkbox element from the second page posted with the form data of the parent page.
Is this possible as my testing hasnt given me any luck.
Am I doing something wrong or do I need to modify the jquery code?
Thanks as always for the assistance.
There is no second page. Really - you're loading HTML content from the second page, but it's being inserted into the parent page. All content, from your browsers perspective, is in the same page. The fields should be included as long as they're inside the DOM inside the element. Use the developer tools for your browser or Firebug for Firefox to make sure that the content is placed within the form element in the DOM. The developer tools should also be able to show you exactly which variables are submitted to the server when the form is submitted.
The 'action' parameter of 'form' will indicate where the content gets submitted (and it seems you've left that part out of your HTML, so it's impossible to say if you've left out or just not included it in the paste.
Am creating a simple cms and i have the following
<input type="submit" value="delete" title="Delete Aricle(s)" id="delete_btn" onclick="delete_confirm()" />
<tbody>
<?php foreach ($articles as $article): ?>
<tr>
<td align="center">
<input id="article_id"name="article[]" type="checkbox" value="<?php echo $article->id; ?>" /></td>
<td align="center"><?php echo anchor('admin/articles/edit/'.$article->id,$article->title); ?></td>
<td align="center"><?php echo $article->author; ?></td>
<td align="center"><?php echo $article->category; ?></td>
<td align="center"><?php published($article->post_status); ?></td>
<td align="center"><?php featured($article->featured); ?></td>
<td align="center"><?php echo $article->views; ?></td>
<td align="center"><?php echo $article->post_date; ?></td>
</tr>
<?php endforeach;?>
</tbody>
The above code is wrapped in a form tag
This my javascript code
function delete_confirm() {
var msg = confirm('Are you sure you want to delete the selected article(s)');
if(msg == false) {
return false;
}
}
This is delete function in my controller
function articles_delete() {
//Load Model
$this->load->model("article_model");
//Assign article id to variable
$checkbox = $_POST['article'];
//loop through selected checkbox
for($i = 0; $i < count($checkbox); $i++) {
//Call delete method of the article model
$this->article_model->delete($checkbox[$i]);
}
$this->session->set_flashdata("message","Article(s) Deleted");
redirect("admin/articles","refresh");
}
But when cancel is clicked in the confirm dialog the form still is still submited and the selected article deleted.
Please advice on what to do
Try changing
<input type="submit" value="delete" title="Delete Aricle(s)" id="delete_btn" onclick="delete_confirm()" />
to
<input type="submit" value="delete" title="Delete Aricle(s)" id="delete_btn" onclick="return delete_confirm();" />
You're running the delete_confirm() function inside of the onclick function. you need to return false from the onclick() function itself.
try onSubmit instead of onClick.
What you'll want to do is something like this:
<form onsubmit="return delete_confirm();" ...>
<input ...>
</form>
When the form is submitted via the type="submit"-button, your delete_confirm()-function is called. If it returns true, the form is submitted. If it returns false, it's not.
Try using AJAX to call the PHP script from within Javascript.
This means set the input button type to button <input type="button" /> and change the onclick such that if cancel is pressed, return false (like what you did) or if OK is pressed, run the AJAX script and redirect the page if required...
You could write the AJAX server-side (PHP) script in a seperate file, and send in the paramters using $_GET.
delete_confirm() must always return false in your case.
You want the delete confirm function to be your onsubmit attribute of the form. Having it on the input element is not enough.
Maybe try to use:
Delete