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'] && $_POST['dept'] == 'staff' ) {
include 'staffForm.php';
} else {
include 'hardwareForm.php';
}
?>
Related
this codes below came from my friend. i want to make a search under this codes . from this codes , it will display all the users to delete. what i want is i want to put search box and search button . so when i insert user id and press search only then it display which user that i want . where should i put this search text , button and php ? i want to put it under this else if codes . help me. im new :(
else if($_SESSION['jawatan'] == 'ADMIN') { ?>
<li> VIEW PROFILE </li>
<li> VIEW REQUEST </li>
<li>LOG OUT</li>
<li><img src="msc.jpg" width = "240" height ="80"></li>
<form method="get">
<input type="text" name="userid" placeholder="search" />
<inpu type="submit" value="Search" />
</form>
<h3>View User</h3>
<table border='1'>
<tr>
<td><b>#</b></td>
<td><b>Nama</b></td>
<td><b>Email</b></td>
<td><b>Division</b></td>
<td><b>Department</b></td>
<td><b>Delete</b></td>
</tr>
<?php
$i = 1;
$whr="";
if(isset($_GET['userid'])){
$whr.= " and mem_id='".$_GET['userid']."'"; // which field you want. If you want to do search name use LIKE instead =(equal to)
}
$result = mysql_query("SELECT * FROM members WHERE mem_role='USER' ".$whr);
if(mysql_num_rows($result)>0) {
while($row = mysql_fetch_array($result)) {
?>
<tr>
<td>
<?php echo $i; ?>
</td>
<td>
<?php echo $row['mem_name']; ?>
</td>
<td>
<?php echo $row['mem_email']; ?>
</td>
<td>
<?php echo $row['mem_division']; ?>
</td>
<td>
<?php echo $row['mem_department']; ?>
</td>
<td>Delete</td>
</tr>
<?php
$i++;
}
} else {
echo '<tr><td>No results found</td></tr>';
}
?>
</table>
<?php
}
HTML
<form method="get">
<input type="text" name="userid" placeholder="search" />
<inpu type="submit" value="Search" />
</form>
PHP
$whr="";
if(isset($_GET['userid'])){
$whr.= " and mem_id='".$_GET['userid']."'"; // which field you want. If you want to do search name use LIKE instead =(equal to)
}
MYSQL
$result = $result=mysql_query("SELECT * FROM members WHERE mem_role='USER' ".$whr);// add whr variable here
I can tell you the steps:
Create search text field and search button ( wrap in form if you want to do with php only)
on click of button ( that would be submit ), fetch that value from search text box and run query based on that
when results appear , fill the table data with that results using foreach.
hope it helps. Ask me for any issue.
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.
I wonder whether someone may be able to help me please.
Firstly, this is something I've never tackled before, so please bear with me if this is something which is straight forward to the more seasoned developer.
The code below is a portion of a script which correctly shows a 'Locations List' for each user.
$i=0;
while ($i < $num) {
$lid=mysql_result($result,$i,"locationid");
$lname=mysql_result($result,$i,"locationname");
$laddress=mysql_result($result,$i,"returnedaddress");
include("admin/link.php");
include("admin/opendb.php");
$fquery = "SELECT COUNT(*) num FROM finds WHERE locationid = '$lid'";
$fcount = mysql_query($fquery) or die('error');
$row = mysql_fetch_assoc($fcount);
$findscount = $row['num'];
mysql_close($connect);
?>
<table width="603" border="0">
<tr>
<th width="156">Location</th>
<th width="302">Address</th>
<th width="131">No. Of Finds Made</th>
</tr>
<tr>
<td><?php echo $lname;?></td>
<td><?php echo $laddress;?></td>
<td><?php echo $findscount;?></td>
</tr>
</table>
<?php
echo'<form name="locations" method="post">';
$i++;
}
if ($num == 0) {
echo"<tr>
<td> </td>
<td> </td>
<td> </td>
<td> </td>
<td> </td>
</tr>
<tr>
<td> </td>
<td colspan='3'><div align='center'><strong>No locations have been added yet. <a href='saveaddress.php'>Click here</a> to add one.</strong></div></td>
<td> </td>
</tr>
<tr>
<td> </td>
<td> </td>
<td> </td>
<td> </td>
<td> </td>
</tr>";
}
echo '<input type="submit" name="submitlocation" id="submitlocation" value="View/Amend Location Details">';
echo '<input type="submit" name="submitfinds" id="submitaddfinds" value="Add Finds">';
echo '</div>'."\n";
if(isset($_POST["submitlocation"])) {
header("updatelocation.php");
}
else if(isset($_POST["submitaddfinds"])) {
header("addfinds.php");
}
?>
What I'm trying to do is add two submit buttons for each Location record, one which takes the user to a page called updatelocation.php and the second which takes the user to a page called addfinds.php. But both must be linked to the Location via the field 'locationid'. So in other words if the user wants to Add Finds for Location 1 upon clicking on the relevant button, they are taken to the Add Finds page for Location 1.
I've read a number of tutorials, but I've obviously misunderstood somewhere along the lines because although I've managed to add the buttons to the record, when I click either button the 'Locations List is refreshed, rather than the user being taken to the respective page.
I just wondered whether someone could possibly have a look at this please and let me know where I'm going wrong?
Many thanks and kind regards
You'd be better off making a seperate page (i.e., forrmaction.php) which processes the logic and sends the user to the correct page. If you do not want to do that, you can also make a seperate form for each button and simply set the action attribute of form to point to either updatelocation.php or addfinds.php
Example of formaction.php:
if (isset($_POST['type']) {
if ($_POST['type'] == 'update')
$url = 'updatelocation.php';
else
$url = 'addthis.php';
header("Location: " . $url);
}
Example of form with two buttons that works with formaction.php:
<form name="locations" method="POST" action="formaction.php">
<input type="submit" name="type" value="update">
<input type="submit" name="type" value="add">
</form>
However, for your question, the problem is that you are not using the correct header. To use a redirect, you must use "Location:" in the header. For example,
header("Location: updatelocation.php");
Be careful, because this will not work if headers have already been sent. You may want to look into output control in that case.
On a page I insert a button that will direct the user to another view (newview.ctp)
in the controller I create a function
function newview()
{
if($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 $this->Html->link('Change Password', '/users/newview'); ?>
This is going to be a meaty question because I am not sure the best way to handle this.
I have a page that contains a number of dojo inline editors, to allow users to change values, when one entry had been changes a save button will appear to prompt the user to save the information.
The page has a number of rows, contained within DIV tags, which relate to a row in a database table.
<?php if($this->userjobdetails != null) : ?>
<?php foreach($this->userjobdetails as $employment) :?>
<div id="employ_<?php echo $this->escape($employment['historyid']);?>">
<table class="employment-table">
<tr>
<td><Strong>
<span dojoType="dijit.InlineEditBox" editor="dijit.form.TextBox" onchange="markEmploymentForUpdate();" id="cmpy_<?php echo $this->escape($employment['historyid']);?>"><?php echo $this->escape($employment['employername']);?></span>
</Strong>
</td>
<td align="left"><input dojoType="dijit.form.FilteringSelect" store="rolestore" searchAttr="name" name="role" id="roleInput_<?php echo $this->escape($employment['historyid']); ?>" value="<?php echo $this->escape($employment['jobrole']);?>"></td>
<td align="left">
<span dojoType="dijit.InlineEditBox" editor="dijit.form.TextBox" onchange="markEmploymentForUpdate();" id="jtitle_<?php echo $this->escape($employment['historyid']);?>"><?php echo $this->escape($employment['jobtitle']);?></span>
</td>
<td width="15px;">
<input type="hidden" value="<?php echo $this->escape($employment['historyid']);?>" name="employid" id="employid_<?php echo $this->escape($employment['historyid']);?>"/>
<img src="<?php echo $this->baseUrl();?>/images/site/msg/small/msg-remove-small.png" border="0" onmouseover="this.style.cursor='pointer';" onclick="removeEmployer('emply_<?php echo $this->escape($employment['historyid']);?>')"/>
</td>
</tr>
</table>
</div>
<?php endforeach;?>
When the user 'saves' the page I want to then using dojo.xhrPost post the data for the elements on the page, so that the database rows are updated.
How would I go about this, having multiple 'rows'??
Thanks
Take a look at dijit.form.Form — the second example shows how to validate a form and do whatever actions you like when user submits it. AFAIK, dijit.form.Form doesn't care how many fields it has, and collects them dynamically.