PHP :: How To Give A Value To Non Selected Checkbox

Multiple Selected="selected" In A List.
I have a database with 1 to many and the many is a list with multiple
selects in a list. When I click on a master record I have as part of
my form the select statement for the multiple choice list. I want the
list to highlight the multiple chosen values. here is part of the
code.

$dresscat = specdresscat($dresstypeid); //this is the query to get the
values in the detail table
$drcatrow = pg_fetch_array($dresscat);// the array for the detail
table.
<select name="stages[]" multiple="multiple">
<?php
$allstages = allstages(); // this is a function to query my database
to get the value for the list.
while ($allstagerows=pg_fetch_array($allstages)) {
$eachstage= $allstagerows["stageid"];
$specstage= $drcatrow["stageid"]; // this is the point I need help
if ($specstage == $eachstage){ ?>
<option value=<?php echo $eachstage;?> selected="selected"><?php echo
allstagerows["stage"]?></option>
<?php }else{ ?>
<option value=<?php echo $eachstage;?>><?php echo
$allstagerows["stage"]?></option>
<?php
}
}
?>
</select>

Why Do Does This Give Me An Error?
I have a file called t.php:

<html>
<body>

<?php
require "http://localhost/~derek/l.php";
rp(3);
?>

</body>
</html>

and l.php:

<?php
echo "define function...";
function rp( $num )
{
echo "rp: $num";
}
echo "defined";
?>

When I load t.php into my browser it prints "define function...defined
Fatal error: Call to undefined function: rp()
in /home/derek/public_html/t.php on line 6"

So, I know the library code is being loaded and executed (because the
strings are echoed) but rp() isn't being defined. Experimentation shows
it's not just functions which I'm not getting defined, as I get the same
sort of error if I just define a variable and try to get it printed in
t.php.

Can someone tell me what I'm doing wrong? I'm using PHP-4.3.4 on SUSE
LINUX-9.1.

This Is A Function That Will Give The Paging
You often need to write function to generate paging, like 1, 2 , 3
etc, na? I also needed. And maintaining all those was a pain, and
every time reinventing the same wheel.

This is a function that will give the paging, while you don't need to
do anything. Just, where you are writing suppose:

$query = mysql_query("select * from tablename where something is
something or anything");

Just write

$query = page("same query here, nothing else");

THe page function will return mysql_query() result. Isn't it great?

Here is the function

//$sql query to execute
//$num=number of rows to show
function page($sql,$num=10){
$pg=mysql_query($sql);
if($pg)
$norow = mysql_num_rows($pg);
$nopage = ceil($norow/$num);
if(!$_POST["pageNo"])$pageNo = 0;
else $pageNo = $_POST["pageNo"];
$sqlq = $sql . " limit ".$pageNo*$num.",$num";//.(($pageNo+1)*$num);
// echo $sqlq;
if($nopage>1){
echo "<scr"."ipt language="javascript">function gopage(num){";
echo " document.forms['pageform'].pageNo.value=num;";
echo " document.forms['pageform'].submit();";
echo "}</script>";
echo "
<table border=0 cellpadding=0 cellspacing=3 width=100%
bgcolor=#DDDDDD><form name="pageform" method="post">
<tr>
<td width="30%"><input type="hidden" name="pageNo">";
$hl = ($pageNo+1)*$num<$norow?($pageNo+1)*$num:$norow;
echo "Listing ".($pageNo*$num+1)."-".$hl." of ".$norow."</td>
<td align=center width="40%">";

for($i=0;$i<5;$i++){
if($tp+$i<$nopage){
if($tp+$i==$pageNo) echo $tp+$i+1;
else echo "<a href="javascript:gopage(".($tp+$i).")">".($tp+$i+1)."</a>";

if($tp+$i+1<$nopage)echo ", ";
else {echo " <a href="javascript:gopage(".($tp+$i).")"> >> </a>
of <a href="javascript:gopage(".($nopage-1).")">".($nopage)."</a>";}
}
else break;
}

echo "</td><td align=right width="30%">";
if($pageNo>0)echo "<a
href="javascript:gopage(".($pageNo-1).")">Previous</a> ";
if($pageNo<$nopage-1)echo "<a
href="javascript:gopage(".($pageNo+1).")">Next</a>";

echo " </td></tr></form></table>";
}
return mysql_query($sqlq);
}

Every 1000 Views Give 1$
Working on a script where everytime a user gets 1000 views they get 1$. Everything works except the if for the amount of views..

How can I make it so it will see if the views is a thouand.. ie.. 1000, 2000, 3000 etc. I don't want to manually have to type 1000 - 500,000!

ie.

<? php
if($views = 1000 || $views = 2000 || $views = 3000){
givemoney($user);
}
?>

How To Trim Parameters And Give Them A Prefix?
// trim the parameters
foreach($_POST as $varname =$value) {$varname = trim($value);}

// give them $p_ prefix
import_request_variables('gp', 'p_');

When I do the above the parameters are not trimmed in the end.

Example:

$lastname = ' ' is trimmed to $lastname = '', but $p_lastname = ' '

What Output Will Give Errors With Cookies
I understand the fact that outputting before sending headers will screw up the whole cookie thing., But what constitutes as output?

this is probably a dumb question, but I am validating form information, and if it is valid I am setting a cookie (or I want to). The problem is that when I am validating the info I am returning true or false. So actually I am returning before I set the cookies.

Would that be the problem?? I am just curious because I thought I had this down...after many hours, and my brain hates me. If returning is the problem, what can I do differently to validate info, and still allow myself to set the cookie after the info has been validated? I tried sending the user to another page (but that was stupid, if i cant set a cookie, then i cant send new header()).

How To Use Sessions In And Give Session Timeouts
I am building a website using the PHP 5 and MySQL .I have set up all the required database ready and the PHP pages are linked with the database.I am able to traverse through my pages by querying through the database and inserting and retrieving the stored data from MySQL database.

Now i am trying to use sessions in my website so that i can limit the users access and implement session timeouts say if the user is not using the web page for 5 minutes then the session is timed out and he is again asked to relogin.

Can You Give Me Example Of Website Using Postgresql Database?
I am planning to use postgresql for my next project (social network). Anyone can point me any website using postgresql database?

Inconsistent IP Address Extracted With PHP - Give Me A Hint, Please!
I'm not new in PHP at all but it's the first time when I used the variable
_SERVER[REMOTE_ADDR].
I need to track for my client, the vistors IP addresses but I got stuck with
this
problem:
I have a cable access at home so I have dedicated ( fixed IP address) and I
tried to record on my server side php script my IP address from home, but I
got always 2 distinct IP addresses: one is x.y.z.204 when I type directly
the PHP script URL from the server into my browser and the second one is
x.y.z.236 when I used a booked link (from my browseer favorites) to the same
PHP URL on the server and also when I use a http link from any html document
towards the same above mentioned PHP URL on the server.
So, I get reported 2 IP addresses of my home computer when I try to record
on a php side script my home computer IP address always the same in the
above mentioned circomstances.
I've tried it for almost 2 days with the same results.

Using Update To Give A Table Default Values
I have a table that is already created but all the fields are set to NULL by default. How would I go about changing this with a mysql_query(), so that if only say 2 of 5 fields are given a value for a column the rest will fill with the default value?

How To Give Personal User Url On A Dating Website Just Like MySpace Does?
I had posted a question regarding how to design high traffic fast
dating website on different PHP groups. I got a lot of useful
responses. I have added links to all the posts below for reference.

I am back with another question regarding the dating website. How do I
go about creating personal user url on the site like it is done on
MySpace.com

The personal url for user1 will look like www.myWebSite.com/user1

The first solution I know is to create a folder for each user and put a
index.php script that show the user profile. But then there would be a
lot of folders. how can I catch the user1 from the url request string
www.myWebSite.com/user1
The ssecond solution would be to set the page not found script in
Apache. But then again I need to capture the 'user1' from the request
string.

Discussion 1-
http://groups.google.com/group/Prof...96e1b11a08c5cf2
Discussion 2-
http://groups.google.com/group/php....5b3b100d70a196e
Discussion 3-
http://groups.google.com/group/PHP-...5531288b538882e
Discussion 4-
http://groups.google.com/group/mysq...8cff7c40c161889
Discussion 5-
http://groups.google.com/group/OOP-...ba2c6e36c3bc006
Discussion 6-
http://groups.google.com/group/PHP-...6c403cec4bc14e8
Discussion 7-
http://groups.google.com/group/comp...712f83d1cdb278d
Discussion 8-
http://groups.google.com/group/alt....14bf3f43c74f633

Php/MySQL Login - Can Anyone Give Me Suggestion As To Why Thiscode Does Not Work ?
I have followed the step by step instructions in the login manual

Created login page
Use LSC01

This works - login and get redirected to the proper page

My problem is the cookie variables don¹t show up (created setsessions.php
page and readsessions.php to test - works)

On the page, I¹ve applied LSC03 * Access by Cookie * this seems to work
since it allows my to enter the page I want

I then applied the LSC05 * User Details -
The correct rsUserDetails is set up

I then include Dynamic Text to the page that is suppose to show the
registered users information but it doesn¹t

Have torn page apart 50 million times I know it¹s one line or something
extremely simple but I can¹t figure it out.

Code for login page

<?php require_once('../../Connections/reg.php'); ?>
<?php
// Buzz inet PHPLSC01 - Login & Set Cookie
$myUsername_rsLogin = "0";
if (isset($HTTP_POST_VARS['email'])) {
$myUsername_rsLogin = (get_magic_quotes_gpc()) ? $HTTP_POST_VARS['email']
: addslashes($HTTP_POST_VARS['email']);
}
$myPassword_rsLogin = "0";
if (isset($HTTP_POST_VARS['password'])) {
$myPassword_rsLogin = (get_magic_quotes_gpc()) ?
$HTTP_POST_VARS['password'] : addslashes($HTTP_POST_VARS['password']);
}
mysql_select_db($database_reg, $reg);
// Verify Login is correct
$query_rsLogin = sprintf("SELECT email, password FROM parent WHERE email=
'%s' AND password = '%s'", $myUsername_rsLogin,$myPassword_rsLogin);
$rsLogin = mysql_query($query_rsLogin, $reg) or die(mysql_error());
$row_rsLogin = mysql_fetch_assoc($rsLogin);
$totalRows_rsLogin = mysql_num_rows($rsLogin);....

Give User A Choice Of Fields To Search With Full Text/Boolean?
I want to give the user the option of searching a combination of the fields $Topic, $Subtopic, $Theswords in Boolean/Full Text. The problem is if I make any of the fields empty prior to the query, I get a MySql syntax error message. The user has the option of selecting fields via checkboxes, but when the field is not selected the user will get the error message because the variable is empty in the query statemtnt.

Are there any code suggestions for avoiding the error message when a field is not selected by the user to be searched? Code:

No Database Selected
what I've got it a simple system that includes pages the reference to which are in a database. This bit worked a treat. until i tried to put something that did some work in one of the included pages. PHP Code:

Selected Box Action
how to set the respond to action when user choose an option inside a <select></select> box? eg. when user select an option in the <select> box then php search for data inside mysql and then display the required data on the <input type=text>.

Php Dropdown - Selected?
i'm using this to pull information from a database and display it in a dropdown list. The problem is I am using the name code when I goto the "Update" page, but i'm not sure how to add the "selected" field, that way what ever I saved in the database will load.

Example: I goto add and it add's the "id" of 7 to the field, when I load the page again it auto's back to 1 instead of clicking in-to 7 as thats whats already there... = ) Code:

Remember Selected Value DropDown
I am attempting to have a drop down menu that displays url from 1 table called links the same table also contains a name for the url. i.e. google.com = name and http://www.google.com = url The value of just the url from links is stored in a separate table called users in a field named search after form is submitted by user. Everything Posts correctly except when go back to edit the record and submit the form the drop down box does not retain the value of what had been previously selected although it is the correct record until form is updated when whatever value is currently in the drop down list will update the record.

I know the code is doing exactly what it is told to do, because I have no functionality in it to recall the value from the field search in the users table. I don't know if it is something I would have to add to the array or option value. I can get it to work if it is all hard coded, but I am attempting not to do this. I am pretty new to php mysql and any help is greatly appreciated.

This is the code I am working with that came mostly from a tutorial. PHP Code:

Resize Only Selected Images
I have an image upload script that is resizing every image that a user uploads.
What I would like to do is have a check box added to the upload form that a user can check to resize the image if they want to or leave it unchecked to just upload the image.

So if the box is checked the original image gets uploaded and a resized image is created as well If the box is not checked just the original image uploaded. I tried adding a check box but every image still was resized - checked or not - Im assiming because it was in the foreach loop so I dont know what to do. Code:

PHP And A Pre-selected Option In A Form
I've got a little text editor thing on my site, and to open a file, you input the file name and choose the folder from a drop-down list and click the Open button. The default filename is "index" and that file keeps track of all the files contained in that folder. It works okay. What's annoying is that regardless of what folder you're messing around in, the drop down list always goes back to showing the default folder name as the page is reloaded. I want it to be preselected to the folder it's in.

It seems simple enough--the folder you're in is $_GET['dir']. But I've read some HTML form how-tos, and I can only see that you can preselect using the tabindex="" or selected attribute.

Radio Button Selected
What I wish to do is have a radio button pre-selected depending on what is pulled from the database.

For example if I had 2 radio buttons:

o Yes o No

One has the value 1 and the other has the value 0

If 1 is selected from the database then Yes is pre-selected, else 0 is pre-selected. How would I do this?

Setting Selected Option Using PHP
I have a drop down list which I want to be set to what was selected when the form is submitted. As the form is processed by the same page, it returns to the preset default. How would I go about making this selection box set the option selected on the next page?

Selected Value In Dropdown Dependent On Database Value
i need a way to make the selected value of a drop down dependend on a result from a database query. PHP Code:

Export MYSQL Selected Rows
Any ideas on how to export selected rows from a mysql table into a schema.sql file?

I want to use mysqldump --opt database table but I don't wanna dump the whole table only results of a select.

PHPMyAdmin does it but I need to do it in shell script code.

Get The Info Depending On The Selected Option
I have a page that is intendend to update a table field on DB i have a drop down menu wich displays all the position fields available on db and a text area where i will insert the new data.

now the problem: I'm using the query SELECT * FROM table WHERE position = $position
the $position is the choosen option from the drop down menu. then have echo "<textarea>$functions</textarea>" where it will show to me the present data available on DB after all this, i have another query UPDATE table SET functions = $functions WHERE position = $position

All is ok and working except the fact that he doesn't displays $functions
when selecting a option from the drop down menu?

1 Record In Database But 2 Records Were Selected?
i'm retrieving a record from database. below the record, i have a yes and no button, where i can let user click and then count the number of yes and no.

everything works fine, but when i start to click the yes or no button, my record ,becomes two records. i only have one record in my database, but two same record was displayed.
this is the code: PHP Code:

Selected Hyperlinks Activate PHP Script
I have a sitemap page with approximately 20 different pages listed on it, 10 of which are password protected, 10 of which are not. I want the 10 non password protected page links on my site map page to execute the following PHP command when they are clicked:

<?
session_unregister("CustomerID");
session_unregister("UserID");
session_unregister("FirstName");
session_unregister("AccessLevel");
session_destroy();
?>

How do I integrate this PHP function into the following code so that, when one of these links is clicked, the PHP script above is executed:

<li class="content_text"><a href="index.php">Home Page</a> - The home page for #*$! </li>
<li class="content_text"><a href="login.php">Login Page</a> - The login page for #*$! </li>

However, I don't want the PHP script to execute when a user clicks one of these links:

Set The Selected Text For Option Box Of A Menu..
Having major issues with this simple task, and I cant work out why its
not doing as it should/expected.

Basically, ive got a drop down box with the added bit of php:

<select name="title" id="title">
<?php
if($_SESSION['title'] != "") {
echo '<option value="'. $_SESSION['title'] .'"
selected="selected"></option>'
}
else {
echo "<option value="" selected></option>";
}
?>

<option>Mr</option>
<option>Mrs</option>
<option>Miss</option>
<option>Ms</option>
<option>Dr</option>
</select>

Basically, if the user submits a form, the title is stored in the
session, this works fine, and then when returned to the page, the value
should be filled in.

Yet, when they return to the page, there value they selected is not
shown in this box, instead, theres just a gap. The source seems
correct, and this is whats shown:

<select name="title" id="title">
<option value="Mr" selected="selected"></option>

Html Pre-selected List Box Scrolled
Is it possible to have a list box with a pre-selected value automatically scrolled to that value when the form is loaded? i.e. if the size of the list box is 3, and the 4th value is pre-selected, I want the list box to appear starting at the 4th value in the form.

Showing What Was Selected In Drop-down Menu
I'm trying to create a form where the user inputs data, i validate it with php, and if some of the input is missing or invalid then i print out errors with the form filled out with the information they entered/selected previously so they can change/add to it.

The problem I'm having is setting my drop down menus to be selected when the form is returned instead of returning their original state. The menu I have is named A ($A). Here's an example of what I'm doing that isn't working: PHP Code:

Keeping Select Option Value Selected
I'm trying to learn how to keep the value "selected" once a form is set to read the values and then f-write them (all within a loop). The part that's giving me a problem is the part which would return the value of the form as "selected". PHP Code:

Display Selected Record That Has Been Passed To URL
I have a page which can list all records from a table in my local database.

For each record that is displayed there is link next to it. When the link is
pressed another page opens up and the ID (automated primary key in my table)
of the record selected is passed to the URL.

So now the URL looks something like this:

http://localhost/Webpages/newpage.php?recordID=3

How do I display the details of the record that has been selected on this
new page?

Multiple Select Box With Selected Option ...
first of all I have three tables (tournament_game, umpire_game, umpires). Updating tournament_game umpires, each game may have from two to four umpires, and those umpires can work from one to n number of games, that's why I'm using connection table umpire_game. I'd like to get selected multiple select box of umpires in game, it could be two to four options, with all umpires listed as options example:

<select name=umpire_game[] size=10 multiple>
<option value=$umpire_id>$umpire_name</option>
<option value=$umpire_id selected>$umpire_name</option>
<option value=$umpire_id selected>$umpire_name</option>
<option value=$umpire_id>$umpire_name</option>
<option value=$umpire_id selected>$umpire_name</option>
<option value=$umpire_id selected>$umpire_name</option>

PHP Code:

How Do You Do Selected In A Multi-value List Box In A Form?
I am populating a list box from a dictionary table and then picking
muliple values to insert into a detail table. How do I get the list box
to have multiple selections highlighted when I pick one of the master
items?

How To Use Selected Data In The Listbox For My Query..
I have a multiple select listbox and I pass the data using array in php. How can I use these records to my query, or in one select statement. For an example:

I have a listbox of item codes; the user selects an item code/s to view it's inventory. The selected itemcodes will be pass on php using array and these records will be used as a query or search items on my database.

Select Pictures And Show Selected
I need to make a webpage (php) where users can select some
excercises(with picture) out of 40, and then display a page with only
the selected excercises (with their picture) that you can print.

Display The Content That Is Selected From The Dropdow
what im trying to do is, i have a drop down that displays all the files that are in the directory when the page loads. when a item is selected from the drop down list i want that file to load in the text area but i cannot get it to work, can someone tell me what im doing wrong? Code:

Calculating The Selected Prices In A Listbox
I am wanting to know how I can add the all of the selected prices in a
listbox? I need to add the selected prices altogether. The listbox
should have only the names of the products. The listbox should be like
below:

How To Get Form To Remember Selected Option?
I am now trying to make a form remember which color (a select list) they selected before pressing any of the submit buttons. I entered this code into the form:

<select name="color">
<option value="red">red</option>
<option value="blue">blue</option>
</select>

The submit is a GET type. For example, if the user selects blue and then clicks on either one of the "submit" buttons, I want the form to remember/show the blue option. Right now the select list always goes back to red, the first choice. I tried doing a <?php echo $red; ?> and <?php echo $blue; ?> for the option value, but this does not work since the select list needs a value.

Select Option Stays Selected
I have a normal drop down menu:

<select name="status" class="db_list_text">
<option value="All" selected="selected">All</option>
<option value="True">Active</option>
<option value="False">Inactive</option>
</select>
The idea for this is for the user to select either true or false and then to click submit to find data that matches either one. The coding is in the same page so the form basically just reloads the page and grabbes whatever info is requested. How can I with PHP, make it that the dropdown stays on the selected option when the page is reloaded?

Dropdown Menu <selected Name="...
I have a dropdown menu like this one:

<select name="country" tabindex="6" id="select_country">
<option value="">- Country List -</option>
<?php
$query = "SELECT * FROM country ORDER BY country ASC";
$result = mysql_query($query) or die('Error, query failed');
while ($row = mysql_fetch_array ($result)) {
$country = $row['country']; ?>
<option value="<?php echo $country ?>"><?php echo $country ?></option>

After I have selected one country and I hit the submit button it go back to the default value in the box, what I want is to show the selected value in box until I selected another value.

If the default value is Australia when I start this page, I then select USA and hit the submit button it reset to the default value "Australia", but I want it to show USA as long I not select anything else. I have tried to use session in selected="<?php echo $_SESSION['country']; ?>" but it didnt work.

Print All Selected Items From Dropdown??
I have a program that allows users to select a 'New Graduate' membership, and for each one selected, they get a free tshirt. I've got that working using a 'is_new_grad' function that I created. So, if they select 2, they see 2 dropdowns to select the sizes of them. It then sends an email w/ the membership info and the shirt sizes selected.

However, when the email is being sent, it's only returning the size of the LAST item selected in from the drop-downs. Example: if I sign up for four of these memberships and select four different tshirt sizes (1 Medium, 1 Lg, 1 XL, and 1 XXL), the info in the email is all XXL: Code:

Echo Error In Selected File
Is there a way of echo'ing a reported error in a selected file, Like if i have a file exists validation can i echo the reported error in my file_exists.php file and place the echo on the page where i want it to show?

Renaming A User Selected File
I have an upload script that allows users to upload images to the server/database. My problem is that if a user uploads a file with the name of a file that already exists on the server, it overwrites the existing file. If I have a script that randomly generates 3 letters/numbers, how would I insert those in between the file name and the file extension? I'm thinking the solution would have something to do with implode, but I'm not sure how I would use it.

I'm using the $_FILES['upfile']['name'] variable.

Associate Radiobuttons With Selected Value From Dropdownbox
I have made the following code to show a dropdownbox filled with data from a mysql-database. It also shows 2 radiobuttons. What I don't know is how to associate the radiobuttons with the selected data in the dropdown.

For example, I select the person 'Bill' from the dropdownbox, then if I select one of the radiobuttons and I submit the values, I want that value to be inserted/updated in the database for the user 'Bill'. Code:

Best Way To Show Selected Option In List Box
I need a way so that we can show selected option in list box without using if condition in loop since some times we have lot of options and we have too much if conditions to check.

Selected Item To Use For Multiple Variables
I've got a list of 'items' and those 'items' have unique 'categories'.
I have 1 table for 'categories' that contains a list of category_ids and their 'category_names'. The other table is the 'item' list. And here's where I'm running into difficulty, I'd like to have the 'item' table contain the category_id from my 'categories' table as a Foreign Key and that category_id's 'category_name' in a 'category' field. I'm getting the Foreign Key, but I can't seem to grab the 'category_name' and assign it to my $category variable in the 'item' table.

I've got a page to insert items and a page to update these items. I was working with the update first. And I'm able to grab the list of categories and select the correct one, but I'm not seeing where to assign the category_name to the $category variable. Here's dropdown list: Code: