PHP :: Re-populating A Form If Errors

Populating Form Text Box.
I have a PHP form for entering data on-line into a mySQL
table..
Because of the intended search facility, I require one field in the form
to be completed with exact and precise item names and spelling.
In testing, I had achieved this with look-up tables.
Ideally I would like the selected item from one of five drop-down lists
to autopopulate the form text field when selected.
Alternatively at least the option of copy and paste.
Copy and paste - for some reason will not work.
Autofill worked fine on a test form with no database connection
and in standard HTML and javascript, however, once the PHP
is added to the page neither method works.
Could anyone please point me in the direction of how to get the
autofill to work?

Populating Form Item From SQL Enum Set Rev#2
$sql = "SHOW COLUMNS FROM table LIKE 'subject' ";
$qry = mysql_query($sql) or die("Query not valid: " . mysql_error());
$res = mysql_fetch_object($qry);
// This returns a row with a field 'Type' containing 'enum(...)'

$res->Type = str_replace('enum('', '', $res->Type);
$res->Type = str_replace('')', '', $res->Type);
//now the string is something like this: one','two','three

$temp = explode('','',$res->Type);
//temp is an array with as much elements as the enum

while ($temp)
print array_pop($temp);

This way is much more easier using the elements since they are elements
of an array. Cleaver isn't it?

Populating A Form Wth Database Info
I've coded a piece of code which populates a form with data read from the database:

$connection=mysql_connec ("localhost", "f2821842", "f2821842");
$result=mysql_select_db("QUERIES");
$query=mysql_query("Select * from Emp_Details where emp_num = '$employnum'");
while($row=mysql_fetch_array($query))
{$empname=$row['emp_name'];}

<form name="webregform" action="webadmin2.php" method="post">
<input name="requiredname" type="text" size="30" value="<?php echo "$empname"; ?>">
</form>

When I echo the $requiredname, I get spaces and no data, and I know that $empname is not a space-it does read a value in a database. 1. How can I get $requiredname to print a value?

Populating Form Fields With Variables
The script itself works fine: mail gets sent. If there are any missing fields the form displays again with a list of the missing fields, but I can't get the form fields of the reloaded form to be populated with teh existing values. PHP Code:

PHP Not Populating Form Variables Passed Through POST Or GET
I"m running:

Windows Xp SP2
Apache2
PHP5
MySQL4

Everything seems fine and dandy, and I can do scripting on server side, but for some reason I cannot get data from a form to be put into a variable.

I have a simple form in an html page which submits, and in the php script that handles the input, none of the data in the form is going into the variables.

My php.ini file does have all the GPC's in place(ie variable_order, the way they get processed).

Populating Form From Database, Then Passing Results To Next Page
I have a multiple select input in a form that's being populated by a row
from my database as such:
<input type="checkbox" name="subm[]" value="$row[ID]">

That part is working fine as I can check the displayed page using View
Source and see that the value is the correct row number from the
database. It is then being submitted on a form by $_POST method to
another page where I want to evaluate the checkboxes and display the
contents of the entire row that corresponds to each value="$row[ID]"
that have been checked. But I can't seem to get it to work. I'm having a
problem passing the selected value. Can someone point me in the right
direction?

$query = ("SELECT * FROM `table`");
$result = mysql_query($query);

print "<p>Data for Selections:";
print "<table border=2><tr><th>You chose:";

foreach ($_POST['subm'] as $value) {
print "<tr><td>";
print "$row[ID];
";
print mysql_field_name($result, 1) . ": " . $row[name]."<br>";
print mysql_field_name($result, 2) . ": " . $row[address]."<br>";
print mysql_field_name($result, 3) . ": " . $row[city]."<br>";
print "</td></tr>";
print "</table>
";
}

if (!isset($_POST['subm'])){
print "<p>No matching entry ";
}

mysql_close();

Populating SELECT Multiple Form Field And Inserting Into Db
I have an update form where I'm trying to populate a SELECT multiple field with a list of 48 categories, from tbl work_cat. And show, as SELECTED, the one or many choices that the user had previously selected from the 48 categories which are stored in tbl cat_relations as $relation_cat.

Then allow the user to update their selections and update the database. But I can't get my form to work. First problem, I can't get the SELECT field on the form to show the categories, and then I don't know where to go from there. Below are my form page and my processing page. Code:

PHP Form Submission Data Errors
One of my variables "link1" that is storing a bunch of urls is having random
lines being duplicated on form submission. I just recently swithed servers
from a PHP 4.3.3 to a 4.2.2 version. This error has only started appearing
on the new machine. The script worked fine on the old server and still does
on the test server. Any ideas? I think it might have to do with the ini
config. i am stumped. here is a link to my code.

Thoughts On The Best Way Of Highlighting Form Input Errors?
I've been creating an application over the last few weeks and generally
pleased with what I have produced but one area is irritating me, form
validation.

At the moment the forms are simply static html templates and the form input
is checked using a validation class. Basically each form field is checked,
every error is stored to an array and at the end of checking of the complete
form, the array is output neatly at the top of the form.

However this is just not sexy enough for me, what I would like to happen is
for the input boxes with errors to be highlighted with a red border (or
something along those lines). I would also like to have the error messages
under the box which the error occured, not altogether at the top of the
page.

So I am currently wondering how to achieve this and what choices are open to
me on methods available? I've had a google around the web and found some
form validation classes but they basically do what I already achieved. I've
thought of creating the forms dynamically but that in itself doesn't address
how to achieve my goal.

I am sure many people have already successfully done this, if you have the
please could give me an outline (the theory) of how you did it. I don't want
or need code, I am simply stuck at the moment on how to approach this.

Populating Form In Parent Window With Value From Popup Window
I have a text box in the parent window.. now i want to populate this text box with a user_id that i will get from query within a popup window.. How do i get the value from the popup window to be passed to the main window form field in php. I know how it can be done in javascript but when i tried combining javascript and php i dont get the desired result.

Validation Errors From Displaying On Form Page Load Instead Of Submit
creating two forms and they work fine. Problem is that the validation errors display immediately on page load instead of on submit. Here's the full code for the form parsing: Code:

Supress Errors At The Page Level? Undefined Index Errors.
I'm creating a simple reply form, and if a form item isn't answered I
get an error:
"Notice: Undefined index: rb_amntspent in
c:inetpubwwwrootmackinawsurvey.php on line 36"
even if in the code I allow for an unselected item. (Code at the
bottom here.)
It works fine otherwise.

Now in the past I've made a change to php.ini to repress errors, but
for this project I don't have access to the php.ini.
Is there a way to supress errors on that page for that session?

Or, is there a way to just not get that error at all?

Here's a sample of the code that gets the form response. If an item is
selected, no error for that question, but if an item is not selected I
get the error above even though I still get a successful echo for the
"else" or "default" option which occurs if nothing is selected:

$rb_visited = $_POST["rb_visited"]; //Have you ever visited Mackinaw
City, Michigan?
if ($rb_visited == "1") {
$rb_visited_reply = "Yes";
} elseif ($rb_visited == "0") {
$rb_visited_reply = "No";
} else {
$rb_visited_reply = "You did not answer this question";
};

$rb_lastvisit = $_POST["rb_lastvisit"]; //When was your last visit to
Mackinaw ity?
switch ($rb_lastvisit) {
case "1":
$rb_lastvisit_reply = "Earlier this year (2003)";
break;
case "2":
$rb_lastvisit_reply = "Last year (2002)";
break;
case "3":
$rb_lastvisit_reply = "2 years ago (2001)";
break;
case "4":
$rb_lastvisit_reply = "3 years ago (2000)";
break;
case "5":
$rb_lastvisit_reply = "4 or more years ago (1999)";
break;
default:
$rb_lastvisit_reply = "You did not answer this uestion";
}

Populating A Select Box
What I want to do for members of my website is to allow them to click a button that will add their name to a list that populates a select box. I can do this on my own, but I was curious if I have to use a database in order for this to be possible.

PHP + MYSQL, Populating Drop Down Box
I am looking to generate a dropdown box from MYSQL data:

db name = h2, table = Working, Column = Home.

Populating Two Combo Boxes
I have one combo box (Category) which is populated by a MySQL table using php. I am trying to use the onchange javascript to submit the value selected so that I can populate the other combo box (sub-category) based on what was previously choosen. So can anybody give me a clue on how to submit the first boxes value so that the second can see what was selected (javascript function).

Populating Dropdown List
I am wanting to know how to populate the <SELECT> dropdown menu with all existing values from the DB and also have the value associated with that id selected. This is an updating area of my admin. Any ideas? Here's the code so far. At the moment it only retrieves the value assigned to that id. PHP Code:

Populating A Drop Down Menu
How would I go about populating a Drop Down menu from data in MySQL? If you can show me wither a URL to learn this, or if you feel like telling me here, that would be great.

Dymanic Populating Checkboxes
i like to retrieve a value from DB,
accordingly i like to display the checkbox, either checked or not.

Now im able to retrieve the value from DB, i can show even show the
checkbox, but i cant make it checked????

the Code i used to display as follows,

#Require the database class
require_once('../dbinfoinc.php');

#Get an array containing the resulting
record
$query = "SELECT * FROM event";
$result = mysql_query($query);

$num=mysql_numrows($result);
mysql_close();

$i=0;
while ($i < $num) {
if(($i % 2) == 0) { $c =
'"TableDetail"' }
else { $c = '"TableDetail2"' }

$event_id=mysql_result($result,$i,"event_id");

$event_name=mysql_result($result,$i,"event_name");

$event_publish=mysql_result($result,$i,"publish");
// if(($event_publish) == 0) {
$event_publish = 'No' }
// else { $event_publish = 'YES'
}
$take_action = 'Delete'
echo "<tr>
<td Class='navText'
align='center' class=$c>$event_id</td>
<td class='navText'
class=$c>$event_name</td>
<td class='navText'
align='center' class=$c><input type='checkbox'
name='publish' if($event_publish==1){'CHECKED'}?></td>
<td class='navText'
align='center' class=$c><a
href='delete_event.php?event_id=$event_id'>$take_action</a></td></tr>";

$i++;
}
?>

Populating A Select Box From A Database
I have a subscription database that allows you to edit a subscribers data. For example, you enter a search term like "Bob" and it returns a list of everyone named Bob. You click the one you want and it goes to a subscriber detail page, where all of his data is populated to a form. You can directly edit this form and click save to overwrite it.

The only issue is the select boxes (drop downs) - say I have four options in the drop down box, red, blue, green and black. On my details page for bob I end up with five, I have those four PLUS whatever was saved for his color, so it shows up twice. If Bob is green, my select box looks like:

Green
Red
Blue
Green
Black

How can I make it so that Green only shows up once?

Populating Drop Down And Table
I have a database with the following fields.
Name | Company | Date

values in each column could be repeated, or not. as in there could be several same names with the same company with different dates, or different names with same company.

How do I populate a table with this info and have drop down boxes, so that I can narrow down the search? For example:
Name | Company | Date
12 | 1 | 1929
13 | 1 | 1929
14 | 1 | 1929
12 | 2 | 1929
12 | 4 | 1929
13 | 1 | 1941
12 | 6 | 1929

So if in the drop down I select '12' under name, only those entries with 12 are shown, and then I can further sort it by selecting only '1' under Company.
Hope I'm clear, I manage to complicate things when I post them.

Populating An Auto_increament Attribute
I have just designed a BD. there are companies with auto increament which are integer. When I try to populate it after i tool value from another form and try to insert it into a table i recieve an error! I don't ask the companyID value from user as it is auto increament. Don't tell me to put NULL for that entry as it didn't work as well! Code:

Populating Table Information
I am trying to dynamically generate bgcolors with to help populate some information for a site. Here is the code:

Auto Populating A Drop Down Box
Basically I'm setting up a website which needs an populated drop down box made up from all fields in a specific column of a table in a mysql db....

Here's the code I've made up using various tutorials....

<?php
$user = "";
$host = ""
$password = ""
$dbName =  ""

/* make connection to database */
mysql_connect($host, $user, $password) OR DIE( "Unable to connect
to database");
mysql_select_db($dbName); //did you forget this line?

$sql = "SELECT model FROM usedVehicles";
$query = mysql_query($sql) or die(mysql_error() . "<br>$sql"); //use the or die(...) part ONLY IN DEVELOPMENT
?>

<form action="action" method="post">
<select name="option">

<?php
while ($row = mysql_fetch_array($result)) {
    echo "<option value="" . $row['model'] . "">" . $row['model'] . "</option>
";
}
?>
</select>
<input type="submit">
</form>

I've left out the connection details for obvious reasons... When I upload and try to test this, jus a blank drop down appears... there are definately fields in the column as I have tried the query on phpMyAdmin.

Populating Fresh Data
Im building a live web chat using php, mysql, javascript, ajax I am fine with sending and retreiving data in real time using ajax as the transfer protocol However, my problem, which is becoming a nightmare, is how to I display it !! Lets say the below is the chat screen,

<div id="messages">
Jamie - Hello
Peter = Hi
Jamie = Nice Day
Peter = I know
</div>

Lets say Jamie's next message is "What are you doing today". I can send that off to the database no problem and I can get peters live chat to pull that message in, however, how do I get it to display underneath Peter = I know

As a practice I was using

document.getelementbyid('messages').innerHTML = document.getelementbyid('messages').innerHTML + "<BR>What are you doing today"

That works, but, lets just say the chat gets to 1000 lines, for each new message those 1000 lines have to be pulled out via javascript, a new line written to it and then printed back in the div

Im not sure but I'd take a good guess at this not being optimal and will cause CPU load on the users machine

What is the best way to do this! Chat messages are deleted after 1 minute, but that doesnt matter because by then they have already been sent the the users chat. Just incase you are thinking of completely repopulating there chat with every message they have sent since the chat started..

Reading Directories, Populating Select Box
Can anybody tell me how to read through a directory, get the file names and populate a select box(pulldown menu) with these names? I know how to loop through a directory using something like this:

while ($files = readdir($handle)) {

$ext=substr($file,-4);

if ($files != "." && $files != ".." && $ext == ".php") {

do some wild and crazy stuff here;

}

}

It's the populating the select box (pulldown menu) while doing this that I need the help on.

Populating Array With Mysql Results
I'm returning a result set of one field in a table, and want to populate an array with the results. I can't for the life of me figure out a simple way to do this without using mysql_fetch_array() to cycle through the results, append that to a var, then explode that into a var and pass that. PHP Code:

Populating Dynamically Generated TextFields
If one assigns something to the value iattribute off an input /text field, that value will be displayed when the form is loaded. My problem is, when building a form dynamically, the only way I can get the text to show is by doing the above, however, as this is an update form, when the form POSTS to processUpdate.php,
both the data entered by the user AND the data in the value attribute are inserted/updated.

So, how does one dynamically build a form where the data is displayed in the text fields but value atttribute is not set. PHP Code:

Populating A Drop Down List With The Year
i'm trying to loop the years from 1960 - 2010 with the function get_year() below:
PHP Code:

Populating DHTML Menu From MySQL
I have a client that would like to have drop down menus added to a nav
bar that is generated from MySQL. Is it possible to have a dynamically
driven DHTML menu from MySQL?

Populating Arrays From MySQL Query
I have the following code: http://pastebin.com/746601

The field 'material' in 'is_material' contains multiple values for each
record in 'is_details'. Because of this I have used
'is_material_lookup' as a reference lookup table containing the
'style_code' and 'material_code' which refer to their full details in
the respective tables.

Currently I have got the script outputting all the details and one
material then in the next block of data, repeating the details with a
different material. What I would like to achieve is having 1 block of
data with a list of all materials in that, instead of the repeat, but
sadly I can't know exactly how to do it.

Populating Dropdown With Mysql Entries
I would like to create a combobox in Flash which is populated with mysql data and programmed with php. For example: There are 3 entries in database .ie. apple, bannana, peach. Now these I want in combobox in Flash MX/flash5.

Populating A Drop Box With Results From Database
I have a MySQL database with a table (category) with two fields, catId (int) and category (char(50)). What I want to do is to get all category names in this database and place all of them into a dropdown box on a web page so that the user can choose from the list of available categories.

Populating Table With Text File
My web hosting service uses phpMyAdmin and at the bottom of the screen
iis an area where I can upload a text file to populate a table.

I have a table named groups with two fields:
groups_idauto-increment primary
groups_name

So how do I create my text file to populate this table?

I'm going to use MS Notepad.

If it's set to auto-increment, do I need to include the number? If
so, does it start at 0 (zero)?

Would the file look like:

0,students
1,faculty
2, staff

or just

students
faculty
staff

Auto Populating Multiple Drop Downs.
im trying to create two dropdowns, i need the first one to be the category and the second one to be the subcategory. The category drop down autopopulates with the correct info from the database. and uses the table "category", the value of each drop down is represented by the "cat" field in the table (cat is basically and integer id number) and "Category" is used as what the user actually sees in the drop down (category is the actual word of the category).

Once the category is selected i would like to have the sub category auto populate with everything that has the same values as the selected category (cat)

Here is a break down of how the tables work.

Table 1 Name: "category"
Fields for Table 1: "cat" (the id number), "category" ( the actual name of the category)

Table 2 Name: "subcategory"
Fields for Table 2: "cat" (corresponds with the cat id from table 1 to pull the correct data), "subc" (the basic id of the subcategory), "subcat" the actual name of the subcategory.

so the way i see it, have a normal drop down populated by a php query. then on change, populate subcategory drop down where cat = cat and display sub category.

Populating The Drop Down And Pressing The Submit Button
1. The first is I have two drop down menus. The first is "year" and the second is "mfr". When a user selects a year from the drop down it then populates the second drop down, mfr, from the MySQL database. Theat is working fine. But the problem I am having is the "submit" button (which I have labeled as "browse"). When I click it. Nothing happens, no action tacks place. I have looked over the code and I can't figure it out. (See Code Box 1 Below).

2. Right now the "mfr" drop down is populated by the MySQL database and reads with a list like "Acr", "Alp", etc. These are abreviations. I need to set up an array to have them instead read the entire mfr name. Example: Instead of "Acr" it needs to be "Acura". Instead of "Alp" it needs to be "Alpine". I need these full names to appear in the drop down. I know I need to do something like this (See Code Box 2 Below) but I can't get my finger on it. Code:

Populating City,state,country Drop Down Menu
In the registration form I have city, state, country fields. I was
wondering if there was a database available on the net which has the
list of states in each of the countries. That way when a user selects
a country I could automatically populate the state drop down menu ...

Auto-populating Multiple Boxes... Php And Pg_fetch_array Problem?
There is a nice tool using jquery/php to populate multiple select boxes out
there, it must be a very stupid question but I?m trying to apply this tool to get data from postgreSQL, but I always get an empty [] value. It means that the...

Dynamically Populating Listboxes Based On Previous Selection
How do I populate the second listbox in this form based on the selection of the first listbox?

The first listbox is populated by this query:
$year_query = "SELECT distinct year FROM bench_data";
The second listbox should essentially be populated by this:
$industry_query = "SELECT distinct comp_desc FROM comp_desc, benc_data where $_POST['year']=bench_data.year and bench_data.comp_key=comp_desc.comp_key";

I know this is alot to ask, but please EXPLAIN the code. I am new and need the explanation elucidated and not just supplied code that will accomplish the goal. Code:

Populating "Month" Dropdown Menu
I can get the current month to display in a text box by inserting
value="<? print strftime("%B"); ?>"> into the code for that
particular text box on a form but how can I populate a drop down with
names of all the months with the current one being 'default' as it
would be in the text box example?

Errors In Php 5.0.4 Not In 4.3.10-16
I use the following statements:

$var = "";
$var = $_POST('varname');

In php 4, when 'varname' was not defined, $var remained an empty string;
In php 5, I suddenly get errors.

I solved these by including the function array_key_exist to check the
occurence of 'varname'
Is this the prefered way or is ther a better solution?

PHP 404 Errors
Some of my visitors keep emailing me about "404 page errors" on my website even though the pages are working perfectly. I have had this problem before , I figured it out and it was because of my "generators" which uses pure PHP codings like this - http://www.generators.hot-lyts.com/linkgen.php.

I had the solution, I made a sub domain and since then I wasn't getting anymore error mails however it occurs again, I have checked the new pages I've added recently and deleted some of them but I still have no luck, I still get tons of mails about this error pages and it is really frustrating because my hits have been dropping big time.

I really have no idea what the problem is with my website. I am not sure if it's my server or some php problems. Code:

Populating A List Box Or A Dropdown List Using Php And Odbc
I need to populate a list box and/or a dropdown list on a form. I have all
the bits and pieces together, all bar the code which takes the result of a
query and creates a list box.

Populating A Second "select" Field
is there a way to populate a second select field in a form based on input for a first select without reloading and using only php? Code:

Ereg_replace - Errors
Whenever I use the ereg_replace function like follows:

Connection Errors ?
host -{ The server }- is blocked because of many connection errors. Unblock with 'mysqladmin flush-hosts'

anyone know why this message show up allot of times and goes away only after restarting mysql and apache ? is there a other way if fixing this ?

Odbc Errors
I am using php and sybase as a backend. To connect to sybase I am using odbc connections. the odbc connection is working fine but when I run an sql statement i get the following error:

" Warning: Supplied argument is not a valid ODBC-Link resource "

I am using an external file which holds my connection function which is:
function db_connect()
{
global $dbhost, $dbusername, $dbuserpassword, $defaultdbname, $dsn;
$myConnection = odbc_connect($dsn, $dbusername, $dbuserpassword) or die("Unable to connect to Server");
}

$myConnection = db_connect();
echo "Connected";

$mySql = "Select * from branch";
$myResult = odbc_prepare($myConnection, $mySql) or die("Could not prepare Statement");
odbc_execute($myResult) or die("Could not execute Statement");

odbc_free_result($myResult);
odbc_close($myConnection);

I get an error after the sql query. Can anyone please tell me what's wrong with this?

PHP Install Errors
I want to install Twig for web email which requires apache and php so I've installed all the right stuff but it still fails. Error_log output:

[Fri Feb 16 11:57:07 2001] [info] Server built: Feb 15 2001 11:38:09
[Fri Feb 16 11:58:46 2001] [error] (8)Exec format error: exec of /var/www/htdocs/index.php3 failed
[Fri Feb 16 11:58:46 2001] [error] [client 134.202.1.235] Premature end of script headers: /var/www/htdocs/index.php3

I configured Apache using the commandline: Code: