PHP - How Do I Change This Code From A Checkbox To A Select Menu

hi guys

i made a ranking for my league,
u can choose different rankings in multiple games and gamemodes

http://www.acidleague.com/League/ffarankings.php

now u have a select menu for the game, and then the gamemode
the select of the mode, contains all the modes for all games

what i want is, u fist select the game, and then u instantly get a dropdown with the modes correspondeding to that game

how do i even start with this?

my select menus are built like this

[php]   $types=mysql_query("SELECT id,name FROM ffa_types");
   while(list($id,$name)=mysql_fetch_row($types)){if ($id == $typeholder)
                        {
                        $typelist.="<option selected value='$id'>$name</option>\n";   
                        }
                     else
                        {
                        $typelist.="<option value='$id'>$name</option>\n";
                        }}
   $games=mysql_query("SELECT id,name FROM ffa_games");
   while(list($id,$name)=mysql_fetch_row($games)){if ($id == $gameholder)
                        {
                        $gamelist.="<option selected value='$id'>$name</option>\n";   
                        }
                     else
                        {
                        $gamelist.="<option value='$id'>$name</option>\n";
                        }
                        }


<form action='ffagamerankings.php' method='post'>
         <select name='game'><option>Choose Game ...</option>$gamelist</select>
         <select name='type'><option>Choose GameType ...</option>$typelist</select>

         <input type='submit' value='Ranks' />[php]

Hey guys, I am having trouble figuring this out. I currently have a database that automatically sorts itself by the id of the data. But, I am wanting the user to be able to change the sort order via a select box. I already have the select box programmed with the options, but I am not sure how to go about coding the options to change the sort order of the displayed data. What would the best way to go about programming this be?

Thank you very much ahead of time!!

There are about 400 records in a database with a field zabp_package = T_SIMT.
They were uploaded in an order where specific category lists were uploaded together.  For example, first 100 dining, then 75 insurance, then 150 health, then 75 cars.  So from an auto increment standpoint they were inserted in that order.
I want to select at random 100 of these 400 and update two fields.  My update statement is:
update  `usersOld` SET  `m_org` = 'ZABP.org Corporate HQ' , `m_orgID` = 'PFL2a96bW' WHERE `zabp_package` = 'T_SIMT'
This works for all, I just need to limit a random selection to 100.
Any Help on this, thanks in advance.

hello i had a look here its prety simple example http://www.formget.com/php-checkbox/     i wish something like this but with many more check boxes about 100,   the senario ls like this   there will be many users in a system,   each user profile will have these kind of questions   for example   Hobby : tenis [] badminton[] footbal[] carrom[]   programming language : perl[] python[] cobol[] as3[]   favorit biscuit : bakers[] subana[] ect[]     ect ect   these checkbox need to reflect for diferent users. administrator can add new category [ ie hobby , programming language] under each category he can add more options such as football, tenis ect   he should be able to have an interface to modify the options entered by user when a user request a change     what is a good way to implement this in php mysql      

Code: [Select]
<form action="checkbox-form.php" method="post">
<input type="checkbox" name="c1" value="1" /> Pending
<input type="checkbox" name="c1" value="2" /> Approved
<input type="checkbox" name="c1" value="3" /> Rejected
<input type="button" name="formSubmit" value="Fetch Results"/>
</form>
<?php
$con = mysql_connect("localhost","root","password");
if (!$con)
  {
  die('Could not connect:' . mysql_error());
  }
  mysql_select_db("Products", $con);
 $result = mysql_query("SELECT * FROM btp_reviews" );
echo "<table border='1' cellspacing='0' cellpadding='0'>";
echo '<tr>
<td>Id</td>
<td>Review</td>
<td>Name</td>
<td>Update</td>
<td>Status</td>
</tr>';
while($row = mysql_fetch_assoc($result))
  {
  echo '<tr>';
  echo '<td>' . $row['id'] . '</td>';
  echo "<td>" . strip_tags($row['review']) . "</td>";
  echo "<td>" . $row['r_name'] . "</td>";
  echo "<td>" ."<form action='radio.php' method='post' ><a href=http://localhost/editt.php?id=".$row['id'].">Edit</a>.</td>
                <td>" ."<input type='radio' name='r1' value='2' />Approve<br/><input type=hidden name=id value='".$row['id']."' />
                         <input type='radio' name='r1' value='3' /> Reject<br/>
                             <input type=submit value=Submit /></form>"."</td>";
                            
  }
echo "</table>";
mysql_close($con);
?>

Hi

I want to add student from a dropdown list to database but I have some problem.

This is my select dropdown menu code
<form name ="student" method = "POST" action ="confirmation.php">
<select name="name">
                                    <option selected>Select Student</option>
                                    <?php
                                    $arrStudent = executeSelectQuery("select * FROM user ");

                                    for ($i = 0; $i < count($arrStudent); $i++) {
                                        $student_result = $arrStudent[$i]['student_id'];
                                        $name_result = $arrStudent[$i]['student_name'];
                                    ?>
                                        <option value="<?php echo $id_result; ?>"><?php echo $id_result; ?>, <?php echo $name_result; ?></option>
                                    <?php
                                    }
                                    ?>
                                </select>
</form>

The output in the dropdown menu look something like this:
1, Alvin
2, Benny
3, Charles
4, Daniel
5, Eva
and so on...

After submitting the form, it will proceed to confirmation.php page.

At the confirmation page, I have the following variable:
$student_result = $_REQUEST['student_id'];
$name_result = $_REQUEST['student_name'];

I want to insert to database with the following insert query
$sql = "INSERT INTO student(student_id, student_name)
    VALUES ('". $student_result . "', '". $name_result ."')";
$insert = executeInsertQuery($sql);

It can insert successfully but, it will not insert the student_name.

May I know where I did wrongly?

Thanks
Ben Chew

Could someone help me I really dont know how to go about coding this, so i would be happy if someone could point me in the right way

Well what I am trying to do is use mysql_num_rows to call up how many rows in the table.

The using how many rows, use a menu with the numbers of rows that are in the table
ex below
mysql_num_rows gets 5 rows
so menu is

<select name="order" >
  <option value="1">1</option>
  <option value="2">2</option>
  <option value="3">3</option>
  <option value="4">4</option>
  <option value="5">5</option>
</select>

Hi, I'm no pro at PHP but I am trying to get a drop down menu to a authenticate before moving to the next part of the form.

What I want is once a selection has been made, ONLY THEN can the user move on, OTHERWISE a message echo appears.

This is the html menu box

<select size="1" name="title">
                    <option>Please Select</option>
                    <option value="Mr">Mr</option>
                    <option value="Mrs">Mrs</option>
                    <option value="Miss">Miss</option>
                    <option value="Ms">Ms</option>
                    <option value="Dr">Dr</option>
                  </select>   

Then this is what I have in the form PHP:

$visitortitle = $_POST['visitortitle'];

if ( HOW DO I GET THIS PART TO AUTHENTICATE AN OPTION HAS BEEN SELECTED? ) {
echo "<p>Please enter a title correctly<br />before you try submitting the form again.</p>\n";
die ( '<a href="pef.html">click here go back and try again</a>' );
echo $id;}

If anyone can help me sort out this part of the form I can move on as the rest is working fine?

Thanks
Gary
 

Hello,

I'm using a dynamically created select menu for a user to make a choice which will then be put in my database with the following code:

<select>
      <?php foreach ($course_number as $row) {
    echo "<option value = '{$row['course_id']}'";
    if ($errors && $_POST["course_id"] == $row['course_id']) {echo 'selected = "selected"';
    }
    echo ">{$row['course_number']}</option>";
    } ?>
Unfortunately, I'm having some problems figuring out how to pull off the selected value.  Right now my database portion looks like this:

$data = array('assignment_name' => $_POST['assignment_name'],

  'due_date' => $_POST['due_date'],

  'course_id' =>  $_POST["row['course_id']"]);

$inserted = $dbWrite->insert('assignments_instructors',$data)

While the assignment_name and due_dates work (they come from text fields), my course_id gives me an Undefined index: row['course_id'] error.  Any help would be appreciated.

Thank you.

I have a test database, I have two names in the database which get returned just fine in my html dropdown. I am trying to figure out how to figure out which item in the list was selected so that I can return information from another table based on the selected id. This is what I am trying now but I don't know how to proceed or if it is correct

Code: [Select]

$result = $mysql->query("SELECT * FROM names") or die($mysql->error);
?>
<select>
<?php
if($result){
while($row = $result->fetch_object()){
$id = $row->nameID;
$name = $row->firstName . " " . $row->lastName;
?>

<option value"<?php $id ?>"><?php echo $name ?></option>

<?php
}?>
</select>
<?php } ?>

I am using jquery .change function to perform an operation when a month is selected from a drop down menu. The change works but I am unable to update the value of the drop down menu with the updated month. My drop down shows the starting value as default even on change. Can anyone help. Following is the code snippet that does change and then the drop down menu form.

Code: [Select]
$("#monthName").change(function()
{
alert($("#monthName").val());
if ($("#post").val() == 1)
{
$("#monthselect").submit();
}
});
Code: [Select]
                   <form id="monthselect" action="<?=$_SERVER['PHP_SELF']?>" method="get">
                        <input id="post" type="hidden" name="post" value="1">
                        <label>SELECT MONTH</label>
                        <select id="monthName" name="monthName">
                                                     
                                <option value="January">January</option>
                                <option value="February">February</option>
                                <option value="March">March</option>
                                <option value="April">April</option>
                                <option value="May">May</option>
                                <option value="June">June</option>
                                <option value="July">July</option>
                                <option value="August">August</option>
                                <option value="September">September</option>
                                <option value="October">October</option>
                                <option value="November">November</option>
                                <option value="December">December</option>

                        </select>
                    </form>
Even after the change, January shows up by default even if I select say June or July. I tried something like following but did not work.

Code: [Select]
$("#monthName option[value=" + $("#monthName").val() +"]").attr("selected","selected") ;

Hi there,
i am relatively new to php, mysql, css etc but learning fast.
My problem is such;
i have a php file which is doing a SELECT mysql_query, WHILE results to strings, then ECHO the resulting rows to produce a list  formatted using  <table> and finally this <table> is inside a <form> which will POST the  changes back to the specific database.tble.row.

I wish to have a drop down menu within the <form><table> which will be populated from a separate database.table.

I have accomplished the drop down menu outside the <?php ?> tags inside  <form><table> which POSTS to a php file but my problem is to add the populating drop down menu inside <?php ?> an already ECHOing resulting rows from the sql query.

 i.e <?php blurb and stuff ?>
<form><table><tr><td>
<select name etc>
<?php
$result = mysql_query("SELECT * FROM tbl WHERE string = tble.rw ORDER BY column");
while($row=mysql_fetch_array($result)){
echo "<OPTION VALUE=".$row['column'].">".$row['column']."</OPTION>";
}
?>
</select> WORKS!!!!

but placing this inside

<?php $x =mysql_query[select] while {strings = conditions; echo ("<form><table><tr><td>
insert populated drop menu here
</td> etc ");
echo"";}?>

doesnt work and just leaves the select drop menu blank

Hoep you understand my problem.
I do not think i can attached the population WHILE loop to a string and just insert the string to the form but maybe i am wrong.

thanks in advance and if you go tthis far reading you must be on lots and lots of coffee

zark

the same page?

hello all, I was hoping someone could help me figure out how to add sizes to my shopping cart.
Right now the items add just fine. The site sells shirts and I'm having trouble displaying the sizes once selected and added to the cart. It will post the size selected, but if the customer tries to add another shirt the size will overwrite the last.

Any help is appreciated.

Heres how I add items to the cart...
Code: [Select]
session_start();
// Process actions
$cart = $_SESSION['cart'];
$action = $_GET['action'];
switch ($action) {
case 'empty':
if($cart) {
unset($cart);
}
break;
case 'add':
if ($cart) {
$cart .= ','.$_GET['product_id'];
} else {
$cart = $_GET['product_id'];
}
break;
The form with the size selection and add to cart button..
Code: [Select]
  <form action="cart.php?action=add&amp;product_id=<?php echo $row_rs_products['product_id'] ?>" method="POST" name="addcart" id="addcart">
      <table width="300" border="0">
        <tr>
          <td><label for="sizes"></label>
            <select name="product_size" id="product_size" title="<?php echo $row_rs_products['product_size']; ?>">
              <?php
do {  
?>
              <?php
} while ($row_rs_sizes = mysql_fetch_assoc($rs_sizes));
  $rows = mysql_num_rows($rs_sizes);
  if($rows > 0) {
      mysql_data_seek($rs_sizes, 0);
  $row_rs_sizes = mysql_fetch_assoc($rs_sizes);
  }
?>
            </select></td>
          <td><input type="submit" name="submit" id="submit" value="add to cart">
function displaying the cart..
Code: [Select]
function showCart() {
global $db;
$cart = $_SESSION['cart'];
if ($cart) {
$items = explode(',',$cart);
$contents = array();
foreach ($items as $item) {
$contents[$item] = (isset($contents[$item])) ? $contents[$item] + 1 : 1;
}
$output[] = '<form action="cart.php?action=update" method="post" id="cart">';
$output[] = '<table>';
// start div
$output[] = '<div id="cart_table">';
//
$output[] = '<tr>';
$output[] = '<td><h4>Product</h4></td>';
$output[] = '<td><h4>Item No.</h4></td>';
$output[] = '<td><h4>Price</h4></td>';
$output[] = '<td><h4>Size</h4></td>';
$output[] = '<td><h4>Quantity</h4></td>';
$output[] = '<td><h4>Price Total</h4></td>';
$output[] = '<td><h4>&nbsp;</h4></td>';
$output[] = '</tr>';
//new row
foreach ($contents as $product_id=>$qty) {
$sql = 'SELECT * FROM products WHERE product_id = '.$product_id;
$result = $db->query($sql);
$row = $result->fetch();
extract($row);
//
$output[] = '<tr>';
$output[] = '<td><a href="product.php?product_id='.$product_id.'">'.$product_title.'</a></td>';
$output[] = '<td>'.$product_plu.'</td>';
$output[] = '<td>$'.$product_price.'</td>';
//
//$output[] = '<td>'.$product_size.'</td>';
$output[] = '<td>'.$_POST['product_size'].'</td>';
//
$output[] = '<td><input type="text" name="qty'.$product_id.'" value="'.$qty.'" size="3" maxlength="3" /></td>';
$output[] = '<td>X&nbsp;$'.($product_price * $qty).'</td>';
$total += $product_price * $qty;
$output[] = '<td><a href="cart.php?action=delete&product_id='.$product_id.'" class="r">Remove</a>&nbsp;</td>';
$output[] = '</tr>';
//end div
$output[] = '</div>';



}
$output[] = '</table>';
$output[] = '<p>Grand total: <strong>$'.$total.'</strong></p>';
$output[] = '<div class="float-right"><button type="submit">Update cart</button>';
$output[] = '</form>';

Hello All   From the begin i have the following code.

<?php
$sql = "SELECT id,phonemodel FROM iphone";
$rows = $conn->sqlExec($sql);

$nr_row = $conn->num_rows;
$meniu ='<ul>';
if($nr_row>=0) {
  foreach($rows as $row) {
     $meniu .= '<li><a href="iphone.php?id='.$row['id'].'">'.$row['phonemodel'].'</a></li>';
	 }
  }
  $meniu .= '</ul>';
  
  echo $meniu;
	 
	 if(isset($_GET['id'])) {
	    $id =  (int)$_GET['id'];
		$sql = "SELECT * FROM iphone WHERE  id = $id";
$rows = $conn->sqlExec($sql);

$nr_rows = $conn->num_rows;
if($nr_rows>0) {
  foreach($rows as $row){
   echo 'Name Tel: '.$row["phonemodel"].' Title : '.$row["titlereparation"].' 
   Pret : '.$row["price"].' Message : '.$row["msj"].' ID : '.$row["id"].'<br />';
   }
  }
  else {
  echo '0 Results';
  }
}
?> 

Basicaly i have a website for phone repairs.And i want to create in iphone.php a menu from DB and when i acces the menu with _GET variable, when i press Iphone 5s (iphone.php?id=id page) the code have to display to me,all reparations for iphone 5s. Until now,i succssed to create the Menu but the code keep add same line in the menu when i add for example a second reparation for 5s.And i don't know how to select all reparation for 5s and display them in a single link like above Iphone 5s (iphone.php?id=id page). Now the script working like this.Create a menu with all phone names. Iphone 5s Iphone 3 Iphone 3s Iphone 4 Iphone 5s Iphone 5s Shoud be like Iphone 5s Iphone 3 Iphone 3s Iphone 4 Other new devices.. And display the reparation in every link from list By ID.i try it to select by phone names,not working. Any body with any ideea please? Thx so much