PHP - Explode? Array? List? Need To Break Values From Msql Text Box For Select List.

I am getting an error on line 3 for the original code, so I need help there, but what I do have a question on is this:

http://kaboomlabs.com/PDI/test2.php

Code: [Select]

 <?php
$dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME)
$result = $mysqli_query("SELECT * FROM comp");

      echo "<SELECT name='comp'>\n";
      while($row = $result->fetch_assoc()) {
         echo "<option value='{$row['name']}</option>\n";   
      }
      echo "</select>\n";

      $result_close();

      ?>

What  I am attempting to do is two fold.

1. Using PHP create a pull down menu that grabs data from the database.

2.  Have something in where after line three there is a dotted break, and then the rest of the list is shown.

There are over 150 entities that go into this database, so the top 3 are going to be the most used, the rest are going to be in alphabetical order.

Now the database has a auto-increment numbering system, the company name, address, phone number, and email if possible. I only want it to show the company name.

Is this possible at all? 

I want to make a list like this:


I store the genres in a array like this:

Code: [Select]
$genre_list = array(
array("ACTION", "First Person Shooter"),
array("ACTION", "Third Person Shooter"),
array("ACTION", "Tactical Shooter"),
array("ACTION", "Fighting"),
array("ACTION", "Arcade"),
array("ADVENTURE", "Adventure"),
array("ADVENTURE", "Platformer"),
array("ADVENTURE", "Point and Click")
);
This is my first attempt to only show the subgenre, i only get a blank list (apart from this <option value="genre">Genre</option> ).

Code: [Select]
<div class="forminput">
<select id="genres" name="genres">
<option value="genre">Genre</option>
<?php for($i=0; $i<count($genre_list[0]); $i++) { ?>
<option value="<?=$i?>"<?=$genre_list[0][$i]?></option>
<?php } ?>
</select>
</div>

I have the following code currently:
Code: [Select]
<?php foreach ((array)$node->field_buy_at as $item) { ?>
<?php print $item['view'] ?>
<?php } ?>
I would like to make the list a drop down with a link so that when a user selects, he goes to a new page. I tried the following:
Code: [Select]
        <select name="select">
      <?php foreach ((array)$node->field_buy_at as $item) { ?>
<?php
            $url = $node->field_buy_at[0]['url'];
            $store = $item['view'];
         ?>
        <? echo "<option value='$url'>$store</option>";?>
<?php } ?>
  </select>
I'm pretty sure it's this "$url = $node->field_buy_at[0]['url'];" that I don't have correct.

Hi i'm new to php.

I want to get all the values of dropdown list on another page wether selected or not.

<?php
 // Daniel URL duplicator.
 // Input links list.
 $linksList = "links.txt";
 // How many times to duplicate the url?
 $manyTimes = 1000;
 // Read in the list.
 for ($x = 0; $x <= $manyTimes; $x++)
 {
  $handle = fopen($linksList, "r");
  $line = fgets($handle);
  echo $line;
  fclose($handle);
 }
?>

Hey Guys,   I'm stuck on this simple bit of code lol what I'm trying to do is load in a list of urls:   site1.com site2.com site3.com etc   For each site that is read, I'm trying to duplicate it X times, above would print to screen the same url 1000 times, then move onto the next print it 1000 times etc until the list is done (or how ever many times I select)   I can't think of the best way to do it!   any help would be appreciated guys!   Graham        

**Disclaimer: It's been a while since I last wrote any code. The quality of my code is likely to be sub-par. You've been warned.**

I have a basic form that's meant to search flat files on our server. The "search engine" I created as two select lists: one for the file names and one for the customer site files come from.

For a reason I can't figure out, whatever option I select from the second select list is never captured when I hit Submit.

However, whatever option I select from the first select list is always captured.

What am I missing? I am sure it's starting right at me.... Any hints welcome. Thank you.

Here's my code:

    Code: [Select]
<HTML>
    <head><title>SEARCH TOOL - PROTOTYPE</title></head>
    <body><h1>SEARCH TOOL - PROTOTYPE</h1>
    <form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
    <fieldset>
    <legend>Filename (one item)</legend><select name="DBFilename" id="DBFilename" size="0">
    <?php $con = mysql_connect("localhost",  "user", "pass"); if (!$con) {  die('Could not connect: ' . mysql_error());}
    mysql_select_db("dev", $con) or die(mysql_error());
    $result = mysql_query("select distinct filename from search_test");
    while ($row = mysql_fetch_array($result))
    { ?>    <option value="<?php echo $row['filename']; ?>"><?php echo $row['filename']; ?></option> <?php } mysql_close($con); ?>
    </select></fieldset>
    <fieldset>
    <legend>Site (one item)</legend><select name="DBSite" id="DBSite">
    <?php $con = mysql_connect("localhost",  "user", "pass"); if (!$con) {    die('Could not connect: ' . mysql_error());}
    mysql_select_db("dev", $con) or die(mysql_error());
    $result = mysql_query("select distinct site from search_test");
    while ($row = mysql_fetch_array($result))
    { ?>        <option value="<?php echo $row['site']; ?>"><?php echo $row['site']; ?></option> <?php } mysql_close($con);
    ?>
    </select></fieldset>
    <input type="submit" name="submit" value="submit" >
    <input type="button" value="Reset Form" onClick="this.form.reset();return false;" />
    </form>
    </body>
    </HTML>
    <?php
    
    if (isset($_POST['submit'])) {
    
            if (!empty($_POST['DBFilename'])) {doFileSearch();}
                    elseif (!empty($_POST['DBSite'])) {doSite();}
    }
    
    function doFileSearch() {
    $mydir = $_SERVER['DOCUMENT_ROOT'] . "/filedepot";
    $dir = opendir($mydir);
    $DBFilename = $_POST['DBFilename'];
    
    $con = mysql_connect("localhost", "user", "pass");
    if (!$con) {die('Could not connect: ' . mysql_error());}
    mysql_select_db("dev", $con) or die("Couldn't select the database.");
    
    $getfilename = mysql_query("select filename from search_test where filename='" . $DBFilename . "'") or die(mysql_error());
    
    echo "<table><tbody><tr><td>Results.</td></tr>";
    while ($row = mysql_fetch_array($getfilename)) {
                    $filename = $row['filename'];
                    echo '<tr><td><a href="' . basename($mydir) . '/' . $filename . '"  target="_blank">' . $filename . '</a></td></tr>';
            }
                    echo "</table></body>";
    }
    
    function doSite() {
    $mydir = $_SERVER['DOCUMENT_ROOT'] . "/filedepot";
    $dir = opendir($mydir);
    $DBSite = $_POST['DBSite'];
    
    $con = mysql_connect("localhost", "user", "pass");
    if (!$con) {die('Could not connect: ' . mysql_error());}
    mysql_select_db("dev", $con) or die("Couldn't select the database.");
    
    $getfilename = mysql_query("select distinct filename from search_test where site='" . $DBSite . "'") or die(mysql_error());
    
    echo "<table><tbody><tr><td>Results.</td></tr>";
    while ($row = mysql_fetch_array($getfilename)) {
                    $filename = $row['filename'];
                    echo '<tr><td><a href="' . basename($mydir) . '/' . $filename . '"  target="_blank">' . $filename . '</a></td></tr>';
            }
                    echo "</table></body>";
    
    }
    ?>

Hi all,
i want my list/menu field values to come from my database. how can i accomplish that?
thanks

i did
Code: [Select]
<select name="select">
                  <option value="0">--select below--</option>
  <option value="1">Me</option>
  <?php
  
                  require_once '../konnect/konex.php';

$result = mysql_query("SELECT * FROM is_clients");

while($row = mysql_fetch_array($result))
  {
  echo "<option value ='2'>".$row"['name']</option>";
  echo "<br />";
  }

?>
                  </select>

Hi all.   how can i make the values show like a list. I tried html line break "<br>" and php \n but all to no avail. It just show all the values in one straigth line.   example of what i want is for the values to appear like this:   1234567890 0987654345 4567890675   instead of :   1234567890 0987654345 4567890675   Thanks

 
<form data-abide method="post" action="">
<div>
<select name="">
  <option value="name">
  <?php
$stmt = $pdo->query("SELECT acct_num FROM table order by id desc");
while ( $row = $stmt->fetch(PDO::FETCH_ASSOC) ) {
echo $row['acct_num'];
}
 
  ?>
</option>
</select>
</div>
<div>
    <label>New Password <small>required</small></label>
    <input type="password" name="password" id="password" required>
    <small class="error">New password is required and must be a string.</small>
</div>
<div>
    <label>Confirm New Password <small>required</small></label>
    <input type="password" name="password2" id="password2" required>
    <small class="error">Password must match.</small>
</div>
 

<input name="submit" type="submit" class="button small" value="Change Password">
</form>

Edited by Mr-Chidi, 13 November 2014 - 01:34 AM.

Hi everyone, I am having trouble passing/displaying the values inside of a selected list.
I created a add/remove list using Jquery and I tried to display the values passed using foreach and for loops but it is still not working. The values I am trying to get are $existing_mID[$j], which is inside of the option value attribute.

Please kindly let me know what should I do in order to get the values and I really appreciate your help.

<?php
$selected = $_POST['selectto'];
if(isset($selected))
{
    echo "something in selected<br />";
    for ($i=0;$i<count($selected);$i++)
    echo "selected #1 : $selected[$i]";
    foreach ($selected as $item)
    echo "selected: item: $item";
}
?>

This is the form

Hi , I have one question .. Can I split showing of content of dynamic list in 2 parts , when I echo list in code ..

Code: [Select]
<?php 
// Run a select query to get my letest 8 items
// Connect to the MySQL database  
include "../connect_to_mysql.php"; 
$dynamicList = "";
$sql = mysql_query("SELECT * FROM products ORDER BY date_added DESC LIMIT 8");
$productCount = mysql_num_rows($sql); // count the output amount
if ($productCount > 0) {
while($row = mysql_fetch_array($sql)){ 
             $id = $row["id"];
 $product_name = $row["product_name"];
 $price = $row["price"];
 $date_added = strftime("%b %d, %Y", strtotime($row["date_added"]));
 $dynamicList .= '<table width="100%" border="2" cellspacing="2" cellpadding="2">
        <tr>
          <td width="17%" valign="top"><a href="product.php?id=' . $id . '"><img style="border:#666 1px solid;" src="inventory_images/' . $id . '.jpg" alt="' . $product_name . '" width="77" height="102" border="2" /></a></td>
          <td width="83%" valign="top">' . $product_name . '<br />
            $' . $price . '<br />
            <a href="product.php?id=' . $id . '">View Product Details</a></td>
        </tr>
      </table>';
    }
} else {
$dynamicList = "We have no products listed in our store yet";
}
mysql_close();
?>


Code: [Select]
<p><?php echo $dynamicList; ?><br />
</p>
It works ok, and putting my files, everything works, but when I put 8 pictures with price and other details, it just show one image with details and another image below with details, and the third image below and so on ..
Can I split dynamic list to show 4 images with details on the left side and 4 on the right side?

Thank you in advance for help , if is possible

Hi,  I'm a php newbie, with some mysql experience. I have a mysql database as follows:
Database=watch, Table=events - fields id, reportno, sdate, comments

What I need is:
1. A dropdown list to display reportno from mysql database.
2. Depending on which reportno I choose, I'd like to open a popup(or separate) page to display the stored information.

Tks in advance for any help

Okay, really newbie question, but for this code...
Code: [Select]
<!-- Gender -->
<label for="gender">Gender:</label>
<select id="gender" name="gender">
<option value="">--</option>
<option value="F">Female</option>
<option value="M">Male</option>
</select>

1.) How do I assign a variable to this?

2.) How do I make this "sticky"?


Here is how I have usually done other form types...
Code: [Select]
<!-- First Name -->
<label for="firstName">First Name:</label>
<input id="firstName" name="firstName" type="text" maxlength="30"
value="<?php if(isset($firstName)){echo htmlentities($firstName, ENT_QUOTES);} ?>" /><!-- Sticky Field -->
<?php
if (!empty($errors['firstName'])){
echo '<span class="error">' . $errors['firstName'] . '</span>';
}
?>


Oh, by the way, at the top of my PHP file I have this code...
Code: [Select]
if ($_SERVER['REQUEST_METHOD']=='POST'){
// Form was Submitted (Post).

// Initialize Errors Array.
$errors = array();

// Trim all form data.
$trimmed = array_map('trim', $_POST);

Thanks,


Debbie

could anyone please help me with the code which is i have already displayed data as a multi select list but now i need to select one or more from them and insert into another database table. would be appreciate your help. thanx

Hi i have this drop down list for date which contain 3 selects DAY MONTH YEAR hwo can i make so that when update form select keeps the same value has before help please

<?php
$months = array('','January','February','March','April','May','June','July','August','September','October','November','December');
echo '<select name="month_of_birth">';
for ($i=1;$i<13;++$i) {
   echo '<option value="' . sprintf("%02d",$i) . '">' . $months[$i] . '</option>';
}
echo '</select>';
echo '<select name="day_of_birth">';
for ($i=1;$i<32;++$i) {
   echo '<option value="' . sprintf("%02d",$i) . '">' . $i . '</option>';
}
echo '</select>';
echo '<select name="year_of_birth">';
$year = date("Y");
for ($i = $year;$i > $year-50;$i--) {
   $s = ($i == $year)?' selected':'';
   echo '<option value="' . $i . '" ' . $s . '>' . $i . '</option>';
}
echo '</select>';
?>

Is there a way to take data that is listed like a,b,c,d,e and turn that into an unordered list?

Basically, I'm going to have a list of schools that will be listed with comma separators.  I'll print that data out in two places, one will be printed as it's in the database.  The other I would like to echo as an unordered list.