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PHP - Echo $form['catcher_id'] Dropdown List Problem

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Echo $form['catcher_id'] Dropdown List Problem	View Content

the line : echo $form['catcher_id'] gives me a dropdown list

when i choose another item from the dropdown i want to do a few things but my code not working:

$selected_catcher = $form['catcher_id'];
foreach($selected_catcher as $val)
{
$catcher_name = $val->getName();
echo $catcher_name." ".$val->getId();

if ($catcher_name = "zed-catcher")
{
echo $form['service_code']->renderLabel();
echo $form['service_code']->renderError();
echo $form['service_code'];

}
}

please help?
thanks

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Populate A Php Form From A Dropdown List

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hi all,i have a php form that i would like to populate from a drop down list,my problem is that i cant place the drop down list at the correct place.what i want is that where the franchisee is the input should be the drop down,please help....this is the code that i am using

Code: [Select]
<form action="<?php echo $_SERVER['PHP_SELF'];?>" method="post">
<p>Add a Bus in the Bus table.</p>
<p>
Fleet Number:<br />
<input type="text" name="fleet_number" size="6" maxlength="10" value="" />
</p>
<p>
Registration Number:<br />
<input type="text" name="registration_number" size="10" maxlength="20" value="" />
</p>
<p>
Model:<br />
<input type="text" name="model" size="10" max length="15" value="" />
</p>
<p>
Franchisee :<br />
<input type="select" name="franchisee_id" size="10" max length="15" value="" />

<select name=dropdown_list>

<?php

    while($row = oci_fetch_array($stid))

    {

echo "<option value='".$row['FRANCHISEE_NAME']."'>";
echo $row['FRANCHISEE_NAME'];
echo "</option> ";

    }

  ?>

</select>
</p>
<p>
<input type="Submit" name="submit" value="Submit !" />
</p>
</form>

Php Dropdown List Populates Form Field

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So I've been attempting to create a dropdown list that will post to mySQL.

The problem I'm running into is the way I populate the dropdown I can't figure out how to return the selected option to mySQL.

Date of Practice:<input type="text" id="dates" name="practdate" datepicker="true" datepicker_format="YYYY-MM-DD" maxlength="100"><br>

<?php

$query="SELECT eoptions,id FROM sm_options";

/* You can add order by clause to the sql statement if the names are to be displayed in alphabetical order */

$result = mysql_query ($query);

echo "<select name=eoptions value=''>Event Options</option>";

// printing the list box select command

while($nt=mysql_fetch_array($result)){//Array or records stored in $nt

echo "<option value=$nt[id]>$nt[eoptions]</option>";

/* Option values are added by looping through the array */

}

echo "</select>";// Closing of list box

mysql_close();

?><br>

Event:<input type="text" name="event" maxlength="100" value="<?php echo "test $output" ?>"><br>

Session Time:<input type="text" name="practtime" id="practtime">min.

Practice Content:<input type="text" name="practnotes"><br>

</form>

I've been moving code around trying to solve the problem so its a bit messy.

Looping Through Dropdown List Problem

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how come the lines:
PHP Code:
echo $form['catcher_id'];

$catcher_names = $form['catcher_id']
foreach($catcher_names as $val)
{
echo "testing loop";
}
produce nothing?? its not doing the foreach.
$form['catcher_id']; is a dropdown list containing catcher names
please help?
thanks

What Day Is It? Buried Dropdown List Option Problem

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Hi, I'll try and keep this brief but it really is a case of searching for needle in a haystack... I've just started working for a UK charity whose main activity is supported by a massive website. Stacks of information sheets, 3 distinct forums (each about the size of Freaks here), numerous blogs, events pages, webinars etc. etc. Essentially, it is an on-line free community that serves the needs of a very sizeable percentage of man and womankind. Like many charities it relies on gratis contributions from those with time or money to spare. I don't have either, but I do believe in what they are doing. The problem is that that the site itself has grown or rather mutated to become a huge data repository, built from a number of contributors with different skills sets and ways of doing things. At the back of it lurks a CMS which staff (a limited number) use to update site content. 95% of the site is in PHP, with a host of other ingredients thrown into the pot (Tiny-MCE, Perl, Ajax, Fancybox, html5, flash, blah and blah). The problem is in the CMS when someone tries to enter a date beyond 2014 there is no option to do so - -.i.e. that's as far as the dropdown goes. I can see the problem but can't find the offending script/files. Anywhere. Here's a snippet from the viewable source for those with keener eyesight:

<form action="index.php?content=data_universal_new&table=events" method="post" ENCTYPE="multipart/form-data" >
<input type="hidden" name="change" value="">
<table summary="" border="0">
<tr>
<td valign="top">Title</td>
<td><input type="text" name="titlex" size="30" maxlength="256" value=""></td>
</tr> <tr>
<td valign="top">Microsite</td>
<td>
<select name="micrositex">
<option value="INF UK" > INF UK</option>
<option value="MTL" > MTL</option>
<option value="ACE BABES" > ACE BABES</option>
</select>
</td>



</tr> <tr>
<td valign="top">Short description</td>
<td><textarea name="short_descriptionx" cols="40" rows="5" ></textarea></td>
</tr> <tr>
<td valign="top">Thumbnail</td>
<td>
<input type="file" name="graphic[thumbnailx]"> <input type="checkbox" name="deletepic[thumbnail]" value="yes"> Delete?
</td>
</tr> <tr>
<td valign="top">Main page</td>
<td>
<div id="tm"><textarea id="elm1" name="main_pagex" style=" height: 400px;width:460px;" class="mceEditor"></textarea></div>
<p>
<script type="text/javascript">


function toggleEditor(id) {
if (!tinyMCE.getInstanceById(id))
tinyMCE.execCommand('mceAddControl', false, id);
else
tinyMCE.execCommand('mceRemoveControl', false, id);
}
</script>
<div><a href="javascript:toggleEditor('elm1');">[Add/Remove editor]</a></div>
<a href="upload.php" title="Upload Documents" class="iframe">Upload Documents</a>
<span class="tooltip" title="Click here to start the document uploader,The documents will be available to link to in the link list drop-down">
<img src="../images/help.png" border="0" width="15" height="15" alt="" align="top" >
</span>
</p>
</td>
</tr> <tr>
<td valign="top">Date</td>
<td>
<select name="date[4]">
<option >1</option>
<option >2</option>
<option >3</option>
<option >4</option>
<option >5</option>
<option >6</option>
<option >7</option>
<option >8</option>
<option >9</option>
<option >10</option>
<option >11</option>
<option >12</option>
<option >13</option>
<option >14</option>
<option >15</option>
<option >16</option>
<option >17</option>
<option >18</option>
<option >19</option>
<option >20</option>
<option >21</option>
<option >22</option>
<option >23</option>
<option >24</option>
<option selected>25</option>
<option >26</option>
<option >27</option>
<option >28</option>
<option >29</option>
<option >30</option>
<option >31</option>
</select>
<select name="date[3]">
<option value="1" >Jan</option>
<option value="2" >Feb</option>
<option value="3" >Mar</option>
<option value="4" >Apr</option>
<option value="5" >May</option>
<option value="6" >Jun</option>
<option value="7" >Jul</option>
<option value="8" >Aug</option>
<option value="9" >Sep</option>
<option value="10" >Oct</option>
<option value="11" selected>Nov</option>
<option value="12" >Dec</option>
</select>
<select name="date[5]">
<option value="2002" >2002</option>
<option value="2003" >2003</option>
<option value="2004" >2004</option>
<option value="2005" >2005</option>
<option value="2006" >2006</option>
<option value="2007" >2007</option>
<option value="2008" >2008</option>
<option value="2009" >2009</option>
<option value="2010" >2010</option>
<option value="2011" >2011</option>
<option value="2012" >2012</option>
<option value="2013" >2013</option>
<option value="2014" selected >2014</option>
</select>
</td></tr> <tr>
<td>&nbsp;</td>
<td><input type="submit" value="Submit"></td>
</tr>
</table>
</form>
</div>
　
　
<div id="man_div">
<p><a href="index.php?content=data_universal_new&add=yes&table=events">Add event</a></p>

Hmm. Any thoughts would be greatly appreciated.
Edited by Ch0cu3r, 25 November 2014 - 07:15 AM.

Foreach To Create Dropdown List Problem

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I have the following code which works (thanks to PHP Freaks members)

    foreach ($swap["sims"] as $swap) 
    		{
                    echo $swap['voice'] . "</br>";
			}

But I can't figure out why the following doesn't work

<select name="sim_for_swap[]">
<?php
    foreach ($swap["sims"] as $swap)  
    		{
?>
<option value="<?php echo $swap['voice'];?>"><?php echo $swap['voice'];?></option>
<?php
}
?>

Can I Split Showing Of Content Of Dynamic List In 2 Parts , When I Echo List

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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

Form Echo Problem

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Having problems with my script echo without form being completed yet. Any help would be appreciated.

The echo that is outputting early is "You must enter a Session Name".

Code: [Select]
<html>
<head>
<title>ReH-0.1--Create a Session</title>
</head>
<body bgcolor="575757">
<center>
<?php

   $error = "Could not connect to the database";
   mysql_connect('localhost','root','') or die ($error);
   mysql_select_db('temp') or die ($error);

   //Set Variables
   $id = $_POST['id'];
   $table = $POST['table'];

   //Check for exsistence
   if (!$id or !$table)
     {
      echo "<font color='#ff0000'>You must enter a Session Name</font>";
     } else {
      echo "Congrats!";
     }

?>
<h2 style="color:#fff">ReH-0.1 Password Encryption</h2><br><br>
<font color="#fff">Before encrypting your password, a session must be started. We need you to enter a personal session name that you will remember so that you password is protected by this session.</font><br><br>
<form action="session_start.php" method="POST">
<label for="id" style="color: #fff;">Session Name:
<input type="text" maxlength="10" size="10" name="id"></label><br><br>
<label for="table" style="color: #fff;">Session Name Confirmation:
<input type="text" maxlength="10" size="10" name="table"></label><br><br>
<input type="submit" value="Create Session">
</form>
</center>
</body>
</html>

Mysql Form Dropdown Box Problem

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Hello, I have been trying to figure this for a while now and reading the tutorials are not helping, I think I'm a little over my head on this one and was hoping someone could help me out with this issue. I am making a User Edit page and would like to have the access level part of the form show the users access current access level thats set in the database when the page loads, and if it needs to be changed you can press the dropdown box and select a new access level. I can't figure out how to show the current access level as default and populate the drop down box with the other access levels in my table. My Tables look like this Users table (users): --------------------------------------------------------------------------------------------------- | id | username | password | flag | realfirst | reallast | dept | --------------------------------------------------------------------------------------------------- 1 loderd 9 test guy Service Auth Table (auth): -------------------------------------------- | id | auth_level | descrip | -------------------------------------------- 1 1 Service Tech 2 2 Office Staff 3 9 Super Admin My SQL Query looks like this

$users = $db->fetch_all_array("SELECT users.id, users.username, users.password, users.realfirst, users.reallast, users.dept, users.flag, auth.auth_level, auth.descrip, auth.id 
							   FROM users LEFT JOIN auth
							   ON users.flag = auth.auth_level
							   WHERE users.id = '".$_GET['id']."'");

I can't seem to figure out how I can do this for the Access Level dropdown box.

                <tr>
                    <td width="19%" align="right" valign="top">Access Level :</td>
                    <td width="1%" align="left">&nbsp;</td>
                    <td width="80%" align="left" valign="top">
						  <?php 
						  echo "<select name='flag' id='flag'>";
						  foreach ($users as $row){
						  if($row[auth_level]==$row[auth_level]){
						  echo "<option value=$row[auth_level] selected>$row[auth_level] - $row[descrip]</option>"; 
						  }else{
						  echo "<option value=$row[auth_level]>$row[auth_level] - $row[descrip]</option>"; 
						  }
						  }
						  echo "</select>";                          ?>
                    </td>
				</tr>

Any help would be greatly appreciated

Echo Day In Dropdown

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I cannot get this to echo out the day from mysql. Basically I need it to say if the field is empty show Day as selected=selected with the list of days, but if it isn't empty, make the day in mysql set to selected=selected.

I have also check the query in mysql so I know the query is correct.

<?php

if (empty($r['dayofbirth']))
{
$isset = isset($_POST['date_of_birth']);

for ($day = 1; $day <= 31; ++$day)
{
echo '<option';

if ($isset && $_POST['date_of_birth'] === $day)
{
echo 'selected="selected"';
}

echo ">${day}</option>";
}
else
{
echo "${r['dayofbirth']}";
}

?>

Echo Dropdown Selection

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I'm trying to do a couple of things using the dropdown list below. I have a table with id, rank, and amount.

1. I need to echo the amount that corresponds to the rank selected from the dropdown list.
2. I also need to put the amount value in a variable so I can do some math with it later on.

I've got the dropdown pulling the rank values from the db just not sure how to make it do the rest I'm trying to do.

Any help would be greatly appreciated

Code: [Select]
<?php
db($host,$db_name,$user,$pass);
function db($host,$db_name,$user,$pass)
{
global $link;
$link=mysql_connect ("$host","$user","$pass");
if(!$link){die("Could not connect to MySQL");}
mysql_select_db("$db_name",$link) or die ("could not open db".mysql_error());
}

$sql="SELECT * FROM dlaw";
$result=mysql_query($sql);

$options="";

while ($row=mysql_fetch_array($result)) {

$rank=$row["rank"];
$options.="<OPTION VALUE=\"$rank\">".$rank.'</option>';
}
?>
<SELECT NAME=rank><OPTION VALUE=0>Select Rank<?=$options?></SELECT>

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If I can get this fixed, I will have completed all but the admin login for this project - my first php/mysql project.

Here is what I need. I have a list_records.php that list all the records in the table 'links' and the category each entry is in from the table 'categories'.

Here are my table structures.
Code: [Select]
-- Table structure for table `categories`
--

DROP TABLE IF EXISTS `categories`;
CREATE TABLE IF NOT EXISTS `categories` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`categories` varchar(37) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=40 ;

-- --------------------------------------------------------

--
-- Table structure for table `links`
--

DROP TABLE IF EXISTS `links`;
CREATE TABLE IF NOT EXISTS `links` (
`id` int(4) NOT NULL AUTO_INCREMENT,
`catid` int(11) DEFAULT NULL,
`name` varchar(255) NOT NULL DEFAULT '',
`url` varchar(255) NOT NULL DEFAULT '',
`content` varchar(255) NOT NULL DEFAULT '',
PRIMARY KEY (`id`),
KEY `catid` (`catid`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=35 ;

On the update.php file, I have a form that lets me make changes to the record.

Here is the codes for update.php

Code: [Select]
<? include "menu.php" ?>

<? include "db.php" ?>
<?php
$id=$_GET['id'];

$sql = "select * from links where id =$id";
$query = mysql_query($sql);

while ($row = mysql_fetch_array($query)){

    $id = $row['id'];
    $catid = $row['catid'];
    $name = $row['name'];
    $url = $row['url'];
    $content = $row['content'];

    //we will echo these into the proper fields

}
mysql_free_result($query);
?>

<table width="65%" align="center">
<tr><td align="left">

<form action="updated.php" method="post">

<input type="hidden" value="<?php echo $id; ?>" name="id"/>

<br>

<b>Website Name:</B><br> Change the name of the website listing.<br>
<input type="text" value="<?php echo $name; ?>" name="name"/>

<br>
<br>
<b>URL:</b><br> Change the URL of the website listing.<br>
<input type="text" value="<?php echo $url; ?>" name="url"/>

<br>
<br>

<b>Description:</b><br> Change the description of the website listing.<br> Limit 255 characters.<br/>
<textarea name="content" cols="45" rows="4" wrap="soft"><?php echo($content);?></textarea>

<br>
<?php
$result = mysql_query("SELECT id, categories FROM categories") or die(mysql_error());
while ($row = mysql_fetch_array($result)) {
$id = $rows["id"];
$categories=$row["categories"];
$options.= '<option value="'.$row['id'].'">'.$row['id'].'-'.$row['categories'].'</option>';
};

?>

<SELECT NAME=catid>
<OPTION VALUE=selected><? echo $catid; ?><? echo $options; ?></OPTION>
</SELECT>

<?php
mysql_close();
?>

<div align="center">
<input type="submit" value="submit changes"/>
</div>
</form>

<br>
</td></tr></table>

The part of he code I need help with is

Code: [Select]
<?php
$result = mysql_query("SELECT id, categories FROM categories") or die(mysql_error());
while ($row = mysql_fetch_array($result)) {
$id = $rows["id"];
$categories=$row["categories"];
$options.= '<option value="'.$row['id'].'">'.$row['id'].'-'.$row['categories'].'</option>';
};

?>

<SELECT NAME=catid>
<OPTION VALUE=selected><? echo $catid; ?><? echo $options; ?></OPTION>
</SELECT>

I want it to default to the category that the entry is in. If you look, you will see in the select portion that I I have Code: [Select]
<$ echo $catid; ?> which echos the proper category ID, but if I use Code: [Select]
<? echo $categories; ?> it echos Writing, which is the last category in the list. Yet, the $options echo the catid and it corresponding category. How can I get the default option to echo BOTH the catid and category name while also listing all the other categories so that the records can be moved to a new category is needed?

Any help will be appreciated. Thank you in advance.