PHP - Moved: How Establish 2 Drop Menu Related?

This topic has been moved to Ajax Help.

http://www.phpfreaks.com/forums/index.php?topic=325435.0

I bought a script not so long back and I am trying to install it on my Linux hosting cPanel with GoDaddy.   I have followed the installation instructions (below) provided by the script author but I am having no luck.   1. Modify server/server.php Line: 282 $master = new ChatBot(“(author's domain)”, 10000);

can anyone  give me an example of how to establish an ODBC connection with PHP. I have a MS ACCESS database i want to open and read from

thanks

Does anyone have an example of establishing two MySQL connections (one local & one external) to pass data from local DB to external using PHP?

Objective: I need to query the table of local data (for email addresses) then the same for external db table and if email not present, insert into that db.

I will use CRON to run the script (will figure that part out later but suggestions welcome if you have that info as well....)

Some of you may have seen one of my many posts about email issues.  Some users don't get them, and I have determined it is probably because we are marked as spam.   We are a service that grades sales team members on their phone skills.  Listening to pre-recorded calls, grading and uploading them to our site, and then another part of our business looks them over and sometimes leaves a message that then get's forwarded to this persons work email.   I have determined there is ways to get marked as spam as default by not having an opt out link.  This is not an option, these sales members employer has opted in, and the emails are going to work related accounts hosted at that employer.  Also, if one of these staff members is not so bright, or disgruntled they may mark us as spam anyways.  The bottom line is that we have very little control over whether we are or are not marked as spam.   So we want to start looking into sending text messages and this is where I start to question how good of an idea this is.   First off, if it was me, and the messages where being sent to a device that my employer did not provide, I would in no way want work related text messages coming to me.  Unless there is a vested interest in getting them.  IE, I'm the boss at this place and am always on the clock.  What if you are on the bottom?  It's just a job for you.   What if it is a pre-paid device, text messages cost money.  What then?  What if they don't even have, or want a cell phone?   The short of it is this.  If I'm at a job that is just another job, and this employer tells me that I have to get these messages.  I'm going to look for another job.  I see the organizations having continuous issues and complaints from their employees.  Thus us as a business having issues keeping clients.   What am I getting into here?  What are your opinions on this matter?  What are your recommendations as to alerting users of something on our site that we can rest assured are being received 100% of the time?   Thanks!   Nick

Hi i currenlty have adrop box filled with companies so the user can select which company they woudl like services from but the default is currently 0 and is to selecvt all firms but im unsure how to do this.

Current code: Code: [Select]
<td>Taxi Firm</td><td> <select name="taxifirm">
<option value="0" selected>All Taxi Firms</option>
<?php
 $sql = mysql_query("SELECT * FROM taxi_Firms"); 
while($row = mysql_fetch_array($sql)){
$uid = $row["Firm_ID"];
$username = $row["Firm_Name"]; 
echo '<option value="'.$uid.'">'.$username.'</option>' ;

}
?>

amny help is welcomed

thanx

im fairly new to php so tend to do trial and error..... more error than trial tbh.

im wondering if it is possible to get a drop down menu to fill from a mysql database and to arrange it alphabetically.   i have created the menu just dont know how to arrange it better.

also  how can i use the items id in drop menu to load other info from that row on the database.

hope  you can help me.

Code: [Select]
<FORM>
<?php $result = mysql_query( "SELECT * FROM movie_info" )
;
echo "<select name= film onChange='submit()' >film name</option>";
while
($nt=mysql_fetch_array($result)){
?>

<?php

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

}
?>
</select>
</FORM>

Hi, i wonder whether someone may be able to help me please.

I am using a combination of PHP and AJAX to create two drop down menus on a HTML form. The data is being pulled from a mySQL database with the options available in the second drop down dependent on the value selected in the first.

The initial drop down menu called 'detectors' and the behaviours for the second drop down menu, 'searchheads' are created with the following AJAX code:

Code: [Select]
Function AjaxFunction(detectorid)
            {
            var httpxml;
            try
            {
            // Firefox, Opera 8.0+, Safari
            httpxml=new XMLHttpRequest();
            }
            catch (e)
            {
            // Internet Explorer
            try
            {
            httpxml=new ActiveXObject("Msxml2.XMLHTTP");
            }
            catch (e)
            {
            try
            {
            httpxml=new ActiveXObject("Microsoft.XMLHTTP");
            }
            catch (e)
            {
            alert("Your browser does not support AJAX!");
            return false;
            }
            }
            }
            function stateck()
            {
            if(httpxml.readyState==4)
            {
           
            var myarray=eval(httpxml.responseText);
            // Before adding new we must remove previously loaded elements
            for(j=document.addfindstolocation.searchheads.options.length-1;j>=0;j--)
            {
            document.addfindstolocation.searchheads.remove(j);
            }
           
           
            for (i=0;i<myarray.length;i++)
            {
            var optn = document.createElement("OPTION");
            optn.text = myarray[i];
            optn.value = myarray[i];
            document.addfindstolocation.searchheads.options.add(optn);
           
            }
            }
            }
            var url="searchheaddetails.php";
            url=url+"?detectorid="+detectorid;
            url=url+"&sid="+Math.random();
            httpxml.onreadystatechange=stateck;
            httpxml.open("GET",url,true);
            httpxml.send(null);
            }
The following code is the file 'searchheaddetails.php' (as highlighted above) which populates the second drop down menu.

Code: [Select]
<?
$detectorid=$_GET['detectorid'];
require "config.php";
$q=mysql_query("SELECT * FROM searchheads WHERE detectorid='$detectorid' ORDER BY 'searchheadname' ASC");
echo mysql_error();
$myarray=array();
$str="";
while($nt=mysql_fetch_array($q)){
$str=$str . "\"$nt[searchheadname]\",";
}
$str=substr($str,0,(strLen($str)-1)); // Removing the last char , from the string
echo "new Array($str)";

?> 
And this is the section of my form that pulls together the two drop down menus.

Code: [Select]
<form name="addfindstolocation" method="post" id="addfindstolocation">
<div align="left">
                                        <select name=detectors id="detectorid" onchange="AjaxFunction(this.value);">
                                            <option value=''>Select One</option>
                                            <?
                                            require "phpfile.php";// connection to database
                                            $q=mysql_query("SELECT * from detectors WHERE userid='1'ORDER BY 'detectorname' ASC");
                                            while($n=mysql_fetch_array($q)){
                                            echo "<option value=$n[detectorid]>$n[detectorname]</option>";
                                            }
                                           
                                            ?>
                                      </select>
                                    </div>
                                </div>
                                <p align="left">
                                    <label></label>
                                    <label>Search Head Used</label></p>
                                <div>
                                    <div align="left">
                                        <select name=searchheads id="searchheadid">
                                        </select>
                                    </div>
The drop down menus work fine, but I'm having a little difficulty with the data that is being saved. For the 'detectors' drop down menu the data being saved upon a selection being made is the 'id' pertient to the relevant detector e.g. 'Detector1' is selected and the 'id' value of '1' is saved which is exactly what I want. However when it comes to the second drop down menu, the value saved is the text value that the user selects, rather than the 'id'. e.g. 'Deep Search Head ' rather than an 'id' of '1'. Could someone perhaps tell me please what I need to change so that the 'id' value is saved rather than the text value.

If it helps, the coding is taken from the following http://www.plus2net.com/php_tutorial/ajax_drop_down_list.php.

Many thanks and kind regards.

Chris

I did search this up but all of them were lists. I want to make a menu drop down like so.... Non-clicked... Clicked...   The grey boxes would be images (unless it is easier to code them). I am a complete noob so please don't use technical terms   Thanks

helo
i a newbie in php. i need some help regarding my system. i want my data from database to display in drop down menu form.anyone have idea how to do that? i really need some help. i have search all over and couldn't find any solution. any suggestions would be helpful 

Hi, i know this is probably very basic but i have been banging my head and looking for tuts.

i have built a mysql php dropdown menu.

all displays fine. now, how do i get the menu to actualy take me to a new url?

the new url should be       www.mysite.com/"menu selection"



Code: [Select]
<?

include_once 'includes/db.php';

$result = mysql_query("select * from crimerate WHERE DISTRICT = 'Limpopo'", $con);

if (!$result) {
    die('Invalid query: ' . mysql_error());

}


$options="";

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

    $id=$row["id"];
    $crime=$row["CRIME"];
    $options.="<OPTION VALUE=\"$id\">$crime</option>";
}
?>


<SELECT NAME=crime>
<OPTION VALUE=0>Choose
<?=$options?>
</SELECT>

Hi I have a temporary web page with a drop down menu. Problem is to get rid of the gap on the drop down menus. Any help please       www.des-otoole.co.uk/top_menu  

Okay, so, here's the scenario.

I have a form that is editing an item that is already in the database. The text fields fill in just fine with that info. However, the drop down menus don't retrieve that info, rather resorting to the defaults, which can be a problem if you don't remember what you originally had.

Is there anyway to make the dropdown menus pull the info from the table and use that rather than resorting back to the default?

I tried using this:

Code: [Select]
<tr><td width="20%">Bonus:</td><td><select name="bonus" value="{{bonus}}">
<option value="Attack" {{bonus1select}}>Add to the attack power of weapon</option>
<option value="Defense" {{bonus2select}}>Add to the defensive power of armor</option>
<option value="None" {{bonus3select}}>No effect</option>
</select><br /></td></tr>

So, it's obvious the "value" portion not working. Any help would be great!!

I'm trying to code a drop-down menu that has four options; one for pie,exp, root 3, and the golden ratio.(all math values)  Beside the drop-down menu, there is an option for user-inputted data, they must input only positive numbers. I must take there selected drop-down menu option and times it by the user inputted number. I'm not sure how to check which option the user chose.

    <form id="s" method="post">
      <select id="math" name="options">
        <option value="nothing">Please select a constant</option>
        <option value="pie">Pi</option>
    <option value="econs">e</option>
    <option value="ratio">Golden Ratio</option>
    <option value="root">Root 3</option>
  

      </select>

    <input type="submit" name="Submit" value="Send">

Everything That ive tried has failed

Hello!

This is my first post here.

I'm usually do programming in Ruby (Rails) but for a special occasion I need PHP.

So hi PHP people! I come in peace   

My question is quite simple. How do you make a drop down menu? Most preferably with database populated fields in it.

In the Rails community, this really helped me make these menus in Rails: http://www.kahfei.com/?p=23

Could someone briefly write something like that too? (The place holders)

Orrr you guys can just explain how to do it. The ones I found on the internet were quite confusing and not explained enough.

Help anyone? Thank you