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PHP - Email Based On Value In Drop Down Menu
I am having issues with my page. I am trying to send a email to a different address based on the entry that my user submits.
Here is my form on my test page. Code: [Select] <form method=POST action=formdata.php> <table width="640" border=0 align="center"> <tr> <td align=right><b>Test</b></td> <td><input type=text name=test size=25></td> <td> </td> <td> </td> </tr> </table> </dl> <div align="center"> <p> <input type=hidden name=FA value=SendMail> </p> <input type=submit value="Submit Form"> </p> </div> </form> here is the php code for my formdata page Code: [Select] <?php $con = mysql_connect("localhost","test","test"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("test", $con); $sql="INSERT INTO formdata (test) VALUES ('$_POST[test]')"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } echo "Your Information Was Successfully Posted"; mysql_close($con); if ($_POST['test'] =="A"){ $to = "email@sample.com"; $subject = "Custom Form"; $message = $_POST['test'] ; $headers = "From: email@sample.com"; $sent = mail($to, $subject, $message, $headers) ; } else if ($_POST['test'] =="B"){ $to = "email2@sample.com"; $subject = "Custom Form"; $message = $_POST['test'] ; $headers = "From: email@sample.com"; $sent = mail($to, $subject, $message, $headers) ; } if($sent) {print "<p>Your mail was sent successfully"; } else {print "<p>We encountered an error sending your mail"; } ?> Similar TutorialsHello my problem here is to have a drop down menu which gathers usernames from the database. Then with a click of a button the information for that specific user selected is shown. I'm close code wise but right now it's just showing me all users. I'm displaying the info in text box's to allow an admin to change the info.
<form action="edit.php" method="post">
<select name="username" id="username">
<?php
// Connects to your Database
$con=mysqli_connect('localhost', 'root', '');
/* check connection */
if (mysqli_connect_errno($con)) {
trigger_error('Database connection failed: ' . mysqli_connect_error(), E_USER_ERROR);
}
$query = "SELECT `username` FROM `bencobricks` . `users`";
$result = mysqli_query($con, $query) or trigger_error("Query Failed! SQL: $query - Error: ". mysqli_error($con), E_USER_ERROR);
if($result) {
while($row = mysqli_fetch_assoc($result)) {
//printf ("%s\n %s\n ",
echo "<label for='username'>Username: </label> <option value='$row[username]'>$row[username] </option><br/>";
echo "<br/><br/> ";
}
}
mysqli_close($con);
?>
</select>
<br />
<input name="send" id="send" type="submit" value="Edit User" />
</form>
Then in the edit.php file i have this code:
<?php
// Connects to your Database
$con=mysqli_connect('localhost', 'root', '');
/* check connection */
if (mysqli_connect_errno($con)) {
trigger_error('Database connection failed: ' . mysqli_connect_error(), E_USER_ERROR);
}
if (isset($_POST['send'])) {
$username = $_POST['username'];
$query = "SELECT * FROM `bencobricks` . `users` ";
$result = mysqli_query($con, $query) or trigger_error("Query Failed! SQL: $query - Error: ". mysqli_error($con), E_USER_ERROR);
// Print the result
if($result) {
while ($row = mysqli_fetch_assoc($result)) {
// $query = "SELECT * FROM `bencobricks` . `users` WHERE `username` = username = $row[username]";
printf ("%s\n %s\n %s\n %s\n %s\n %s\n %s\n %s\n %s\n %s\n %s\n",
"<label for='username'>Username: </label> <input type='text' name='username' value='$row[username]'>" . "</input><br/>",
"<label for='email'>Email: </label> <input type='text' name='email' value='$row[email]'>" . "</input><br/>",
"<label for='membership'>Membership: </label> <input type='text' name='membership' value='$row[membership]'>" . "</input><br/>",
"<label for='firstName'>First name: </label> <input type='text' name='firstName' value='$row[firstName]'>" . "</input><br/>",
"<label for='lastName'>Last name: </label> <input type='text' name='lastName' value='$row[lastName]'>" . "</input><br/>",
"<label for='gender'>Gender: </label> <input type='text' name='gender' value='$row[gender]'>" . "</input><br/>",
"<label for='dateOfBirth'>Birthdate: </label> <input type='text' name='dateOfBirth' value='$row[dateOfBirth]'>" . "</input><br/>",
"<label for='date'>Date: </label> <input type='text' name='date' value='$row[date]'>" . "</input><br/>",
"<label for='sets'>Sets: </label> <input type='text' name='sets' value='$row[sets]'>" . "</input><br/>",
"<label for='checkbox'>Checkbox: </label> <input type='text' name='checkbox' value='$row[checkbox]'>" . "</input><br/>",
"<label for='admin'>Admin? </label> <input type='text' name='admin' value='$row[adminFlag]'>" . "</input><BR>",
"<br/><br/> ");
}
}
}
mysqli_close($con);
?>
Any help is greatly appreciated.
Hi all, I'm currently rewriting our contact us form using the php Pear Mail package. It allows both html based emails and text based emails, and displays the relevant version based on the end recipients broswer capabilities. I have the html version working perfect, however the text based version spits out the whole message without line breaks. Since some of the computers in our building only except text emails (due to security) it needs to be laid out so that it is legible. I've searched around and tried various options but the emails are still arriving without breaks. So as an example: A User Inputs the following message using a textarea on the php form: Hi there, Great website, need help with such and such though. Regards Bob The email message gets collected and included in a html message. Then a duplicate text based message is made like so (Code so far): Code: [Select] $plaintxt = '-----------------------------------------' . "\n"; $plaintxt .= Website Enquiry' . "\n"; $plaintxt .= '-----------------------------------------' . "\n"; $plaintxt .= 'A visitor to our website has submitted the following enquiry:' . "\n\n"; $plaintxt .= 'Name: ' . $visitor_name . "\n"; $plaintxt .= 'Company Name: ' . $visitor_companyname . "\n"; $plaintxt .= 'Email Address: ' . $visitor_email . "\n"; $plaintxt .= 'Telephone Number: ' . $message_telephone . "\n\n"; $plaintxt .= 'Subject of Enquiry: ' . $msg_subject . "\n"; $plaintxt .= 'Enquiry: ' . "\n\n"; $plaintxt .= $message_body . "\n\n"; $plaintxt .= '-----------------------------------------' . "\n\n"; $plaintxt .= 'User IP Address:' . $ip . "\n"; $plaintxt .= 'User Browser Info:' . $httpagent . "\n"; $plaintxt .= 'Date/Time Submitted:' . $time . "\n"; $crlf = "\n"; $headers = array('From' => $contact_from_email, 'Return-Path' => $contact_from, 'Subject' => $subject, 'To' => $contact_to); // Creating the Mime message $mime = new Mail_mime($crlf); // Setting the body of the email $mime->setTXTBody($htmlmessage); //Text version of the email $mime->setHTMLBody($plaintxt); // HTML version of the email "\n"; - These were suggested by users on the pear mail manual. Then once sent and opened you'd expect to see: Quote ----------------------------------------- Website Enquiry ----------------------------------------- A visitor to our website has submitted the following enquiry: Name: Bob Company Name: Someplace Ltd Email Address: bob@someplace.com Telephone Number: 01234 567890 Subject of Enquiry: help Enquiry: Hi there, Great website, need help with such and such though. Regards Bob ----------------------------------------- User IP Address: 123.456.789.012 User Browser Info: Mozilla Date/Time Submitted: 29th September 2011 09:56 However when opened in all the email applications I have (Outlook, Gmail, Live Mail, Hotmail, Yahoo) it looks like this: Quote -----------------------------------------Website Enquiry-----------------------------------------A visitor to our website has submitted the following enquiry:Name: BobCompany Name: Someplace LtdEmail Address: bob@someplace.comTelephone Number: 01234 567890Subject of Enquiry: helpEnquiry: Hi there,Great website, need help with such and such though.RegardsBob-----------------------------------------User IP Address: 123.456.789.012User Browser Info: MozillaDate/Time Submitted: 29th September 2011 09:56 Yet if you look at the source of the email. The line breaks are there. Am I missing something? Hello, i want to reproduce a content switcher based on drop-down menu (example: http://www.infocercetare.ro/index.php) anyone can help me with the logic of the function the rest i think i can handle. best regards Hi, I am trying to create a drop down menu that will select a value that is stored in the database - right now the code creats a dropdown (with nothing selected) - hope someone can help. in the database, the values are stored as --null- Option1 Option2 Option3 my code is Code: [Select] $instruction = $_GET['instruction']; <?php <select id="instruction" name="instruction"> <option value="" <?php if (!empty($instruction) && $instruction == '' ) echo 'selected = "selected"'; ?>></option> <option value="Option1" <?php if (!empty($instruction) && $instruction == 'Option1') echo 'selected = "selected"'; ?>>Option1</option> <option value="Option2" <?php if (!empty($instruction) && $instruction == 'Option2') echo 'selected = "selected"'; ?>>Option2</option> <option value="Option3" <?php if (!empty($instruction) && $instruction == 'Option3') echo 'selected = "selected"'; ?>>Option3</option> </select> <? Hi Guys, I have successfully managed to get my select drop down boxes to populate from a DB which i am very proud of as I'm still learning PHP but I really need some help on the next part please... I want to display a price based on the drop down selections so in my DB i have this... id colours size quantity sides price stock variations 1 1 A6 0-99 SINGLE 200 100g GLOSS 1 2 1 A6 100-199 SINGLE 300 100g GLOSS 1 What would i need to add to my PHP so that it checks the selections made by the user and displays the relevant price in the price filed in my page????? Hopefully someone can help. thanks Craig I have a search form that has a drop down Code: [Select] <select name="radius" id="radius"> <option value="5">5 mi.</option> <option value="10">10 mi.</option> <option value="15">15 mi.</option> <option value="20">20 mi.</option> <option value="50">50 mi.</option> <option value="100">100 mi.</option> </select> I'm trying to limit the results in a foreach loop within what was selected. Meaning, if someone selects 10, results within 10 miles will show The foreach is Code: [Select] foreach ($stores as $k=>$v) { $output = "<h3 style='margin:0;padding:0'><b>".$storeinfo[$k]['MktName']."</b><br>(approx ".$v." miles)</h3>"; $output .= "<p style='margin:0 0 10px 0;padding:0'>".$storeinfo[$k]['LocAddSt']."<br>"; $output .= $storeinfo[$k]['LocAddCity'].", ".$storeinfo[$k]['LocAddState']." ".$storeinfo[$k]['zipcode']."</p>"; print_r($output); } $v being the distance. So I need to show only the results of $v that are less than $r. How would I go about doing this? Right now, $v displays numbers like 5.04, 173.9 and so forth. Can anybody help me out? Thanks in advance. Hello,, I have looked for numerous examples, and have only got part of the answer. I have successfully populated a drop down list by reading the database. It populates with a list of quest titles. Once the user selects the quest. I want a textbox to populate with the Quest Description. I found an example that uses AJAX/PHP to do this, but it doesn't access a db to read/populate. The main code is in index.html The original example you select a country, and it populates a text field with the currency code. It passes a variable called $country to another file called find_ccode.php. I changed the first case in find_ccode.php to access my database, and populate the text box with quest description. Ideally I will like to make this case dynamic based off QuestID, would populate textbox with corresponding Quest Description (QDescrip). I then tried the other way in index.php I dynamically populated the dropdown list, and gave the option values the QuestIDs to be passed to $country, but something isn't working. I have a write files command to let me know what the $country variable is passing, and it never writes. It works in index.html though. I apologize in advance for my amateurish coding. I still have a lot to learn. I have included the code below, but if you know a more elegant way of doing it. I am creating a role-playing game, as you probably have figured out by now. I hope to one day put it on the web, but right now using it as a way of learning html, and PHP. I really want to know how to catch the value of the dropdown box. I would like to know what it is before sending over. also what is this.value. I assume this is what the value of the dropdown box is when you have selected something. I have been beating my head for days trying to figure this out. Any help is very much appreciated! Index.html <html> <head> <title>Changing textbox value based on dropdown list using Ajax and PHP</title> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1"> <script> // Developed by Roshan Bhattarai // Visit http://roshanbh.com.np for this script and more. // This notice MUST stay intact for legal use //fuction to return the xml http object function getXMLHTTP() { var xmlhttp=false; try{ xmlhttp=new XMLHttpRequest(); } catch(e) { try{ xmlhttp= new ActiveXObject("Microsoft.XMLHTTP"); } catch(e){ try{ xmlhttp = new ActiveXObject("Msxml2.XMLHTTP"); } catch(e1){ xmlhttp=false; } } } return xmlhttp; } function getCurrencyCode(strURL) { var req = getXMLHTTP(); if (req) { //function to be called when state is changed req.onreadystatechange = function() { //when state is completed i.e 4 if (req.readyState == 4) { // only if http status is "OK" if (req.status == 200) { document.getElementById('cur_code').value=req.responseText; } else { alert("There was a problem while using XMLHTTP:\n" + req.statusText); } } } req.open("GET", strURL, true); req.send(null); } } </script> </head> <body style="font: 12px Verdana, Arial, Helvetica, sans-serif;"> <form style="text-align:center" method="post" action="" name="form1"> <p style="color:#000099 ">When you change the dropdown list, the respective currency code of the country will be displayed in the textbox which is fetched from PHP using Ajax. </p> <p>Country : <select name="country" onChange="getCurrencyCode('find_ccode.php?country='+this.value)" <?php mysql_connect('localhost', '$user', '$password'); mysql_select_db('sok'); $result = mysql_query('select QuestID,QTitle, QDescrip from QuestList'); $options=""; while ($row=mysql_fetch_array($result)) { $QuestID=$row["QuestID"]; $QTitle=$row["QTitle"]; $options.="<OPTION VALUE=\"$QuestID\">".$QTitle; } ?> <option value="">Select Country</option> <option value="1">USA</option> <option value="2">UK</option> <option value="3">Nepal</option> </select><br/><br/> Currency : <input type="text" name="cur_code" id="cur_code" ></p> </form> </body> </html> Index.php <html> <head> <title>Changing textbox value based on dropdown list using Ajax and PHP</title> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1"> <script> // Developed by Roshan Bhattarai // Visit http://roshanbh.com.np for this script and more. // This notice MUST stay intact for legal use //fuction to return the xml http object function getXMLHTTP() { var xmlhttp=false; try{ xmlhttp=new XMLHttpRequest(); } catch(e) { try{ xmlhttp= new ActiveXObject("Microsoft.XMLHTTP"); } catch(e){ try{ xmlhttp = new ActiveXObject("Msxml2.XMLHTTP"); } catch(e1){ xmlhttp=false; } } } return xmlhttp; } function getCurrencyCode(strURL) { var req = getXMLHTTP(); if (req) { //function to be called when state is changed req.onreadystatechange = function() { //when state is completed i.e 4 if (req.readyState == 4) { // only if http status is "OK" if (req.status == 200) { document.getElementById('cur_code').value=req.responseText; } else { alert("There was a problem while using XMLHTTP:\n" + req.statusText); } } } req.open("GET", strURL, true); req.send(null); } } </script> </head> <body style="font: 12px Verdana, Arial, Helvetica, sans-serif;"> <form style="text-align:center" method="post" action="" name="form1"> <p style="color:#000099 ">When you change the dropdown list, the respective currency code of the country will be displayed in the textbox which is fetched from PHP using Ajax. </p> <p>Quest : <select name="country" getCurrencyCode('find_ccode.php?country='+this.value)">' <?php mysql_connect('localhost', '$user', '$password'); mysql_select_db('sok'); $result = mysql_query('select QuestID,QTitle, QDescrip from QuestList'); $options=""; while ($row=mysql_fetch_array($result)) { $QuestID=$row["QuestID"]; $QTitle=$row["QTitle"]; $options.="<OPTION VALUE=\"$QuestID\">".$QTitle; } $myFile = "test_catchoption.txt"; $fh = fopen($myFile, 'w') or die("can't open file"); fwrite($fh, $options); fclose($fh); ?> Choose your Quest <?=$options?> </SELECT> <br/><br/> Quest Description : <input type="text" name="cur_code" id="cur_code" ></p> </form> </body> </html> find_ccode.php <?php // Developed by Roshan Bhattarai // Visit http://roshanbh.com.np for this script and more. // This notice MUST stay intact for legal use $country=$_REQUEST['country']; $myFile = "testFilefindcc.txt"; $fh = fopen($myFile, 'w') or die("can't open file"); fwrite($fh, $country); fclose($fh); switch($country) { case "1" : mysql_connect('localhost', '$user', '$password'); mysql_select_db('sok'); $result = mysql_query("select * from QuestList where QuestID = '1'"); $row = mysql_fetch_assoc($result); echo $row['QDescrip']; break; case "2" : echo "GBP"; break; case "3" : echo "NPR"; break; } ?> I'm not totally new to php but I'm no guru either. I'm building an intranet and have three seperate user roles, user - manager - admin. There are some menu items I don't want to allow simple users to see. This will expand later to give them different views of a page as well (some can view/others can edit). As it stands the login validation is holding they're user level in $SESSION. To give you an idea take a look at what I'm trying to do: Code: [Select] function usermenu($usermenu) { if($user_level=0) echo ("<ul id="gooeymenu2" class="solidblockmenu"> <li><a href="main.php">Home</a></li> <li><a href="forms.php">Forms</a></li> <li><a href="/support/index.php" target="_new">Support</a></li> <li><a href="documents.php">Documents</a></li> <li><a href="admin/index.php">Admin</a></li> <li><a href="logout.php">Logout</a></li> </ul> <script> gooeymenu.setup({id:'gooeymenu2', selectitem:1, fx:'swing'}) </script>" "); else($user_level=1,2) echo ("<ul id="gooeymenu2" class="solidblockmenu"> <li><a href="main.php">Home</a></li> <li><a href="forms.php">Forms</a></li> <li><a href="/support/index.php" target="_new">Support</a></li> <li><a href="documents.php">Documents</a></li> <li><a href="new.php">New Adviser</a></li> <li><a href="admin/index.php">Admin</a></li> <li><a href="logout.php">Logout</a></li> </ul> <script> gooeymenu.setup({id:'gooeymenu2', selectitem:1, fx:'swing'}) </script>" "); Any help would be great, thanks in advance. Jason 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> 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 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
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 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
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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!! 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> 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> <input type="submit" name="Submit" value="Send"> Everything That ive tried has failed 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 I have a mysql table which will store users email addresses (each is unique and is the primary field) and a timestamp. I have added another column called `'unique_code' (varchar(64), utf8_unicode_ci)`. What I would very much appreciate assistance with is; a) Generating a 5 digit alphanumeric code, ie: 5ABH6 b) Check all rows the 'unique_code' column to ensure it is unique, otherwise re-generate and check again c) Insert the uniquely generated 5 digit alphanumeric code into `'unique_code'` column, corresponding to the email address just entered. d) display the code on screen. What code must I put and where? **My current php is as follows:** Code: [Select] require "includes/connect.php"; $msg = ''; if($_POST['email']){ // Requested with AJAX: $ajax = ($_SERVER['HTTP_X_REQUESTED_WITH'] == 'XMLHttpRequest'); try{ if(!filter_input(INPUT_POST,'email',FILTER_VALIDATE_EMAIL)){ throw new Exception('Invalid Email!'); } $mysqli->query("INSERT INTO coming_soon_emails SET email='".$mysqli->real_escape_string($_POST['email'])."'"); if($mysqli->affected_rows != 1){ throw new Exception('You are already on the notification list.'); } if($ajax){ die('{"status":1}'); } $msg = "Thank you!"; } catch (Exception $e){ if($ajax){ die(json_encode(array('error'=>$e->getMessage()))); } $msg = $e->getMessage(); } } Hey guys im looking to do a drop down menu which is pretty basic however, in this drop down menu it uses an array of 4 values eg Orange, Apples, Bananas, Strawberry's. For this menu if a user has previously chosen an Orange which is stored on the db then then when the drop down menu is loaded it should move Orange to the top of the list, the same applies for the other values. How could I do this? I know how to do the select statement and place the values in an array. from there I dont know quite how to approach it. () Thank you. 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 |
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