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PHP - Need To Capture & Echo Back Drop Down Menu Selection
Need some help to define the php logic to echo back a selection if/when chosen from the drop down menu.
Note: I'm using just a small subset of the USA states to show setup and what I'm trying to accomplish - it will be the same for all the other drop down menus as well.... Here's the example code: Code: [Select] <form name="form1" method="post" action="<?php $_SERVER[PHP_SELF]; ?>"> <select name="address_state" id="address_state" tabindex="25"> <option value="" selected> - Select State -</option> <option value="AK">Alaska</option> <option value="AL">Alabama</option> <option value="AR">Arkansas</option> <option value="AZ">Arizona</option> <option value="CA">California</option> <option value="CO">Colorado</option> <option value="CT">Connecticut</option> <option value="DC">District of Columbia</option> </select> <input name="submit" type="submit" value="submit" tabindex="900" id="submit" /> </form> Issue: I have validation (logic) define to check all the (other) text, radio, and check boxes of the form and that all works great! However, any time one of those particular fields fail validation, any selected "drop down" item is lost - it defaults back to no selection and the user has to make all drop down menu selections again. Can someone provide method to echo back - in php being the page reloads to itself - the drop down selection as in the example above? Thanks! Similar TutorialsMy problem is when the entries in the form are validated and some fields with incorrect entries are found the fields filled out stay on the re-load of the form, all of them except the US States drop down menu which reverts back to Alabama every time. I would like to see that the user selection stays in place even after a re-load of the form. and with my limited experience I am having an issue trying to figure this out for a school project. Any help at all would be greatly appreciated, like what variables to create/call on and where exactly I insert this code. Much appreciated. here is the display code: <?php
/* Program name: exercise_4_form.inc
* Description: Defines a form that collects a user's
* name and mailing address.
*/
$rows = array( "first_name" => "First Name (optional)",
"last_name" => "Last Name",
"phone" => "Phone Number",
"city" => "City",
"state" => "State",
"address" => "Address",
"zip_code" => "ZIP Code",
"e_mail" => "E-Mail");
$states_list = array('AL'=>"Alabama",'AK'=>"Alaska",'AZ'=>"Arizona",
'AR'=>"Arkansas",'CA'=>"California",'CO'=>"Colorado",'CT'=>"Connecticut",
'DE'=>"Delaware",'DC'=>"District Of Columbia",'FL'=>"Florida",'GA'=>"Georgia",
'HI'=>"Hawaii",'ID'=>"Idaho",'IL'=>"Illinois", 'IN'=>"Indiana", 'IA'=>"Iowa",
'KS'=>"Kansas",'KY'=>"Kentucky",'LA'=>"Louisiana",'ME'=>"Maine",
'MD'=>"Maryland", 'MA'=>"Massachusetts",'MI'=>"Michigan",'MN'=>"Minnesota",
'MS'=>"Mississippi",'MO'=>"Missouri",'MT'=>"Montana",'NE'=>"Nebraska",
'NV'=>"Nevada",'NH'=>"New Hampshire",'NJ'=>"New Jersey",'NM'=>"New Mexico",
'NY'=>"New York",'NC'=>"North Carolina",'ND'=>"North Dakota",'OH'=>"Ohio",
'OK'=>"Oklahoma", 'OR'=>"Oregon",'PA'=>"Pennsylvania",'RI'=>"Rhode Island",
'SC'=>"South Carolina",'SD'=>"South Dakota",'TN'=>"Tennessee",'TX'=>"Texas",
'UT'=>"Utah",'VT'=>"Vermont",'VA'=>"Virginia",'WA'=>"Washington",'WV'=>"West Virginia",
'WI'=>"Wisconsin",'WY'=>"Wyoming");
$submit = "Submit mailing information";
?>
<html>
<head>
<style type='text/css'>
<!--
body {
background: #42413C;
color: #000;
font-family: Tahoma, Geneva, sans-serif;
font-size: 90%;
}
form {
margin: 1em 2 2 2;
padding: 1;
width: 1100px;
margin-left: 35%;
}
.field {padding-bottom: 1em;
}
label {
font-weight: bold;
float: left;
width: 15%;
margin-right: 1em;
text-align: right;
background-color: #9F6;
padding-right: 5px;
}
select {margin-bottom: 1em;}
#submit {
margin-left: 15%;
}
h3 {
width: 500px;
margin-left: 35%;
}
-->
</style>
</head>
<body>
<h3>Please enter your mailing information below:</h3>
<?php
/* loop that displays the form */
echo "<form action='$_SERVER[PHP_SELF]' method='POST'>";
foreach($rows as $field => $label)
if($field == "state")
{
echo "<label for='$field'>$label</label>
<select id='$field' name='$field'>\n";
foreach ( $states_list as $abbr => $sname )
{
echo "<option value='$abbr'>$sname</option>\n";
}
echo "</select>\n";
}
else
{
echo "<div class='field'><label for='$field'>$label</label>
<input id='$field' name='$field' type='text' value='".@$$field."'
size='25%' maxlength='65' /></div>\n";
}
echo "<div id='submit'>
<input type='hidden' name='submitted' value='yes'>
<input type='submit' value='$submit'></div>";
?>
</form>
</body>
</html>
Here is the Logic on the combined php file that displays and has the statements on it for the form:<?php
/* Program name: exercise_4_form_verifications.php
* Description: Program checks all the form fields for
* blank fields and verifies them
*/
if(isset($_POST['submitted']) and $_POST['submitted'] == "yes")
{
foreach($_POST as $field => $value)
{
if(empty($value))
{
if($field != "first_name")
{
$blank_array[] = $field;
}
}
else
{
$good_data[$field] = strip_tags(trim($value));
}
}
if(@sizeof($blank_array) > 0)
{
$message = "<p style='color: black; margin-bottom: 0;font-weight: bold; margin-left: 35%'>
You didn't fill in one or more required fields.
You must enter:
<ul style='color: yellow; margin-top: 0; margin-left: 35%;
list-style: none' >";
foreach($blank_array as $value)
{
$message .= "<li>$value</li>";
}
$message .= "</ul>";
echo $message;
extract($good_data);
include("exercise_4_form.inc");
exit();
}
foreach($_POST as $field => $value)
{
if(!empty($value))
{
$name_verify = "/^[A-Za-z' -]{1,50}$/";
$phone_verify = "/^[0-9)(xX -]{7,20}$/";
$city_verify = "/^[A-Za-z' -]{1,50}$/";
$address_verify = "/^[A-Za-z0-9 .,'-]{1,50}$/";
$zip_verify = "/^[0-9]{5}(\-[0-9]{4})?$/";
$e_mail_verify = "/^.+@.+\\..+$/";
if(preg_match("/last/i",$field))
{
if(!preg_match($name_verify,$value))
{
$error_array[] = "$value is not a valid last name";
}
}
if(preg_match("/phone/i",$field))
{
if(!preg_match($phone_verify,$value))
{
$error_array[] = "$value is not a valid phone number";
}
} // endif phone format check
if(preg_match("/city/i",$field))
{
if(!preg_match($city_verify,$value))
{
$error_array[] = "$value is not a valid city";
}
}
if(preg_match("/address/i",$field))
{
if(!preg_match($address_verify,$value))
{
$error_array[] = "$value is not a valid address";
}
}
if(preg_match("/zip/i",$field))
{
if(!preg_match($zip_verify,$value))
{
$error_array[] = "$value is not a valid ZIP code";
}
}
if(preg_match("/e_mail/i",$field))
{
if(!preg_match($e_mail_verify,$value))
{
$error_array[] = "$value is not a valid e-mail address";
}
}
}
$clean_data[$field] = strip_tags(trim($value));
}
if(@sizeof($error_array) > 0)
{
$message = "<ul style='color: yellow; margin-top: 0; margin-left: 35%;
list-style: none' >";
foreach($error_array as $value)
{
$message .= "<li>$value</li>";
}
$message .= "</ul>";
echo $message;
extract($clean_data);
include("exercise_4_form.inc");
exit();
}
else
{
echo "<ol>";
foreach($_POST as $field => $value)
{ echo "<h3><li> $field = $value</li></h3>";
}
echo "</ol>";
}
}
else
{
include("exercise_4_form.inc");
}
?>
Edited by mac_gyver, 15 November 2014 - 12:14 PM. code tags around posted code please I'm trying to do a couple of things using the dropdown list below. I have a table with id, rank, and amount. 1. I need to echo the amount that corresponds to the rank selected from the dropdown list. 2. I also need to put the amount value in a variable so I can do some math with it later on. I've got the dropdown pulling the rank values from the db just not sure how to make it do the rest I'm trying to do. Any help would be greatly appreciated Code: [Select] <?php db($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> Hey, I made a register page and its all working great. I want to have a "Back" button when the user submits the form if password1 & password2 do not match or username allready exists. But trying to put one in after the echo command isn't working for me. Can anyone help? Heres the error I recieve: Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /home/jamiew90/public_html/Staff/register.php on line 49 Heres line 49 & the surrounding lines incase needed, line 49 is the one beginning with echo: //none were left blank! We continue... if($password != $cpassword) { // the passwords are not the same! echo "<br><br><b>Passwords do not match</b> <p> <FORM><INPUT TYPE="button" VALUE="Back" onClick="history.go(-1);return true;;"> </FORM>";"; }else{ // the passwords are the same! we continue... Is it possible to even add a back button after echo? Thanks for the help - Jamie Hello everyone, i'm kinda stuck here although i had this working before, i don't know what's wrong here. Basically i've got a form with a drop-down menu, that menu has 3 values, x,y and z. Now that form is working with post, sending me to the next page where i have a set of arrays and a echo which should print some text based on the option chosen from the drop down menu. my code in the form is this: Code: [Select] <select name="value1" id="value1"> <option value="x">x</option> <option value="y">y</option> <option value="z">z</option> </select> the array on the next page: Code: [Select] $arrayvalue = array("x"=>"test1", "y"=>"test2", "z"=>"test3"); and the echo: Code: [Select] <?php echo $arrayvalue["value1"];?> That leaves me with an empty page. What am i doing wrong here ? am i missing something ? any help is greatly appreciated. Sabrina Hi All, I have 2 tables: one CarMake - CarMakeID - CarMakeDesc two CarModel - CarModelID - CarModelMake - CarModelDesc Depending on what the user selects in the first dropdown (carmake) the possible selection in the second dropdown (model) needs to be limited to only the models from the selected carmake. in the second table (Carmodel : the 'CarModelMake' = CarMakeID, to identify the make) How do I limit the dropdown 'CarModel' based on the selected CarMake in the first dropdown. link : http://98.131.37.90/postCar.php code : -- -- -- Code: [Select] <label> <select name="carmake" id="CarMake" class="validate[required]" style="width: 200px;"> <option value="">Select CAR MAKE...</option> <?php while($obj_queryCarMake = mysql_fetch_object($result_queryCarMake)) { ?> <option value="<?php echo $obj_queryCarMake->CarMakeID;?>" <?php if($obj_queryCarMake->CarMakeID == $CarAdCarMake) { echo 'selected="selected"'; } ?> > <?php echo $obj_queryCarMake->CarMakeDesc;?></option> <?php } ?> </select> </label> <label> <select name="carmodel" id="CarModel" class="validate[required]" style="width: 200px;"> <option value="">Select MODEL...</option> <?php while($obj_queryCarModel = mysql_fetch_object($result_queryCarModel)) { ?> <option value="<?php echo $obj_queryCarModel->CarModelID;?>" <?php if($obj_queryCarModel->CarmodelID == $CarAdCarModel) { echo 'selected="selected"'; } ?> > <?php echo $obj_queryCarModel->CarModelDesc;?></option> <?php } ?> </select> </label> I'm a novice.. and appreciates all the help ! I am creating a form that will allow the user to select the make of vehicle "FORD" for example. If that make of vehicle is selected among different makes of vehicles, then another box will appear, with all the models for that particular model "Fiesta" for example. What type of code accomplishes this setup in my web page? I do not want to list 500 models in one drop down list, but just those for each make in the first drop down list. Thanks much! guys i have this dropdown menu echo "<form method=\"post\" action=\"\">"; echo "<center><select name=\"mydropdown\" size=\"0\" style=\"height:4em; width:15em;\">"; echo "<option value=\"Milk1\">{$_SESSION['tem1']}</option>"; echo "<option value=\"Milk2\">{$_SESSION['tem2']}</option>"; echo "<option value=\"Milk3\">{$_SESSION['tem3']}</option>"; echo "</select></center>"; echo "</form>"; and i wand to update a variable every time i select one of the contents... how can i done this??? If I have a drop-down list... Code: [Select] <select name="attendees"> <option value="">--</option> <option value="1">1</option> <option value="2">2</option> <option value="3">3</option> <option value="4">4</option> </select> Where does the value that the user selects get stored? Does it get stored in $_POST by default (if a Form is used)? What variable would hold the selected value? Debbie hello fellas, need some help please if possible. i have created a date of birth section in my form where the user selects his/her date of birth from the dropdown menu. they would first select the day then month then year of their birthday. how would i setup the database to get this to work? i currently have: Code: [Select] day VARCHAR( 2 ) NOT NULL , month VARCHAR( 4 ) NOT NULL , year VARCHAR( 4 ) NOT NULL , is this correct? many thanks I am trying to make a Web Form for people to fill out that has two drop down menus where i want the first drop down menu's selection changes the values of the selections in the second drop down. ((IE. If the first drop down value is "product 01" then the second drop down shows values "Red, Green, Blue" while if the first drop down value is "product 02" then the second drop down shows values "Yellow, Green")) Anyone have any ideas? Thanks in Advance. 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 This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=341998.0 I have a dropdown menu in a form where users must select their instrument: here's the dropdown: Code: [Select] <label for="instrument"><span class='red'>*</span>Your main instrument</label> <select name="instrument" id="instrument"> <option value="select" selected="selected">Select your instrument…</option> <option value="bassoon">Bassoon</option> <option value="cello">Cello</option> <option value="clarinet">Clarinet</option> <option value="double_bass">Double Bass</option> <option value="flute">Flute</option> <option value="french_horn">French Horn</option> <option value="oboe">Oboe</option> <option value="percussion">Percussion</option> <option value="trombone">Trombone</option> <option value="trumpet">Trumpet</option> <option value="tuba">Tuba</option> <option value="viola">Viola</option> <option value="violin">Violin</option> <option value="other">Other</option> </select> I want to validate to ensure that the user has selected an instrument from the list. I was going to do something like this: Code: [Select] // validate for instrument selection if(!isset($_POST['instrument'])) { $serrors[] = 'Please select your instrument.'; } else { $_SESSION['instrument'] = $_POST['instrument']; } The problem is, the session variable gets set if the user bypasses the menu altogether, leaving the default 'Select your instrument...' option (at the top) as it is. How can I ensure a selection other than the 'Select your instrument...' at the top of the list? Thanks in advance. Hi, I wonder if I somebody can help me I have three drop down lists which are populated from data thats stored in my sql database. I also have a table which is echoed from the database all on the same page. What i want to do is allow users to select values in the drop down lists which will then echo different results below in the table. Is this possible. So on page load it will show all of the info in the html table. Then when the user changes the drop downs and clicks submit it will change the query to something SELECT dropdown1 value FROM users where Dropdown2 = 1 AND Dropdown3 = 2 etc. Is this possible to display this information on the same page using the dropdown list or does it need to load a different page as the client side is trying to edit a server side script? If its possible please can someone give me an example. Thanks in advance Edd
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Untitled Document</title>
</head>
<body>
<?php
//MySQL Database Connect
include 'sqlconnect.php';
?>
<select name="aktivitet" id="aktivitet">
<?php
//MySQL Database Connect
include 'sqlconnect.php';
$sql = mysql_query("SELECT title FROM aktiviteter");
while ($row = mysql_fetch_array($sql)){
echo "<option value=\"owner1\">" . $row['title'] . "</option>";
}
?>
</select>
</body>
</html>
why dosnt this work?, if you cant see what i want to happen, ill post an explaination
Please, take a look to the following code.After clicking Next it goes to overview.php.Why when I click back on my browser to return to this page again, it is not returning back? When I click back I receive "Confirm Form Resubmission" message. After refreshing page it loads page. I guess problem is in "session_start();" part. Something to do with cookies. Please, help me it is very urgent for me. <?php session_start(); echo "<html> <head> <title>Hello World</title> <meta http-equiv='Content-Type' content='text/html; charset=Windows-1252'/> </head>"; require_once ('functions.inc'); if(!isset($_POST['userid'])) { echo "<script type='text/javascript'>"; echo "window.location = 'index.php'"; echo "</script>"; exit; }else{ session_register("userid", "userpassword"); $username = auth_user($_POST['userid'], $_POST['userpassword']); if(!$username) { $PHP_SELF = $_SERVER['PHP_SELF']; session_unregister("userid"); session_unregister("userpassword"); echo "Authentication failed " . "Please, write correct username or password. " . "try again "; echo "<A HREF=\"index.php\">Login</A><BR>"; exit; } } function auth_user($userid, $userpassword){ global $default_dbname, $user_tablename; $user_tablename = 'user'; $link_id = db_connect($default_dbname); mysql_select_db("d12826", $link_id); $query = "SELECT username FROM $user_tablename WHERE username = '$userid' AND password = '$userpassword'"; $result = mysql_query($query) or die(mysql_error()); if(!mysql_num_rows($result)){ return 0; }else{ $query_data = mysql_fetch_row($result); return $query_data[0]; } } echo "hello"; echo "<form method='POST' name='myform' action='overview.php'>"; echo "<input type='submit' value='Next'>"; echo "</form>"; ?> 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
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
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