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PHP - Drop Down Menu Input Help Needed
I am brand new to php and I don't know where I am going wrong with this.
I would like to have drop down menu which is populated from mySQL and which would allow user to select one option. The selection would be used on the following page once the form is submitted. Right now I have the following script and it populates the dropdown menu but it does not return value when selection is made. <form action="databaseinputcheck2.php" method="post"> <?php mysql_connect("localhost", "xxxxxxxxxxxx", "xxxxxxxxxxxxxx") or die(mysql_error()); echo "Connected to MySQL<br />"; ?> <p>Lotteries: <?php mysql_select_db("wpawlowski") or die(mysql_error()); // Get all the data from the "LotteryNames" table $lotterynameresult = mysql_query("SELECT id, Name FROM LotteryNames") or die(mysql_error()); ?> <select name="LotteryName" size="1"> <option selected value="">Select One</option> <option value="">----------</option> <?php // keeps getting the next row until there are no more to get while($currentlottery = mysql_fetch_array( $lotterynameresult )) { $lid=$currentlottery['id']; $lname=$currentlottery['Name']; // Print out the contents of each row into drop down menu echo "<option value='$lid'>$lname</option>\n"; //$options.="<OPTION VALUE=\"$lid\">".$lname; //Selection input line } ?> <input type="submit" value="GO" /> </form> I looked on the internet and it is my understanding that if I comment out the line starting withc "echo" and uncomment the one below the drop down menu should still be populated and the selection would be stored in the $options variable which in turn could be transfered to databaseinputcheck2.php with $_POST to be used there, but when I do uncomment that line and comment the echo line out my drop down menu is not populated, what am I doing wrong? Similar TutorialsAs a complete newbie to php and webdesigning i have a following problem.I would like to retrieve the data from database and display it in a drop down menu.Then i should allow the user to select the values from drop down list along with other details,in other words i have to embed the drop down output as the form input for the user and store the form data in another table.I am running a xampp server and i am using php 5.4 version.Please help.My code is as follows.In this case project_name is displayed as the drop down output.but how do i use the same drop down output as a input in the form. <html> <head></head> <body> <?php error_reporting(E_ALL ^ E_DEPRECATED); include 'connect.php' ; $tbl_name="projects"; $sql="SELECT project_name FROM $tbl_name "; $result=mysql_query($sql); if($result === FALSE) { die(mysql_error()); } ?> <form name="resources" action="hourssubmit.php" method="post" > <?php echo "<select name='project_name'>"; while ($row = mysql_fetch_array($result)) { echo "<option value='" . $row['project_name'] ."'>" . $row['project_name'] ."</option>"; } echo "</select>"; ?> </form> </body> </html> I'm working on a fix for cURL that replaces all relative urls with absolute. Here's the code: <?php //Get web address //FORMAT: url=site.com, url=site.net, url=site.org $page = $_GET['url']; //Format web address $http = 'http:\/\/'; $www = 'www.'; if(preg_match('/'.$http.'/', $page)){preg_replace('/'.$http.'/', '', $page);} if(preg_match('/'.$www.'/', $page)){preg_replace('/'.$www.'/', '', $page);} $page = rtrim($page, '/'); $page = 'http://www.'.$page.'/'; //cURL // create curl resource $ch = curl_init(); // set url curl_setopt($ch, CURLOPT_URL, $page); //return the transfer as a string curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); // $output contains the output string $output = curl_exec($ch); // close curl resource to free up system resources curl_close($ch); //Convert relative URL to absolute $output = preg_replace('/src="/', 'src="'.$page, $output); $output = preg_replace('/href="/', 'href="'.$page, $output); $output = preg_replace('/action="/', 'action="'.$page, $output); echo $output; ?> As you can see it's pretty basic. In many cases it fixes broken styles, links, images and form actions. I am looking for any ideas as to how I can add some more intelligence to this script. 1. What else should it do 2. Where is it not doing it's job 3. Can it do what it's already doing better Any input offered is much appreciated. I'm not looking for someone to write code, but if you are intrigued and want to add a snippet to it, that's cool! Feel free to keep a copy of your own if you like the idea to build off of. Thanks Peeps, E I have two basic dropdown lists in a form, my question is this: After selecting an item from the 1st dropdown list, the second dropdown list should automatically load only fields that are connected with the first selected item. (ex Site1 has Tank01 and Tank06, but Site2 has Tank03, Tank04 and also a Tank06). What is the script needed / or how do I change my script to accommodate this? Here is the basic script I use for the second dropdown list: <?php include("../xxx.php"); $cxn = mysqli_connect($host,$user,$password,$dbname); $query = "SELECT DISTINCT `diesel_tank_id` FROM `diesel_tank` ORDER BY `diesel_tank_id`"; $result = mysqli_query($cxn,$query); while($row = mysqli_fetch_assoc($result)) { extract($row); echo "<option value='$diesel_tank_id'>$diesel_tank_id</option>\n"; } ?> Hello people, thank you to everyone that has helped me on this forum. You have been terrific. I am submitting a form to a database and am having a small problem with the validation scripts. Basically if the user doesn't put something into one of the fields they get an error message. However, I am keeping the information already submitted in the input box with the following code Code: [Select] <input type='text' name='username' size = '40' maxlength='30' value = '<?php echo $username; ?>'> This is so that the user doesn't have to input the same data again if he forgot one box. How do I do the same for "drop down" boxes? Here is the code from the form for the "ppr" (aviation terms 'Prior Permission Required') Code: [Select] <select name = "ppr"> <option value = "Yes">Yes </option> <option value="No">No </option> </select> The problem I have is that there are quite a number of these drop down boxes on the form I have created. And it always goes back to the first "option value" when there is an error. I do have a variable for $ppr, please could someone tell me how to incorporate it so that the user doesn't have to select the correct option value all over again. Hope my explanation is clear. Thanks in advance. Hi All, I have a snippet of code (below) which I use to create a dynamic menu from a MySQL database for a community website. Fields used in the database a ID parent (what the parent page is) page_order (what is says on the tin) short_title (a short title for the page for the URL) title (the page title - used as the text for the link) content (the content for the page) I have got the dynamic list to display fine, but what I would like to be able to do is give each list item a 'class' tag when it has been selected (and thus is the page being shown). Can anyone suggest how to amend the below code to allow this? Code is: Code: [Select] <!-- ##### START LIST_INFO_PAGES ##### --> <?php include("../config/config.php"); ?> <?php $sql = "SELECT * FROM `PCNET_$filename` WHERE parent = 'council' ORDER BY page_order ASC"; $sql_result = mysql_query ($sql, $connection ) or die ('request "Could not execute SQL query" '.$sql); while ($row = mysql_fetch_assoc($sql_result)) { print ''; if ($row["status"]<>'') { echo '<li><a href="../council-indiv.php?id='; echo ($row["id"]).'&short_title='; echo nl2br(stripslashes(utf8_decode($row["short_title"]))).'" title='; echo nl2br(stripslashes(utf8_decode($row["title"]))).'>'; echo nl2br(stripslashes(utf8_decode($row["title"]))).'</a></li>'; }; }; ?> <!-- ##### END LIST_INFO_PAGES ##### --> Help will be welcome! Neil Hi All, I am trying create 4 dropdowns and load the values based on the value selected in the previous dropdown. I am able to do this for 3 dropdowns, but for the fourth dropdown I am no able to populate the value. Could someone help me out with this please? --------------------------------------- ajaxData_1.php
<?php
// Include the database config file
include_once 'dbconfig_1.php';
if(!empty($_POST["solutions_id"])){
// Fetch state data based on the specific country
$query = "SELECT * FROM releases WHERE solutions_id = ".$_POST['solutions_id'];
$result = $db->query($query);
if($result->num_rows > 0){
echo '<option value="">Select release</option>';
while($row = $result->fetch_assoc()){
echo '<option value="'.$row['releases_id'].'">'.$row['releases_name'].'</option>';
}
}
else{
echo '<option value="">Release not available</option>';
}
}
elseif(!empty($_POST["releases_id"])){
$query = "SELECT * FROM versions WHERE releases_id = ".$_POST['releases_id']."";
$result = $db->query($query);
if($result->num_rows > 0){
echo '<option value="">Select version</option>';
while($row = $result->fetch_assoc()){
echo '<option value="'.$row['version_id'].'">'.$row['supported_version'].'</option>';
}
}
else{
echo '<option value="">Version not available</option>';
}
}
elseif(!empty($_POST["versions_id"]))
{
$query = "SELECT * FROM platforms WHERE version_id = ".$_POST['version_id']."";
$result = $db->query($query);
if($result->num_rows > 0){
echo '<option value="">Select Version</option>';
while($row = $result->fetch_assoc()){
echo '<option value="'.$row['platforms_id'].'">'.$row['platforms_name'].'</option>';
}
}
else{
echo '<option value="">Platform not available</option>';
}
}
?>
<?php
if(isset($_POST['submit']))
{
echo 'Selected Solution ID: '.$_POST['solution'];
echo 'Selected Release ID: '.$_POST['release'];
echo 'Selected Version ID: '.$_POST['version'];
echo 'Selected Platform ID: '.$_POST['platform'];
}
?>
Index_3.php
<!DOCTYPE html>
<html>
<head>
<title style="color: 000000">Solution Compatibility Matrix</title>
<style>
select
{
padding: 12px;
min-width:280px;
margin-top:10px;
}
.san{
width:280px;
margin:0 auto;
margin-top:90px;
background-color:#FFFFFF;
padding:55px;
color: 000000;
}
label
{
color:#000000;
margin-bottom:25px;
}
html{
background-color: #3399CC;
}
</style>
</head>
<body>
</body>
</html>
Thanks, Arun Edited April 23, 2020 by cyberRobotAdded code tags I have a search function in php where the text characters are matched to characters in a tables field--- works perfectly.... I need to make the input box have a droplist of words from database, this is also easy for me to do. the problem here is there is no definitive value! the options list always outputs a blank in the url--- its supposed to search a matching value and then output the matching value to url... Here is the droplist code: $output['RATESTITLE']='<input class="inputbox" type="text" size="24px" name="ratestitle" value="'.$sch->filter['ratestitle'].'" onfocus="if (this.value ==\''.$output['LANGUAGE_SEARCH_RATESTITLE'].'\') {this.value = \'\'}" />'; this outputs a input text box--- i want to have a droplist of options to populate this text box... If you must know this is the third day im at it... 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 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!! 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> 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
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
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'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 all !
While it is clear that the input in an input text box requires to be filtered or sanitized, yet it is not clear to me if and why would the input of a dropdown menu / checkboxes / radio require to be filtered or sanitized.
Can someone tell me if these inputs require sanitization? if yes, can you please explain how these would pose a security threat if left un-sanitized.
Thanks !
I have this function completely written in my class file that I am working on. The point to this function is to be able to check the login of a user or administrator for either of the control panels associated with my site. It will check the session intime as well as the page / module referenced. Once it passes all those checks, it will check and ensure the emailaddress/password stored in the current session still holds true and the account is still active... if the account is still active it will update the lastActivity as well as update all of the session variables with what is currently in the database. What I am looking for is basically a look at the function, see if it looks good.. If there is any part to it that could create security holes for the site just off the login function itself... Usage: $q->validUser($_SESSION['user'], $_mod); <?php function validUser($sess, $p) { if ($sess['inTime'] == '' && $p != 'login' && $p != 'logout') { session_destroy(); $login = '0'; $_int = ''; return $login; } else if ($sess['inTime'] < time()-3600 && $p != 'login') { $sess['inTime'] = ''; session_destroy(); $this->check_login($sess, $p); } else { $this->user = $sess['emailAddress']; $this->pass = $sess['password']; $login = $this->sql_query("SELECT * FROM users WHERE emailAddress = '".$this->user."' AND password = '".$this->pass."' AND status = '1' LIMIT '1'"); if ($login = $this->sql_numrows($login) < 1) { $sess['inTime'] == ''; session_destroy(); $login = '0'; } else { // logged in, lets update the database for last_activity AND the session. $this->sql_query("UDATE users SET lastActivity = '".now()."' WHERE emailAddress = '".$this->user."'"); $login = $this->sql_query("SELECT * FROM users WHERE emailAddress = '".$this->user."' AND password = '".$this->pass."' AND status = '1' LIMIT '1'"); $login = mysql_fetch_assoc($login); foreach ($login as $key => $value) { $sess[$key] = $value; } $sess['inTime'] = time(); $login = '1'; } return $login; } } ?> That is the main function, sql_query and sql_numrows is: <?php function sql_query($query = "", $transaction = FALSE) { unset($this->query_result); if ($query != "") { $this->num_queries++; if ($transation == BEGIN_TRANSACTION && !$this->in_transation) { $result = mysql_query("BEGIN", $this->db_connect_id); if (!$result) { return false; } $this->in_transaction = TRUE; } $this->query_result = mysql_query($query, $this->db_connect_id); } else { if ($transaction == END_TRANSACTION && $this->in_transaction ) { $result = mysql_query("COMMIT", $this->db_connect_id); } } if ($this->query_result) { unset($this->row[$this->query_result]); unset($this->rowset[$this->query_result]); if ($transaction == END_TRANSACTION && $this->in_transaction ) { $this->in_transaction = FALSE; if (!mysql_query("COMMIT", $this->db_connect_id)) { mysql_query("ROLLBACK", $this->db_connect_id); return false; } } return $this->query_result; } else { if ($this->in_transaction ) { mysql_query("ROLLBACK", $this->db_connect_id); $this->in_transaction = FALSE; } return false; } } function sql_numrows($query_id = 0) { if(!$query_id) { $query_id = $this->query_result; } return ($query_id) ? mysql_num_rows($query_id) : false; } ?> Any insight that can help to benefit these functions would be appreciated. I am trying to get the drop downs to work in this order on the second <li> 903,896,898,1513 but they won't. I think it's picking up ascending order maybe? Is there another php/include I should use? It's a WordPress site. Code: [Select] <div id="navigation"> <div id="navbar_menu"> <ul id="nav_menu"> <li><?php wp_list_pages('include=3&title_li=&depth=2');?></li> <li><?php wp_list_pages('include=2,903,896,898,1513&title_li=&depth=2');?></li> <li><?php wp_list_pages('include=23,911,913&title_li=&depth=1');?></li> <li><?php wp_list_pages('include=25,921,925,927&title_li=&depth=1');?></li> <li><?php wp_list_pages('include=27&title_li=&depth=2');?></li> <li><?php wp_list_categories('include=3&title_li=&depth=2');?></li> <li><?php wp_list_pages('include=407&title_li=&depth=2');?></li> </ul> </div> </div> Thanks in advance! [attachment deleted by admin] 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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