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PHP - Fetching Selected Value From One Dropdown To Populate Another Dropdown
Hi freaks,
I'm new to php first of all. I'm dynamically binding a dropdownlist with mysql database . After the user selects an item from it , I want to match that item with another table so as to populate another database. The code I'm using to populate dropdown: Code: [Select] <?php $con = mysql_connect("localhost","root",""); if(!$con) { die ('Can not connect to : '.mysql_error()); } mysql_select_db("ims",$con); $result=mysql_query("select cat_id,cat_name from category"); echo "<select name=cat>"; while($nt=mysql_fetch_array($result)) { echo "<option value=$nt[cat_id]> $nt[cat_name] </option>"; } echo "</select>"; mysql_close($con); ?> Now after the user selects any one of the item , I want to bind another dropdown on the same page using such query like $result=mysql_query("select subcategory.sc_id,subactegory.sc_name from subcategory,category where subcategory.sc_id=$nt[cat_id]"); Please anyone tell me the logic and code to do it. Also tell me do I need an intermediate page to post the 1st dropdown value and then continue with 2nd dropdown. I couldn't figure out the concept anyhow. Help on this will be highly appreiable . (Tell me if I'm not clear with my question) Similar TutorialsI have a form with dropdown list that is populated with values from Sql Server table. Now i would like to use this selected item in SQL query. The results of this query should be shown in label or text field. So when a user selects item from dropdown menu, results from SQL query are shown at the same time. I have two dropdown list at the moment in my form. First one gets all values from a column in table in SQL Server. And the second one should get a value from the same table based on a selection in first dropdown list.
When i load ajax.php i get 2 error mesages: This is my code so far. I have tried to do it with this Ajax script. But i can only get first dropdown to work. The second dropdown(sub_machinery) does not show values, when first dropdown item is selected. The second dropdown should show values from databse table with this query( *$machineryID* is first dropdown selected item): SELECT MachineID FROM T013 WHERE Machinery=".$machineryID. Index.php
<!doctype html>
<?PHP
$server = "server";
$options = array( "UID" => "user", "PWD" => "pass", "Database" =>
"database");
$conn2 = sqlsrv_connect($server, $options);
if ($conn2 === false) die("<pre>".print_r(sqlsrv_errors(), true));
echo " ";
?>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
</head>
<body>
<section id="formaT2" class="formaT2 formContent">
<div class="row">
<div class="col-md-2 col-3 row-color remove-mob"></div>
<div class="col-md-5 col-9 bg-img" style="padding-left: 0;
padding-right: 0;">
<h1>Form</h1>
<div class="rest-text">
<div class="contactFrm">
<p class="statusMsg <?php echo
!empty($msgClass)?$msgClass:''; ?>"><?php echo $statusMsg; ?></p>
<form action="connection.php" method="post">
<div>machinery</div>
<select id="machinery">
<option value="0">--Please Select Machinery--</option>
<?php
// Fetch Department
$sql = "SELECT Machinery FROM T013";
$machanery_data = sqlsrv_query($conn2,$sql);
while($row = sqlsrv_fetch_array($machanery_data) ){
$id = $row['Id'];
$machinery = $row['Machinery'];
// Option
echo "<option value='".$id."' >".$machinery."</option>";
}
?>
</select>
<div class="clear"></div>
<div>Sub Machinery</div>
<select id="sub_machinery">
<option value="0">- Select -</option>
</select>
<input type="submit" name="submit"
id="submit" class="strelka-send" value="Insert">
<div class="clear"> </div>
</form>
</div>
</div>
</div>
</div>
</section>
</script>
<script type="text/javascript">
$(document).ready(function(){
$("#machinery").change(function(){
var machinery_id = $(this).val();
$.ajax({
url:'ajaxfile.php',
type: 'post',
data: {machinery:machinery_id},
dataType: 'json',
success:function(response){
var len = response.length;
$("#sub_machinery").empty();
for( var i = 0; i<len; i++){
var machinery_id = response[i]['machinery_id'];
var machinery = response[i]['machinery'];
$("#sub_machinery").append("<option
value='"+machinery_id+"'>"+machinery+"</option>");
}
}
});
});
});
</script>
</body>
</html>
Ajaxfile.php
<?php
$server = "server";
$options = array( "UID" => "user", "PWD" => "pass",
"Database" => "database");
$conn2 = sqlsrv_connect($server, $options);
if ($conn2 === false) die("<pre>".print_r(sqlsrv_errors(), true));
echo " ";
$machineryID = $_POST['machinery']; // department id
$sql = "SELECT MachineID FROM T013 WHERE Machinery=".$machineryID;
$result = sqlsrv_query($conn2,$sql);
$machinery_arr = array();
while( $row = sqlsrv_fetch_array($result) ){
$machinery_id = $row['ID'];
$machinery = $row['MachineID'];
$machinery_arr[] = array("ID" => $machinery_id, "MachineID" =>
$machinery);
}
// encoding array to json format
echo json_encode($machinery_arr);
?>
Edited May 6, 2019 by davidd
Hey Guys, I know it may seem pretty simple, but im having trouble populating a drop down list. Here is my code at the moment, but what it's doing is displaying the names all in one value, where it should be in separate select values. *Note that i have only done it to the first one. See attachment. 'AntonMatt' are next to each other, they should be separate select values. Code: [Select] <? $id = $_GET['id']; $selectplayers="SELECT * FROM players WHERE club='$club' AND team='$team'"; $player=mysql_query($selectplayers); ?> <table class='lineups' width="560" cellpadding="5"> <tr> <td colspan="2">Starting Lineup</td> <td colspan="2">On the Bench</td> </tr> <tr> <td width="119"> </td> <td width="160"> </td> <td width="69"> </td> <td width="160"> </td> </tr> <tr> <td>Prop</td> <td><select name="secondary" style="width: 150px"> <option value='' selected="selected"><? while($rowplayer = mysql_fetch_array($player)) { echo $rowplayer['fname']; } ?></option> </select></td> <td>16.</td> <td><select name="secondary16" style="width: 150px"> <option value='' selected="selected">Secondary Position</option> </select></td> </tr> <tr> <td style="padding-top: 8px;">Hooker</td> <td style="padding-top: 8px;"><select name="secondary2" style="width: 150px"> <option value='' selected="selected">Secondary Position</option> </select></td> <td style="padding-top: 8px;">17.</td> <td style="padding-top: 8px;"><select name="secondary17" style="width: 150px"> <option value='' selected="selected">Secondary Position</option> </select></td> </tr> </table> </div> </div> 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> This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=351349.0 Hello forum,
So I've been developing an app mostly in PHP, but am rather afraid of JS. Hope to fix that.
I have an AJAX dropdown using JQuery to search locations. It works great. However, I want to make it similar to what is seen on this site:
http://placefinder.com/
As you can see, the dropdown, when clicked populates a box. Then the user submits the form and the data is used in the application.
I have no clue how to make the form populate with data from the DB (I'm using mySQL) when clicked. So far, I've only been able to make it clickable as a URL (not what I want, obvioiusly!)
Is there a way to do this on a really small, simple script for starters? I'm certain their is, but don't even know where to begin.
Any help appreciated
I must be missing something simple. I've got this little script that pulls rows from the database to populate a dropdown. If one of them matches a predetermined variable, then I want it to show selected. It's... almost working. The dropdown prints, and shows the options. But the selected item shows separate, printed just below the dropdown? Here's what I've got: Code: [Select] <?php $quer3=mysql_query("SELECT discount_id, discount_name,discount_amount FROM tbl_discount order by discount_amount"); echo " <select name='discount_id'><option value=''>Select</option>"; while($row = mysql_fetch_array($quer3)) { if($row[discount_id]==$discount_id){echo "<option selected value='".$row[discount_id]."'>".$row[discount_name]." / $".$row[discount_amount]."</option>";} else{echo "<option value='$row[discount_id]'>$row[discount_amount]</option>";} echo "</select>"; } ?> Hi I have been trying to get a value to be selected in a mysql populated dropdown list but can't get it to work and was hoping someone could help I have a database with user info in it and this is an update page where they can update their details. The code i have (which doesn't work) is: <select name="agency"> <? $query1 = mysql_query("SELECT * FROM agents ORDER BY agent ASC",$connect); while($myrow = mysql_fetch_assoc($query1)){ $agent = $myrow['agent']; echo "<option"; if ($agent == $agency) { echo "selected='selected'"; } echo ">$agent</option>"; } ?> </select> The $agency value is the current agency which is stored in the users profile and the value does exist in the list which is being populated (also, i have define $agency further up in my code) so i don't know why the selected value won't display. No value is displayed in the dropdown list on the page - but the values are in the list if i remove the selected='selected' part of the code. Any help yould be greatly appreciated. Merry Christmas Andy The rest of my coding works however this part does not and I'm trying to figure out why. I'm sure my syntax isn't right so I hope someone can correct my mistake. $contentpageID = $_GET['id']; $query = "SELECT contentpages.contentpage, contentpages.shortname, contentpages.contentcode, contentpages.linebreaks, contentpages.backlink, contentpages.showheading, contentpages.visible, contentpages.template_id FROM contentpages WHERE contentpages.id = '" . $contentpageID . "'"; <label for="template">Template</label> <select class="dropdown" name="template" id="template" title="Template"> <option value="0">- Select -</option> <?php $query = 'SELECT * FROM templates'; $result = mysql_query ( $query ); while ( $template_row = mysql_fetch_assoc ( $result ) ) { print "<option value=\"".$template_row['id']."\" "; if($template_row['id'] == $row['template_id']) { print " SELECTED"; } print ">".$template_row['templatename']."</option>\r"; } ?> </select> i skipped solving this the other day with an easier way but now i am stuck with this stumble again in another area and i have no way out.... so here it goes, i have an ajax dropdown box...i need to get the value that is selected by the user when it is clicked and then pass this value to a new pop window by appending to its url....any suggestions? mysql.php <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("ckeditor",$con); ?> --------------------------------------------------------- add.php <?php include("mysql.php"); if(isset($_POST["button2"])) { $sql="INSERT INTO cktext (section) VALUES ('$_POST[select2]')"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } ?> ----------------------------------------------------------------- home.php <form id="form1" name="form1" method="post" action="add.php"> <tr> <td>Section:</td> <td><select name="select2" id="select2"> <option selected="selected" value="MALE">Male</option> <option selected="" value="FEMAIL">FEMAIL</option> </select></td> </tr> <input type="submit" name="button2" id="button2" value="Upload" /> </form> In DATABASE :- cktext table attribute "section" is varchar type. BUT IT RETURN ME BLANK OUTPUT. friend the code display all the managers name
<?phpand the variable below displays the result $Manager= $row['Manager'] ; but how do i do it on the above dropdown list the it shows which result is in the $Manager variable? I'm just attempting to learn how PHP handles things and I can't quite wrap my head around how to apply Selected to the final Option and show the Traits for the Character based on the Selected Option. I understand this might need POST, if it does, I would appreciate a bit of help on how I would set this up as POST since I didn't think a drop down required a submit button. Code: [Select] $character= array (array(Name=>"Barry","Class"=>"Fighter",Level=>1,Str=>10,Dex=>10,"Int"=>10),array(Name=>"Lindehar","Class"=>"Ranger",Level=>1,Str=>10,Dex=>10,"Int"=>10),array(Name=>"Verelden","Class"=>"Mage",Level=>1,Str=>10,Dex=>10,"Int"=>10)); print "Select a Character:<br /><select>"; foreach($character as $array_num) { foreach($array_num as $char_trait=>$trait_value) { if($char_trait==Name) { $selected_value=""; $generated_chars="<option selected=".$selected_value." value='".$char_trait."'>".$char_trait.": ".$trait_value."</option>"; print $generated_chars; if($selected_value=/".$char_trait.": Barry"/") { foreach($char_trait=="Barry") { print "<h4>".$char_trait.": ".$trait_value."</h4>"; } } } } } print "</select><p />"; i have the following code: <td width='100px'>Suppliers <select name="supplier"> <?php $catcher_id = $service->getCatcherId(); $supplier_names = LpmAdnetworkPeer::getByName($catcher_id); foreach($supplier_names as $row) { ?> <option><?php echo $row->getName(); ?></option> <?php } ?> </select> </td> then on the same form i have a submit button that takes me to the next form..the problem now is how can i access the ID of the item seleted in the dropdown on the NEXT form please? i can get the name from the list by $_POST['supplier'] on the next form thanks Hey Everyone... First off, I am only a young web developer and i'm working on a school project and am making a text-based game online... Now what i'm having trouble with... I want a drop-down list that has a list of characters classes Clubber Mixer Sauceror Tamer And I want whatever is selected to be placed into the database along with the username/password (THIS ALL WORKS FINE JUST NOT THE DROP DOWN LIST) All help appreciated Yesterday, I created a topic about how I could update records and I managed to achieve that successfully. Now I have another dilemma. When I have a specific record I want to update, I want to change a category ID of an product (e.g. change it from 1 to 2) but how do I go about doing this? Here is my code thus far: Code: [Select] <?php require_once ('./includes/config.inc.php'); require_once (MYSQL); $id=$_GET['prodID']; $results = mysqli_query($dbc, "SELECT * FROM product WHERE prodID=".$_GET['prodID'].""); $row = mysqli_fetch_assoc($results); ?> <form action="" method='POST'> Product ID: <input type="text" value="<?php echo $row['prodID'];?>" name="prodID" /> <br /> Product: <input type="text" value="<?php echo $row['product'] ;?>" name="product" /> <br /> Product Description: <input type="text" value="<?php echo $row['prod_descr'] ;?>" name="prod_descr" /> <br /> Category: <select name="category"> <option value="<?php echo $row['catID'];?>"></option> </select> Price: <input type="text" value="<?php echo $row['price'] ;?>" name="price" /> <br /> In Stock: <input type="text" value="<?php echo $row['stock'] ;?>" name="stock" /> <br /> <br /><input type="submit" value="save" name="save"> </form> <?php if(isset($_POST['save'])) { $id = $_POST['prodID']; $product = $_POST['product']; $descr = $_POST['prod_descr']; $price = $_POST['price']; $stock = $_POST['stock']; // Update data $update = mysqli_query($dbc, "UPDATE product SET product='$product', prod_descr='$descr', price='$price', stock='$stock' WHERE prodID=".$_GET['prodID'].""); header( 'Location: update_save.php' ) ; } ?> I'm new to this form and php/mysql so sorry if this isn't the right place to post this. This is what is in the first dropdown box. Code: [Select] $selValues['john'] = "a, b, c, d, e"; These are the different lists I want it to put in the second drop down box depending on what they choose in the first. Code: [Select] $selValues['list1']= " a1, a2, a3, a4, a5"; $selValues['list2']= " b1, b2, b3, b4, b5"; This is how I made an attempt to make it work before posting here. Along with plently of other ways. Code: [Select] if ($selValues['john']=a) { $chan_input = selectBuilder($selValues['$list1'] }; else if ($selValues['john']=b) { $chan_input = selectBuilder($selValues['$list2'] }; Basicly how I need it to work is there are two drop down boxes. First they will chose between a,b,c,d,e. If they chose a then on the second drop down box I only want them to be able to select a1,a2,a3,a4,a5. Folks, I have a dropdown, once values are selcted, these values should be put in a URL structure so that it matches with the one MoD_rewrite rule in my .htaccess. Here is my Mod_Rewrite Rule: RewriteRule ^(.*)/([^/]*)\.html$ search.php?q=$1&sc=$2 [QSA,L] Here is my Dropdown Code: <div id="search" > <form id="searchform" method="get" action="search.php"> <label>Search By Brand/ Manufacturer: </label> <select name="q"> <option value="SelectBrand">Select Brand</option> <?php if(isset($this->search->options)): ?> <?php foreach($mfg as $lolachild): ?> <option value="<?php echo $lolachild; ?>"><?php echo ucwords($lolachild); ?></option> <?php endforeach; ?> <?php endif; ?> </select> <select name="sc"> <option value="All"><?php eprint(LangAll); ?></option> <?php if(isset($this->search->options)): ?> <?php foreach($this->search->options as $cat): ?> <option value="<?php echo $cat->value; ?>"><?php echo $cat->name; ?></option> <?php endforeach; ?> <?php endif; ?> </select> <input type="submit" value="<?php eprint(LangSearch); ?>" /> </form> </div> Problem is, upon Submit, it goes to this link structu http://mydomain.co.uk/search.php?q=fisher&sc=302 It should rather be: http://au2.co.uk/fisher/302.html What am i missing or doing wrong? Cheer Natasha I am not a developer but I can modify code to work for me. The following code works on my test machine (Windows 10, IIS, PHP 7.4) but doesn't work on my website (cPanel, Some version of Linux, PHP 7.4). The two dropdowns are for State and City. You are supposed to be able to select the state and then select a city from that state then bring up a report for craft breweries in the city. When selecting State from the first dropdown, the page refreshes, the URL is correct with the reports.php?cat=<STATE> so $cat is being set, but the first dropdown no longer has the state selected and the second dropdown is populated with All cities and not just one ones from the selected state. Any ides why this is working fine on one machine and not the other? Selected code from reports.php
<?php
?>
/////// for second drop down list we will check if State is selected else we will display all the cities/////
echo "<form method=post action='brewerylistbycity.php'>";
}
////////// Starting of second drop downlist /////////
//// End Form /////
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