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PHP - Dropdown Box With Two Fields, One Hidden???
Hi I have an index.php with three drop down boxes on, each drop down uses an SQL statement to pull seatNumber from 'seats' each seat number has a UI and a forgein key field called SrideID.
I have the buttons on the dorp down boxes going to dd-check.php, here the seat chosen is posted and a simple echo(just testing it works atm) for me to do any queries I also need the SrideID within the dropdown box, i tried adding it within the <option></option> but it only made the drop down duplicate. 1 1 2 2 3 3 ect..... what i would like is for the SrideID field to be included with in the dropdown box but hidden so i can post it to the dd-check.php and then show the time and price of that particular ride. here is the table and the code so far. Thanks for any help. CREATE TABLE `ride` ( `RrideID` tinyint(1) default NULL, `name` varchar(20) default NULL, `time` tinyint(1) default NULL, `price` varchar(4) default NULL ) INSERT INTO `ride` VALUES (1, 'Swinging Ship', 7, '2.50'); INSERT INTO `ride` VALUES (2, 'Roller Coaster', 5, '3.75'); INSERT INTO `ride` VALUES (3, 'Ice Blast', 4, '3.00'); CREATE TABLE `seats` ( `seatID` int(8) NOT NULL default '0', `SrideID` tinyint(1) default NULL, `seatNumber` int(2) default NULL, PRIMARY KEY (`seatID`) ) INSERT INTO `seats` VALUES (1, 1, 1); INSERT INTO `seats` VALUES (2, 1, 2); INSERT INTO `seats` VALUES (3, 1, 3); INSERT INTO `seats` VALUES (4, 1, 4); INSERT INTO `seats` VALUES (5, 1, 5); INSERT INTO `seats` VALUES (49, 2, 1); INSERT INTO `seats` VALUES (50, 2, 2); INSERT INTO `seats` VALUES (51, 2, 3); INSERT INTO `seats` VALUES (52, 2, 4); INSERT INTO `seats` VALUES (53, 2, 5); INSERT INTO `seats` VALUES (69, 3, 1); INSERT INTO `seats` VALUES (70, 3, 2); INSERT INTO `seats` VALUES (71, 3, 3); INSERT INTO `seats` VALUES (72, 3, 4); INSERT INTO `seats` VALUES (73, 3, 5); code for index.php <table width="200" border="1"> <tr> <h1>Swinging Ship</h1> <?php echo '<form method="post" name="f1" action="dd-check.php">'; $query_s = "SELECT * FROM `seats` WHERE SrideID='1' ORDER BY seatID"; $result = mysql_query($query_s); echo"<select name='seatNumber'><option value=''>Select Seat</option>"; while( $row = mysql_fetch_array($result) ) { echo '<option>'.$row['seatNumber'].'</option>'; } echo '</select>'; mysql_free_result( $result ); echo "<input type='submit' value='Submit'>"; echo "</form>"; ?> </tr> <tr> <h1>Roller Coaster</h1> <?php echo '<form method="post" name="f2" action="dd-check.php">'; $query_r = "SELECT * FROM `seats` WHERE SrideID='2' ORDER BY seatID"; $result = mysql_query($query_r); echo"<select name='seatNumber'><option value=''>Select Seat</option>"; while( $row = mysql_fetch_array($result) ) { echo '<option>'.$row['seatNumber'].'</option>'; } echo '</select>'; mysql_free_result( $result ); echo "<input type='submit' value='Submit'>"; echo "</form>"; ?> </tr> <tr> <h1>Ice Blast</h1> <?php echo '<form method="post" name="f3" action="dd-check.php">'; $query_i = "SELECT * FROM `seats` WHERE SrideID='3' ORDER BY seatID"; $result = mysql_query($query_i); echo"<select name='seatNumber'><option value=''>Select Seat</option>"; while( $row = mysql_fetch_array($result) ) { echo '<option>'.$row['seatNumber'].'</option>'; } echo '</select>'; mysql_free_result( $result ); echo "<input type='submit' value='Submit'>"; echo "</form>"; ?> </tr> </table> code for dd-check.php <?php $result=$_POST['seatNumber']; echo " <p> the seat choosen is : $result <p> "; //TEST PULL ALL QUERY $testPullAll = mysql_query("SELECT * FROM ride"); echo "<table border='1'> <tr> <th>Name</th> <th>Duration</th> <th>Price</th> </tr>"; while($row = mysql_fetch_array($testPullAll)) { echo "<tr>"; echo "<td>" . $row['name'] . "</td>"; echo "<td>" . $row['time'] . "</td>"; echo "<td>" . $row['price'] . "</td>"; echo "</tr>"; } echo "</table>"; ?> the end result will be a query in dd-check.php that says something like if [hidden SrideID = 1] echo 'the name, time and price of the ride' if [hidden SrideID =2 echo name, time and price if number 3 then echo the rest of the row. any help would be great, i've been using trial and error so far (since yesterday afternoon) with no success =( ps sorry for the long post, i wanted to try and explain everything the best i can. =) Similar TutorialsIs there anything wrong (or insecure) with using hidden form fields? I am working on a page where the user can choose one of 4 different subscription options. The approach I was going to use is to have a separate form for each plan, and when the user chooses one, submit a hidden form value so my script knows which subscription plan to grab out of the database. Thoughts? Hi I've got a form that users enter an identification pin in for their first question as well as clicking a radio button(Q1.php) . They click submit and the answers are put through a couple of multiplication function and then placed in a table (alongside the pin) via Q2.php. On the Q2 page I want to place the 2nd question that the user needs to answer. Rather than making them retype the pin I was hoping to process it as a hidden field - being picked up from their Q1.php entry. This 'hidden pin echo ' process will continue throughout Q2- Q10 pages. However - I can't seem to include any php in the page after the Q1 form has been processed. Have I got my php tags messed up? <?php $con = mysql_connect("localhost","ca","d"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("candango", $con); $q1 =$_POST['q1']; $pin =$_POST['pin']; $ans1 = $q1 * 3; $ans2 = $q1 * 5; $ans3 = $q1 * 2; $enter_sql= "INSERT INTO aapcm2 (pin,ans1, ans2, ans3) VALUES ('$pin','$ans1','$ans2','$ans3')"; $enter_query =mysql_query($enter_sql) or die (mysql_error()); ?> <body> <p>Thank you - You have successfully entered your answers.</p><form action="q2.php" method="post"> <p>What do you think the answer to this one is? <input type="radio" name="q2" id="q2" value="-3"> <label for="q2"></label> <input name="Hidden" type="hidden" id="Hidden" value="$pin" /></p> <p> <input type="submit" name="button" id="button" value="Submit"> </p> </form> <p></p> </body> Is there any way using only PHP? The problem is that I don't have access to the form processor, or to MySQL. Hey guys,
Thank you in advance... here is my situation, I have a form with three (3) fields in it, the 'student name' is unlimited textfield with an "add more" button to it and I have two select fields ('number of shirts' and 'trophies') that depend on the number of entries for 'student name'...
I want to create the select fields based on this math, for as many 'student name' entries:
1- i want to have the select form for 'number of shirts' to be 0 up to that number... so if there are 6 'student name' entries, the select options will be 0,1,3,4,5,6,7
2- I want to have the select form for 'trophies' to be 5 'student name' entries to 1 'trophies', for example if there are 6 'student name' entries, the select options will be 0,1... if there are 13 entries, options will be 0,1,2... So if there are less than 5 'student name' entries, the select field will not show (hidden)
of course if there are no 'student name' entries, these two fields won't show up (hidden)
let me know if that make sense and ANY help or directions will be GREATLY APPRECIATED.
Thanks guys!
I 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
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) 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 /////
Code: [Select] <td > <form action="" method="post"> <b> <?php echo $likes_count; ?> </b> <b> | </b> <b> <?php echo $dislikes_count; ?> </b> <button class="LikeButton" type="submit" name="likes" value="+1">Likes</button> <button class="DislikeButton" type="submit" name="dislikes" value="-1">Dislikes</button> <input type="hidden" name="hidden_con_id" value="<?php echo $con_id; ?>" /> </form> <?php echo $con_id; ?> </td> I have a table with a voting system and the problem I am experiencing is that the hidden id, which I call the hidden contribution id is not set thus the numeric array changes. I know that it is not best practice to use a numeric array for this, though I learned that afterwards. This is the array: Code: [Select] // POST BUTTONS inside the table if (isset($_POST['likes'])) { $likes = $_POST['likes']; } if (isset($_POST['dislikes'])) { $dislikes = $_POST['dislikes']; } if (isset($_POST['likes']) || isset($_POST['dislikes'])) { $hidden_con_id = $_POST['hidden_con_id']; } // $array = array ($likes, $dislikes, $con_id, $user_id); if (isset($likes)) { $array[] = $likes; } if (isset($dislikes)) { $array[] = $dislikes; } if (isset($hidden_con_id)) { $array[] = $hidden_con_id; } if (isset($dnuser_id)) { $array[] = $dnuser_id; } As said the problem I have is that the hidden_con_id variable is not set and that consequently array[2] becomes the user_id variable. Any suggestions why it is not becoming set? hi , when I use hidden input tag and save there information to send with the form , is that secure? I mean , if the user or hacker or anything can change the value of the hidden tag? thanks , Mor. how would i retrieve a hidden field from this form Code: [Select] $checkoutBtn .= '<form action="checkout.php" method="POST" id="checkform"> <input type="hidden" name="item_name_' . $x . '" value="' . $eventname . '"> <input type="hidden" name="amount_' . $x . '" value="' . $studentsprice . '"> <input type="hidden" name="quantity_' . $x . '" value="' . $each_item['quantity'] . '">;[ <input type="hidden" name="custom" value="' . $product_id_array . '"> <input type="submit" class="checkoutbtn" name="checkoutbtn" value="CHECKOUT" /> </form>'; /code] I have a while loop that fetches data from the database and prints it out organized in a table. Now I want to implement a voting functionality, the problem I'm encountering is, once I've printed out a list of tables one table after other with the while loop I need to figure out a way to tell the query TO WHICH of those tables to ADD or SUBTRACT the vote. I thought of implement a hidden id field into the while loop of contributions, the id field would be fetched from the auto_increment field in the contribution table in the MySQL database. Since the id field is unique there can be no misunderstandings to which table to add the vote. My question how can I do exactly that? How can I add a hidden id field to the while loop with the table WHICH I then can pass along to the voting script. Here's the while loop: while ($row = mysqli_fetch_array($data)) { echo "<table padding='0' margin='0' class='knuffixTable'>"; echo "<tr><td width='65px' height='64px' class='avatar_bg' rowspan='2' colpan='2'><img src='$avatar_path' alt='avatar' /></td><td class='knuffix_username'><strong>" . $user_name; echo "</strong><br />" . $row['category'] . " | " . date('M d, Y', strtotime($row['contributed_date'])) . "</td></tr><tr><td>"; echo "<form action='' method='post'> <input type='submit' name='plusVote' value='Y' /> <input type='submit' name='minusVote' value='N' /> </form></td><td class='votes'>Y[ - ] | N[ - ]</td></tr>"; echo "<tr><td class='knuffix_name' colspan='3'><strong>" . htmlentities($row['name']) . "</strong><br /></td></tr>"; echo "<tr><td colspan='2' class='knuffix_contribution'><pre>" . $row['contribution'] . "</pre><br /></td></tr>"; echo "</table>"; } If I simply add $row['con_id'] it's obviously going to be echo'd out, what is the right practice to have it hidden? This is just a snippet of my code. The php script itself is called test.php and so it calls itself once the form is submitted. I keep having problems retrieving the data back correctly; I am testing in retrieving the data on the same script from the same page before retrieving the POST data from another webpage. If $decimalSum is a variable that assigns a value that is hard-coded then I would be able to retrieve the same value. However, if the value is computed I can not retrieve it unless I click on the "Find" button twice. This is such a strange behavior and I don't know why. Does anyone have any suggested solutions to this? <!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" xml:lang="en" lang="en"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title></title> </head> <body> <?php echo '<form enctype="multipart/form-data" method="post" action="test.php">'; echo ' <label for="_firstName">First name : </label>'; echo ' <input type="text" id="_firstName" name ="_firstName">'; echo ' <input type="checkbox" value="1" name="_firstNameChecked"/>'; echo ' <label for="_middleName">Middle name : </label>'; echo ' <input type="text" id="_middleName" name ="_middleName">'; echo ' <input type="checkbox" value="1" name="_middleNameChecked" />'; echo ' <label for="_lastName">Last name : </label>'; echo ' <input type="text" id="_lastName" name ="_lastName">'; echo ' <input type="checkbox" value="1" name="_lastNameChecked" />'; echo '<br />'; echo '<input type="submit" name="Find" value="Find" />'; $firstNameChecked = intval(($_POST['_firstNameChecked'])); $middleNameChecked = intval(($_POST['_middleNameChecked'])); $lastNameChecked = intval(($_POST['_lastNameChecked'])); $decimalSum = (int)((2*2*2)*$firstNameChecked + (2*2)*$middleNameChecked + (2*1)*$lastNameChecked); //$decimalSum = 28; echo '<br />'; echo '$decimalSum = ' . $decimalSum . '<br />'; echo '<input type="hidden" name="_decimalSum" value = "' . $decimalSum . '" />'; $decimalSum2 = ($_POST['_decimalSum']); echo '$decimalSum2 = ' . $decimalSum2 . '<br />'; echo '</form>'; echo '</body>'; echo '</html>'; ?> I have created a form which allows people to add data to a database. However, I want to replicate the form across several pages and give each page a category. For example, if I have a website based on sport. I have a page which enters information into the baseball category, a page which enters information into the ice hockey category, a page which enters information into the soccer category etc. Can someone advise how I would do this? Hello everyone, I have a question. I'm trying to figure out a way to make it so that when a user visits a page for a certain file, they are given the file to download but can't see the link to that file. Example: There is a file on http://serverB.com/file.rar A person goes to http://serverA.com/?file=file.rar They can download http://serverB.com/file.rar without seeing the link to that file. Is there any way to do this? Thanks. I have a menu that i want to be added to every page of my coding using a hidden variable, but i cannot get it to work. I using this with a few if conditions. the index page should navigate every page. can anyone help? I have attached the files to illustrate the coding i have done so far. [attachment deleted by admin] Hey all, i have a little problem. I have a form, and when submitting, i need to use the ID that i put in the $_GET array earlier. When you're entering my site, and clicks a certain link, the $_GET will store show=1, and when you're submitting the form on the page, the value of 'show' is needed on the redirect page.. I thought that it would be smart to transfer it with a hidden textfield, but somehow it fails? <form id="comment-form" action="comment_profile.php" method="POST"> <fieldset> <div class="field"> <textarea name="comment-field" rows="2" cols="66"></textarea> <input type="hidden" name="pofileid-field" value="<?php echo $_GET['show']; ?>" /> // <--- Here is the hidden field. <input type="submit" name="submit-field" class="submit" value=""/><i> (HTML tags kan bruges)</i> </div> </fieldset> </form> Above you can see a piece of my code, and i marked the hidden field aswell. What i do not understand is, why isn't the hidden fields value return the value of show when i check in comment_profile.php, as that is where it returns? Thanks in advance -Niixie Hello, My question is how to send values over and over without using forms or even hidden forums. I have tried to use sessions but it didn't work out for me. Any suggestions! Hi Guys, Appreciate any help here please... Basically I want to hide some inputs on my registration page and populate them will default values. Everything works okay, apart from the input below: Code: [Select] <input type="hidden" name="seek_location" value="Anywhere"> When I use the above, the MYSQL database field for seek_location is left empty... I've noticed that the code below pulls the location into a dropdown list. You know the kind of thing, Australia, UK, USA etc etc: Code: [Select] $seek_location = $wcr[$seek_location]; Code: [Select] <? $p=0;asort($wcr);reset($wcr); while (list ($p, $val) = each ($wcr)) { echo '<option value="'.$p.'">'.$val; } ?> So, all I want to do is have: Code: [Select] <input type="hidden" name="seek_location" value="Anywhere"> Populate seek_location with Anywhere Any help appreciated Thanks Rob Here's my PHP form code: Code: [Select] <form method='POST' action="<?php basename($_SERVER['PHP_SELF']);?>" onSubmit="return stripInputBoxes(1)"> <?php echo "<div id = 'purchaseOrderRow1' style = 'display:none;border: 1px solid black;'>"; echo $integrityBuildingProductsPDF; echo '<input type = "hidden" id = "pdf1" value = "" name = "pdf1" />'; echo '<button name = "savePurchaseOrder" type = "submit">Save Purchase Order</button>'; echo "</div>"; ?> </form> Here's my Javascript: Code: [Select] function stripInputBoxes(pdfNumber) { if (!document.getElementsByTagName) return; for (i = 0; i < document.getElementById('tableBody').getElementsByTagName('tr').length - 3; i++) { document.getElementById(i).parentNode.innerHTML = document.getElementById(i).value; } document.getElementById("pdf" + pdfNumber).value = "ljkasdkljf"; return true; } Will this set the value of the hidden variable? I'm trying to echo the value of $_POST['pdf1'] but it comes out to nothing when the page reloads. How do I do this? What I'm trying to do here is display a purchase order on the screen and the user can edit the quantities in the purchase order. They click SAVE to save the new updated quantities and it auto generates a PDF that's saved on the server with those new quantities. The problem is I need to run the stripInputBoxes function to take the <input> tags out of the HTML code before creating the PDF because the PDF generator doesn't recognize the tags. |
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