I have a register form (for events) in which the number of fields input are not same. It depends upon the events. Different events limit different numbers of members for registration. So, in my events TABLE I've a row members which store the number of members.
So,finally my registration form goes like this:
if ($row['members']>1){
echo "Hello ".$sess_name.", you can add ".($row['members']-1)." more members to register for ".$row['name']."<br>";
echo "On successful registration of the event, all the team members will receive an email on their registered email ids<br>";
$mem=array();?>
<form action="" method="post">
<?php for ($i=1;$i<$row['members'];$i++){
echo '<label>Insert ID of Member '.$i.' :</label>';
echo '<input type="text" id="id" size="20" name="'.$mem[].'"><br>';
}
echo '<button type="submit" id="submit" value="Register" name="register1">Register</button>';?>
</form>
I'm storing the ID that are being input in an array($mem)
But I don't know how to pass it ($mem) in another page to store it in MySQL table.
Add an hidden input field
<form action="" method="post">
<?php for ($i=1;$i<$row['members'];$i++){
echo '<label>ID of Member '.$i.' :</label>';
echo '<input type="text" id="email" size="20" name="'.$mem[].'"><br>';
echo '<input type="hidden" id="userID" name="userID[]" value='.$i.'>';
}
echo '<button type="submit" id="submit" value="Register" name="register1">Register</button>';?>
</form>
Related
please correct me here.
i have created multi page form where i want to pass data from each pages to final pages and then submit those on email. first one is apply.php, there are many input fields, but i have listed some of those, here when some enters passport number in passport field, i want this to be passed in everypage of the form and print this at couple of places on each page. here getting some issues when passing some of these fields.
this is first page ( apply.php )
<?php
// Start the session
session_start();
?>
<form name="search_form" method="post" onSubmit="return chk();" action="apply2.php">
<input name="passportno" id="passportno" type="text" maxlength="20" placeholder="Enter Passport No." size="43" >
<input name="birthdate" type="date" class="textBoxDashed" size="43" id="birthdate" datepicker="true" datepicker_min="01/01/1900" datepicker_max="21/11/2017" maxlength="10" datepicker_format="DD/MM/YYYY" isdatepicker="true" value="">
<input name="button1" type="submit" value="Continue">
this is apply2.php . here there is some issues, i am not able to find, as you can see below codes, i am able to print date of birth but not able to print passport no ( input from form1 ). Please correct where i am wrong here.
<?php
session_start();
$msg="";
////include("connect.php");
if (isset($_POST['button1']))
{
extract($_POST);
$code=strtolower($_POST['captcha_code']);
$sess=strtolower($_SESSION["code"]);
if ($sess==$code)
{
$appid=time().rand();
$result=mysqli_query($con,"select *from registration where email='$email'");
if (mysqli_fetch_row($result)>0)
{
?>
<script>
alert("This email is already exist");
</script>
<?php
}
else
{
$query="insert into registration values('$appid','$passportno','$birthdate','$email')";
if (mysqli_query($con,$query))
$msg="Data saved Successfully.Please note down the Temporary Application ID $appid";
else
echo "not inserted".mysqli_error($con);
if (!isset($_SESSION["appid"]))
{
$_SESSION["appid"]=$appid;
}
}
}
else
{
?>
<?php
}
}
?>
<form name="OnlineForm" method="post" onsubmit="return OnSubmitForm();" action="apply3.php">
<input name="applid" id="applid" value="<?php echo $_SESSION["appid"];?>">
<input type="hidden" name="birthdate" value="<?php echo $birthdate;?>"><b><?php echo $birthdate;?>
<input name="passportno" type="text" class="textBoxDashed" id="passportno" value="" size="43" maxlength="14" value="<?php echo $passportno;?>">
input name="sc" type="submit" class="btn btn-primary" id="continue" value="Save and Continue" onclick="document.pressed=this.name">
Don't use extract. Also do some checking to see if the data is set. As for not getting the the data try $_POST['passportno'] and if you want to pull the values and put them back into the input boxes simply use <?php echo isset($_POST['passportno'])?$_POST['passportno']:'' ?> to return nothing if it is not defined.
Also you need to do add some protection to your inputs.
You can add protection by using $passportno = mysqli_real_escape_string($con, $passportno);
Hi I'm quite new to php and I'm currently making a webpage similar to the ones used by supermarkets for stock management control as part of an assignment. I have the following form where the cashier would enter the product Id and the quantity of the item being purchased.
The form will then call another php file named cashsale.php which will take these inputs and update the tables in my database so that levels of stock on shelves in supermarkets are up to date with the new amounts (i.e. older qty - qty entered) and management can be advised when reorder is needed.
As it is the form works well, however I was advised to edit it in a way that a cashier can enter multiple products and quantities before submitting (i.e. the form will sort of show itself again) and allow the user to edit or remove any items before actually submitting the values to cashsale.php to manipulate the tables. I seem to be at a loss as to how this can be done.
I wanted to create a new button named "Add" which would display the form again, i.e. allow the user to check in more items, but I am confused as to how this can be done and also as to how I will be able to update tables then since I would be having more then just 2 inputs.
Can anyone help me on this please? Thanks in advance. The following is my html form:
center form action="cashsale.php" method ="post"
Product ID: <input name= "id" type="int" > <br>
Quantity:<input name="qty" type="int">
<input type="button" name = "Add" onclick="add">
<input type="Submit" name ="Submit" value = "Submit">
form center
I was not allowed to use html tags for form and center so I removed the < >. The following is some of the modifications done in the cashsale.php file just to give a clearer example.
$result = mysql_query("SELECT * FROM shelfingdetails where prodId =' " .$id. " '");
if (!$result){
die (mysql_error());
}
while ($row =mysql_fetch_array($result))
{
$qtyOnShelf= $row ['QtyOnShelf'];
$max=$row['max'];
$newQtyShelf=$qtyOnShelf-$qty;
}
$update=mysql_query("UPDATE shelfingdetails SET QtyOnShelf ='". $newQtyShelf. "' where prodId = '". $id. "';");
I hope someone can help. Thanks!
You just have to pass an array. For this you have to generate the inputs with PHP or javascript (I'm gona use jQuery to keep the code nice and simpe).
If you use PHP:
// PHP
<?php
if(isset($_POST['submit']) && $_POST['submit']){
//save data
} elseif(isset($_POST['Add']) && $_POST['Add']) {
$max = (int)$_POST['max']
} else { $max = 1; }
?>
<form action="" method="post">
<?php
for($i = 0;$i < $max;$i++){
?>
Product ID: <input name="id[]" type="int" > <br>
Quantity: <input name="qty[]" type="int">
<?php
}
?>
<input type="hidden" name="max" value="<?= $i; ?>" />
<input type="submit" name="Add" />
<input type="submit" name="Submit" value="Submit" />
</form>
If yo use Javascript:
//HTML
<form action="" method="post" id="form">
<div id="add">
Product ID: <input name="id[]" type="int" > <br>
Quantity: <input name="qty[]" type="int">
</div>
<input type="hidden" name="max" value="<?= $i; ?>" />
<input type="button" name="Add" onclick="addRow();" />
<input type="submit" name="Submit" value="Submit" />
</form>
// jQuery
function addRow(){
$("#add").append("<br />Product ID: <input name='id[]' type='int' >" +
"<br />Quantity: <input name='qty[]' type='int'>");
}
you have to use id[] and qty[] to pass them as an array and with the add button generate as many of them as you need. Like so:
Product ID: <input name="id[]" type="int" > <br>
Quantity: <input name="qty[]" type="int"> <br>
Product ID: <input name="id[]" type="int" > <br>
Quantity: <input name="qty[]" type="int"> <br>
Product ID: <input name="id[]" type="int" > <br>
Quantity: <input name="qty[]" type="int">
<input type="button" name = "Add" onclick="add">
Then on the backand use for loop to save all the data.
$max = count($_POST['id']);
$id = $_POST['id'];
$newQtyShelf = $_POST['qty'];
for($i = 0;$i < $max;$i++){
$update=mysql_query("UPDATE shelfingdetails
SET QtyOnShelf ='". (int)$newQtyShelf[i]. "'
WHERE prodId = '". (int)$id[i]. "';");
}
I just wanted to show you the idea, please don't use this specific code, because you should use mysqli instead of mysql and also mysqli_escape_string to make sure that the user not submits incorrect data.
I want to edit a dynamically-generated form (meaning: I don't know how many rows will be generated). This content is generated within a while loop, and the HTML generated has creates buttons of input-type=submit, generating as many identically-named buttons as there are iterations in the loop.
Of the generated buttons, I want to know which submit button has been clicked, in order to provide the user the same form for which it has been clicked. Disregard the name of the database and password to connect it; the connectivity is fine.
Feel free to suggest any new method to achieve the desired functionality.
The code is as follows:
echo "you have reached your travel details page. your recent travelling details are as follows".'</br>';
$dbc=mysqli_connect('localhost','xyz','xyz','abc') or die("connection to DB failed");
$query="SELECT * FROM travel_details WHERE emailid='{$_SESSION['username']}' ORDER BY dep_date DESC";
$result=mysqli_query($dbc,$query) or die("error in querying the DB");
?>
<h1>Your travel details are:-</h1>
<form name="showtraveldet" METHOD="POST" action="edittraveldet.php">
<table border="1">
<tr>
<th>Starting point</th><th>Ending point</th><th>No of passengers</th><th>Expected fare</th><th>Departure date</th>
<th>Departure time</th><th>Arrival Date</th><th>Arrival Time</th><th>Car Model</th><th>Car number</th>
<th>Who is driving</th><th>Driver's license number</th>
</tr>
<?php
while ($row=mysqli_fetch_array($result))
{
$tid=$row['travel_id'];
echo "the value of tid is '{$tid}'";
echo'<tr><td>'.$row['start_point'].'</td><td>'.$row['end_point'].'</td><td>'.$row['no_of_pass'].'</td><td>'.
$row['exp_fare'].'</td><td>'.$row['dep_date'].'</td><td>'.$row['dep_time'].'</td><td>'.$row['arr_date'].'</td><td>'.$row['arr_time'].'
</td><td>'.$row['car_model'].'</td><td>'.$row['car_no'].'</td><td>'.$row['who_is_driving'].'</td><td>'.$row['driver_license_no'].'</td>
<td><input type="submit" name="edit" value="Edit"></td></tr><input type="hidden" name="travelid" value="'.$row['travel_id'].' ;?>">';
}
edittraveldet.php :-
$travelid=$_POST['travelid'];
echo "the travel id in the variable is $travelid and got the value from '{$_POST['travelid']}'";
$dbc=mysqli_connect('localhost','xyz','xyz','abc') or die("connection to DB failed");
$query="SELECT * FROM travel_details WHERE travel_id='{$travelid}'";
$result=mysqli_query($dbc,$query) or die("error in querying the DB");
mysqli_close($dbc);
$row=mysqli_fetch_array($result);
?>
<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>" onsubmit="return validatewheregoing()" name="wheregoing">
<h1> Enter your travelling details so that other travellers can join you</h1>
<fieldset>
<legend> Travelling details </legend>
Start Point: <input type="text" name="start" value="<?php echo $row['start_point']; ?>"/><br />
End point: <input type="text" name="end" value="<?php echo $row['end_point']; ?>"/><br />
Passengers allowed: <input type="number" name="noofpass" value="<?php echo $row['no_of_pass']; ?>"/><br />
Expected Fare per passengers in rupees:<input type="number" name="fare" value="<?php echo $row['exp_fare']; ?>"/><br />
Departure Date:<input type="date" name="depdate" value="<?php echo $row['dep_date']; ?>"/><br/>
Departure time:<input type="time" name="deptime" value="<?php echo $row['dep_time'] ;?>"/><br/>
Arrival Date:<input type="date" name="arrdate" value="<?php echo $row['arr_date']; ?>"/><br/>
Arrival time at destination:<input type="time" name="arrtime" value="<?php echo $row['arr_time']; ?>"/><br/>
Car Model and name:<input type="text" name="cardet" value="<?php echo $row['car_det']; ?>"/><br/> <!--make this as a dropdown box for better database matching-->
Car Number:<input type="text" name="carno" /><br/><input type="checkbox" name="taxi" value="check this box if pooling a taxi">
Is the car self driven or driven by driver?<input type="radio" name="drivedet" value="Selfdriven" checked=""/>Self Driven<input type="radio" name="drivedet" value="driverdriven" />Driver driven<br />
Driver's License number<input type="text" name="licence_no"/></br>
<input type="checkbox" name="taxi" value="check this box if pooling a taxi"></br>
<input type="hidden" name="travelid" value="<?php echo $travelid ;?>" />
<input type="submit" value="invite travellers" name="editwheregoing"/>
</fieldset>
</form>
If only you can change your code, I would suggest you put the form tag itself in the while loop, each having the same action pointing to the same url but submitting different information to the destination page. This way you don't have to worry about the button clicked
while($row=mysqli_fetch_array($result))
{
//<form action="sameactionurl.php" name="form_1">
//<input type="hidden" name="travelid" value="$row['travelid']" />
//</form>
}
Another solution, if you don't want to change your code is use JavaScript to set a common hidden field to the value of the current ID before submitting the form
Name your submit buttons in a standard fashion and append a 3-digit number to the end, "button_XXXX_###" where ### is the number and XXX was the original name of your button.
After submission, check your request parameters for all variables starting with name "button_XXXX", and split the actual name "button_XXXX_####" by '_' character, and the "###" suffix will reveal the number of the button pressed.
It might be easier to create a form for each row instead, though.
I have a page with several forms which are dynamically generated using PHP. I am validating them using the jQuery Validation plugin. The forms are all the same, but relate to different items so I have given all of the forms the same class so they can be validated by one function (each form also has a unique ID). But I'm having some problems:
I would like the error messages to appear by the correct form items, but if I'm just using the form's class, the validator won't know which form the item is from.
I have a hyperlink to submit the form (and a regular submit button in <noscript> tags), and would usually use jQuery to submit the form, but again, how will jQuery know which submit link I've clicked, and which form to submit?
The easiest thing I can think of is to pass the form ID to the validate some how. Is that possible?
The forms look like this:
<?php while($row= pg_fetch_row($groups)) { ?>
<p class="error" id="error-<?php echo $row[0] ?>"></p>
<form action="../scripts/php/groups-process.php" method="post" id="editgroup-<?php echo $row[0] ?>" class="editgroup">
<label for ="edit-<?php echo $row[0] ?>" >Edit group name:</label>
<input type="text" class="text" size="20" maxlength="30" name="edit" id="edit-<?php echo $row[0] ?>" value="<?php echo $row[1] ?>" />
<noscript><input type="submit" name="editgroup" value="Submit" /></noscript>
<div id="submitcontainer-<?php echo $row[0] ?>"></div>
</form>
<?php } ?>
I would normally validate the form like this:
$(document).ready(function(){
$("#editgroup").validate({
rules: {edit: {required: true, maxlength: 30}},
messages: {edit: {required: 'Please enter a group name', maxlength: 'Please enter a shorter group name'},
errorContainer: "p#error",
});
$("#submitcontainer").html('<a class="button" href="javascript:void();" id="submitlink" name="submit">Submit</a>');
$("#submitlink").click(function() {
$("#editgroup").submit();
});
});
give the same class to all the form than try this,
<form id="1" class="common" method="post" action="page.php">
<input type="text" class="common_input_class" size="20" maxlength="30" name="edit" id="whatever" value="whatever" />
<input type="submit" name="submit" class="submit_this_form" value="submit" />
</form>
<form id="2" class="common" method="post" action="page.php">
<input type="text" class="common_input_class" size="20" maxlength="30" name="edit" id="whatever" value="another value" />
<input type="submit" name="submit" class="submit_this_form" value="submit" />
</form>
<script type="text/javascript">
$(".common").submit(function(){
var form_id = $(this).attr('id');
var input_val = $(this).children('.common_input_class').val();
if (input_val == '')
{
alert("input field is required");
return false;
}
});
</script>
I ended up iterating through my result twice, so I create a form validator for each form dynamically, and then dynamically create the forms. This was the best way I could think of to give me the control I wanted, although obviously it's slower and produces more code - not an ideal solution but it will do for this situation.
I am now developing a travel agency website, in this site when the user reserves a trip he/she have to enter trip related details. This information is collected in a forma. The user also enters the number of people who are travelling.
My question is, how do I gather the same information for everyone who is travelling? Basically I need the form to be generated many times depending upon the number of people of same family ,, so I can capture all their data. How do I do this?
well there are my Code;plllz help me am so confused and tried lots of things to solve it;
thanks in advance
<form action = "insertpassenger.php" method = "POST">
<center>Enter all the information below</center>
<?php for ($i=0;$i<$pplno;$i++) : ?>
people<?php echo $i+1 ; ?>
<input type="text" name="cpr" size="9" value="<?php echo $cpr;?>" maxlength="9">CPR
<input type="text" name="pplno" size="30" maxlength="25">Number Of People
<input type="text" name="gcpr" size="9" maxlength="9">dad CPR
<input type="reset" value="clear" name="clear">
<input type="submit" value="join" name="join">
<?php endfor; ?>
</form>
Your code jumps in and out PHP rather a lot.
Just declare the item names as array entries, either with implicit or explicit numbering:
for ($i=0;$i<$pplno;$i++) : ?>
<input type="text" name="cpr[]" size="9" value="<?php echo $cpr[$i];?>" maxlength="9">CPR
<input type="text" name="pplno[]" size="30" maxlength="25">Number Of People
<input type="text" name="gcpr[]" size="9" maxlength="9">dad CPR
<input type="reset" value="clear[]" name="clear">
<input type="submit" value="join[]" name="join">
<?php endfor;
or...
for ($i=0;$i<$pplno;$i++) {
print "<input type=\"text\" name=\"cpr[$i]\" size=\"9\" value=\"$cpr[$i]\" maxlength="9">CPR";
....
}
I would personally use a session variable and count down how many times the form needs to be completed. Unfortunately that would cause to page to reload after each form entry, but this allows you to have as many amount of forms as your user requests, without a screen scrolling down a few pages to create all the forms on one page.
At the start of your code before you displaying anything to the browser :
<?php
session_start ();
?>
And where you receive your count for looping:
<?php
if (!isset($_SESSION['yourAppName']))
}
$_SESSION['yourAppName'] = $pplno;
} else {
$_SESSION['yourAppName']--;
}
if ($_SESSION['yourAppName'] > 0) {
?>
<form action=''>
<input type="text" name="cpr" size="9" value="<?php echo $cpr;?>" maxlength="9">CPR
<input type="text" name="pplno" size="30" maxlength="25">Number Of People
<input type="text" name="gcpr" size="9" maxlength="9">dad CPR
<input type="reset" value="clear" name="clear">
<input type="submit" value="join" name="join">
<input type="submit" value="Proceed">
</form>
<?php
} else {
// code when all forms are filled in
}
?>
remember to have your form return to the same page. This code is only to guide you, don't expect it to work without some editing. :)