I have a table of items and I would like to place an order button at the end of the table. The button would then take user to "order.php". Where based on the "Picture ID" they would then order the selected item. I've had a look at similar questions online however haven't been able to get it working with what I currently have.
if ($result->num_rows > 0)
{
echo "<table>\n";
while($row = $result->fetch_assoc())
{
echo "<tr>\n";
echo "<td>". $row["Picture ID"]."</td>\n";
echo "<td>". $row["name"]."</td>\n";
echo "<td>". $row["doc"]."</td>\n";
echo "<td>". $row["width"]."</td>\n";
echo "<td>". $row["height"]."</td>\n";
echo "<td>". $row["price"]."</td>\n";
echo "<td>". $row["description"]."</td>\n";
echo "<td> <form action='order.php' method='post'> <input type='button' name='id' value= ></form>"
echo "</tr>\n";
}
echo "</table>\n";
}
You have to pass the picture id as hidden input text and use the "submit" type for the button:
echo "<td> <form action='order.php' method='post'> <input type='hidden' name='id' value='" . $row["Picture ID"] . "'/> <input type='submit' name='submit' value='Buy'/></form>"
In order.php you can get the Picture ID from $_POST:
if (!empty($_POST['submit'])) { // submit button was pressed
echo 'Picture ID: ' . $_POST['id'];
}
Make sure to escape the $_POST['id'] properly when using it in SQL queries.
Related
I have search google, here and w3schools but this answer i cant find anyway. Not even in the "questions that may already have your answer".
I am trying to learn a bit more about AJAX and i have come to a hold at this guide W3schools AJAX database
All the guide i can get to work but when i try to suit it to my needs it goes wrong. What i want is that when i get to "getuser.php" i want to be able to update db in this file. If possible without me leaving this page with the result i have found. I choose from a dropdown table before this site. The php files which is supposed to update the db works (tried them on a normal page, and all is good). My current workaround is to add a button which opens a second window to update the info.
When i get to this point:
<?php
$q = intval($_GET['q']);
include 'db.php';
$con = new mysqli($servername, $username, $password, $dbname);
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"webhelp");
$sql="SELECT * FROM advisors WHERE id = '".$q."'";
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_assoc($result)) {
echo "<table><tr><td>Phone</td><td>" . $row['phone'] . "</td>
<td><form action='addphone.php' method='post'>
<input type='hidden' name='id' value='".$q."'>
<td><input type='text' name='phone'></td>
<td><input type='submit' value='Update'></td>
</form></td></tr></table>";
echo "<tr><td>LoB</td><td>" . $row['lob'] . "</td>
<td><form action='addlob.php' method='post'>
<input type='hidden' name='id' value='".$q."'>
<td><select name='lob'>
<option value='". $row['lob'] ."'>" . $row['lob'] . "</option>".
$sql = "SELECT * FROM lob";
$result = $con->query($sql);
while($row = $result->fetch_assoc()) {
echo "<option value='" . $row["lob"] . "'>" . $row["lob"] . "</option>"; }
"</select></td>
<td><input type='submit' value='Update'></td>
</form>
</tr>";
echo "<tr><td>Country</td><td>" . $row['country'] . "</td>
<td><form action='addcountry.php' method='post'>
<input type='hidden' name='id' value='".$q."'>
<td><select name='country'>
<option value='". $row['country'] ."'>" . $row['country'] . "</option>".
$sql = "SELECT * FROM country";
$result = $con->query($sql);
while($row = $result->fetch_assoc()) {
echo "<option value='" . $row["country"] . "'>" . $row2["country"] . "</option>" ; }
"</select></td>
<td><input type='submit' value='Update'></td>
</form>
</tr>";
}
echo "</table>";
mysqli_close($con);
?>
The "Update" buttons doesnt work. It doesnt matter where i place the files (same folder, different folder) ect. However if i add a button with a link outside of the then that button work. But as soon as it is inside a table PLUS also method="post" is in the form it mess it up.
What am i doing wrong?
Alternatively is it possible to make a button here which carries the $id over to a small popup window? (I can open it in a new window but I can't choose how big the window should be)
You want to do this for all your 3 forms. I give you example for first one.
Form
echo "<form class='addPhoneForm' action='addphone.php' method='post'>
<table>
<tr>
<td>Phone</td><td>" . $row['phone'] . "</td>
<td><input class='phoneID' type='hidden' name='id' value='".$q."'></td>
<td><input class='phoneNumber' type='text' name='phone'></td>
<td><input class='submitme' type='submit' value='Update'></td>
</td>
</tr>
</table>
</form>";
AJAX
$(document).ready(function(){
$(".submitme").click(function(){
//collect variables from input
var phoneID = $(".phoneID").val();
var phoneNumber = $(".phoneNumber").val();
// store in a string
var dataAddPhone = 'phoneID='+ phoneID + '&phoneNumber='+ phoneNumber;
// send to database
$.ajax({
type: "POST",
url: "addphone.php",
data: dataAddPhone,
cache: true,
//if success
success: function(response){
//display message
$(".displayMessage").html(response);
//and reset form
$(".addPhoneForm").trigger("reset");
}
});
return false;
});
});
I'm trying to make a shopping page. I've kept checkboxes for selection and a number type field for quantity. As checkboxes are array I took the quantity field also an array... I know that we can check the checkbox state with checked method, now what I'm trying to achieve is to fetch corresponding quantity value and store it in a session varaible.
<?php
session_start();
if(isset($_POST['sub'])){
$_SESSION['cbb']=$_POST['cb'];
if(isset($_POST['qnty'])){
$_SESSION['qtty']=$_POST['qnty'];
header("location:userlogin.php");
}
}
?>
<html>
<head>
</head>
<body topmargin="20">
<font color='#7FFF00'>
<h1 align='center'>ONLINE SHOPPING</h1>
</font>
<form name="onshop" align="right" action="<?=$_SERVER["PHP_SELF"]?>"
method="post" onsubmit="return validate()">
<input type="submit" value="Add to cart" name="sub" style="background-
color:#7FFF00" />
<input type="reset" value="Reset" style="background-color:#7FFF00"/>
<?php
$db=mysqli_connect("localhost","root","","onlineshop");
if(!$db){
echo "Connection failure.";
}
$sql="SELECT * FROM stock";
$result=mysqli_query($db,$sql)or die("Invalid query ".mysqli_error($db));
echo "<TABLE width=50% height=70% align='center'>";
echo "<TR><TH>ITEM NAME</TH><TH>PRICE</TH><TH>IMAGE</TH><TH>Do you like?
</TH><TH>QUANTITY</TH></TR>";
while($myrow=mysqli_fetch_array($result))
{
echo "<TR><TD align='center'>";
echo $myrow["item"];
echo "</TD>";
echo "<TD align='center'>";
echo $myrow["price"];
echo "</TD>";
echo "<TD align='center'>";
$image=$myrow["image"];
echo '<img height="100" width="100" src="data:image/jpeg;base64,'.base64_encode( $image ).'"/>';
echo "</TD>";
echo "<TD align='center'>";
$itm_id=$myrow["id"];
echo "<input type='checkbox' name='cb[]' value='$itm_id'>";
echo "</TD>";
echo "<TD align='center'>";
echo "<input type='number' name='qnty[]' value='1' max='50'
min='1'>";
echo "</TD>";
echo "</TR>";
}
echo "</TABLE>";
mysqli_close($db);
?>
</form>
<script type="text/JavaScript">
function validate(){
var count=0;
var i;
var cbelements = document.getElementsByName("cb[]");
if(count==0){
for(i=0;i<cbelements.length;i++){
if(cbelements[i].checked){
count++;
}
}
if(count<1){
alert("Select atleast 1 item.");
return false;
}
}
}
</script>
</body>
</html>
Can the onchange event be useful in anyway? or the associative array ?
I have a PHP form as below. Once user selects a particular row/rows I need to send the corresponding data to my PHP backend script. Here the externalID is unique. I face two issues in my code
Currently I get the entire form data to my PHP backend script,I need to data corresponding to only the rows I selected.
I have two buttons in the page and how can I call different php scripts on click of the button. Currently the click of "send MTDATA" calls the php script in form action.
src code:
<form target="iframe_b" action="/php_src/first.php" method="POST"
echo "sending data">
<fieldset>
<br><br>
<input type="button" id="activate_device" name="activatedevice" value="Activate_device">
<input type="submit" value="SEND MTDATA">
<table border="1">
<tr>
<th><input type="checkbox" id="selectall"/></th>
<th>External ID</th>
<th>Status</th>
</tr>
<?php
$dbhost = 'localhost:3036';
$dbuser = 'root';
$dbpass = 'xxxx';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if (!$conn) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db("ApplicationServer") or die(mysql_error());
// Get all the data from the "example" table
$result = mysql_query("SELECT EXTERNAL_ID,CONNECTION_STATUS FROM DEVICE_DETAILS") or die(mysql_error());
while ($row = mysql_fetch_array($result)) {
$extID = $row['EXTERNAL_ID'];
$conStatus = $row['CONNECTION_STATUS'];
echo "</tr><tr>";
echo "</td><td>";
echo "<input type=\"checkbox\" class=\"case\" name=\"checkBox".$extID."\" />";
echo "</td><td>";
echo "<input type=\"text\" class=\"classextID\" value=\"$extID\" name=\"textExID".$extID."\" />";
echo "</td><td>";
echo "<input type=\"text\" class=\"classConnection\" value=\"$conStatus\" name=\"textcon".$extID."\" />";
}
?>
</table>
</fieldset>
</form>
output:
array(9) { ["checkBox123456#mydomain_com"]=> string(2) "on" ["textExID123456#mydomain_com"]=> string(19) "123456#mydomain.com" ["textcon123456#mydomain_com"]=> string(16) "request accepted" ["checkBox1234#mydomain_com"]=> string(2) "on" ["textExID1234#mydomain_com"]=> string(17) "1234#mydomain.com" ["textcon1234#mydomain_com"]=> string(16) "request accepted" ["checkBox53278#mydomain_com"]=> string(2) "on" ["textExID53278#mydomain_com"]=> string(18) "53278#mydomain.com" ["textcon53278#mydomain_com"]=> string(16) "request accepted" }
All html inputs - except unchecked checkboxes - are sent to the server so the only way to avoid that is to use javascript to remove the unselected rows from the form completely before the form is sent.
However, if the total amount of data is not the problem but you just want to be able to easily select the checked rows, you should use arrays in your html:
echo "<input type=\"checkbox\" class=\"case\" name=\"checkBox[".$extID."]\" />";
^ array ^
echo "</td><td>";
echo "<input type=\"text\" class=\"classextID\" value=\"$extID\" name=\"textExID[".$extID."]\" />";
echo "</td><td>";
echo "<input type=\"text\" class=\"classConnection\" value=\"$conStatus\" name=\"textcon[".$extID.]"\" />";
Now your $_POST['checkBox'], etc. variables will be arrays in php as well so you can easily loop over them as the key is your $extID value. And as I mentioned before, only the checked checkboxes are sent to the server:
// Loop over the sent-in / selected checkboxes
foreach (array_keys($_POST['checkBox']) as $key) {
var_dump($_POST['textExID'][$key]);
// etc.
}
Also note that your html is probably not valid as you are not closing the last row but I don't think that would be causing any problems with the form.
One way is to collect the input in an array.
<?php
echo "</tr><tr>";
echo "</td><td>";
echo "<input type='checkbox' class='case' name='$extID[ checkBox ]' />";
echo "</td><td>";
echo "<input type='text' class='classextID' value='$extID' name='$extID[ textExID ]' />";
echo "</td><td>";
echo "<input type='text' class='classConnection' value='$conStatus' name='$extID[ textcon ]' />";
?>
Then if you check the $_POST you will find an array within an array, loop the first array and collect the details for each $extID.
As for the two buttons, you could name the submit button and then check to see which has been pressed.
<input type="button" id="activate_device" name="activatedevice" value="Activate_device">
<input type="submit" value="SEND MTDATA" name="submit_button">
<?php
if( isset( $_POST[ 'submit_button' ] ) )
{
// Process submit
}
elseif( isset( $_POST[ 'activatedevice' ] ) )
{
// Process device
}
This is my first post here so I apologize in advance if the formatting is wrong.
I am working on a form that pulls data from MySQL using a loop and outputs it to HTML page. The user then has the option to approve or deny the entries, and based on user selection validation should be required or optional. My current code will validate correctly, but only for the first row being outputted from the loop. I am trying to validate all rows. I have tried using a while loop and foreach statement with no success. Any help would be greatly appreciated!
My Loop & Form:
//connect to the database
$db=mysqli_connect('localhost','root','') or die ('I cannot connect to the database because: ' . mysql_error());
//-select the database to use
$mydb=mysqli_select_db($db,"my_db") or die(mysql_error());
//-query the database table
$sql="SELECT id, date, client_name, client_number, date_completed, status FROM clients WHERE client_name LIKE '%" . $search ."%' OR status LIKE '%" . $search ."%' ";
//-run the query against the mysql query function
$result=mysqli_query($db, $sql);
//-count results
$rows=mysqli_num_rows($result);
if($rows=mysqli_num_rows($result)) {
echo "<h2 style='margin-left: 10em;margin-bottom: -0.4em;'><br><br><br>" . $rows . " result(s) found for " . $search . "</h2><br />";
}elseif($rows=mysqli_num_rows($result) == 0) {
echo "<h2 style='margin-left: 10em;margin-bottom: -0.4em;'><br><br><br>0 result(s) found for " . $search . "</h2><br />";
}
//-create while loop and loop through result set
while($row=mysqli_fetch_assoc($result)){
$id=$row['id'];
$date=$row['date'];
$client_name=$row['client_name'];
$client_number=$row['client_number'];
$date_completed=$row['date_completed'];
$status=$row['status'];
echo "<form method='post' enctype='multipart/form-data' action=''>";
echo "<table border='0'>";
echo "<tr>\n";
echo "<td>Timestamp</td><td>Client Name</td><td>Client Number</td>Status</td><td>Date Completed & Returned</td><td>Upload Zip Files</td>\n";
echo "</tr>";
echo "<tr>\n";
echo "<td readonly class='date'>$date</td>\n";
echo "<td><input readonly type='text' id='client_name' name='client_name' value='$client_name'></td>\n";
echo "<td><input readonly type='text' id='client_number' name='client_number' value='$client_number'></td>\n";
echo "<td><select id='status' name='status' aria-invalid='false'>
<option value=''>Select an option</option>
<option value='Denied'>Denied</option>
<option value='Approved'>Approved</option>
</select></td>\n";
echo "<td><input type='date' id='date_completed' name='date_completed_returned' value='$date_completed'></td>\n";
echo "<td><input type='file' id='upload' name='upload'></td>";
echo "<td class='submit'><input type='hidden' id='hidden' name='hidden' value='$client_name'><input type='submit' id='save' name='save' value='Save'></td>\n";
echo "</tr>";
echo "</table>";
echo "</form>";
}
My jQuery code:
I am trying to make date_completed and upload fields required if user selects "Approved" under status field, and optional if he selects "Denied".
<script>
$('#status').on('change', function() {
if ( this.value == 'Approved')
$("#date_completed").prop('required',true)
}).trigger("change"); // notice this line
$('#status').on('change', function() {
if ( this.value == 'Approved')
$("#upload").prop('required',true)
}).trigger("change"); // notice this line
</script>
Thanks to KevinB I was able to find a solution to my problem.
I added class names for the input fields (status, date_completed, upload) I was trying to validate:
Updated code below:
echo "<form method='post' enctype='multipart/form-data' action=''>";
echo "<table border='0'>";
echo "<tr>\n";
echo "<td>Timestamp</td><td>Client Name</td><td>Client Number</td>Status</td><td>Date Completed & Returned</td><td>Upload Zip Files</td>\n";
echo "</tr>";
echo "<tr>\n";
echo "<td readonly class='date'>$date</td>\n";
echo "<td><input readonly type='text' id='client_name' name='client_name' value='$client_name'></td>\n";
echo "<td><input readonly type='text' id='client_number' name='client_number' value='$client_number'></td>\n";
echo "<td><select id='status' name='status' class='status' aria-invalid='false'>
<option value=''>Select an option</option>
<option value='Denied'>Denied</option>
<option value='Approved'>Approved</option>
</select></td>\n";
echo "<td><input type='date' id='date_completed' name='date_completed' class='date_completed' value='$date_completed'></td>\n";
echo "<td><input type='file' id='upload' name='upload' class='upload'></td>";
echo "<td class='submit'><input type='hidden' id='hidden' name='hidden' value='$client_name'><input type='submit' id='save' name='save' value='Save'></td>\n";
echo "</tr>";
echo "</table>";
echo "</form>";
<script>
$('.status').on('change', function() {
if ( this.value == 'Approved')
$('.date_completed').prop('required',true)
}).trigger("change"); // notice this line
$('.status').on('change', function() {
if ( this.value == 'Approved')
$('.upload').prop('required',true)
}).trigger("change"); // notice this line
</script>
I have a problem. I need to get the value from a select tag then use it in php for my sql. Here is my code
<div class="form-group">
<label> ROOMS </label>
<?php
echo "<select value= 'TRoom1' id ='TRoom1' class='form control'>";
echo "<option>Select Room Type</option>";
while ($row1 = mysql_fetch_array($result2))
{
echo "<option>" . $row1['Room_type'] . "</option>";
}
echo "</select>";
?>
this is for the sql command
<div class="modal-body">
<div class="container">
<?php
$selectedValue = $_POST['TRoom1'];
$sql = "SELECT RoomNumber FROM rooms Where Room_type = '$selectedValue' ";
$result = mysql_query($sql);
echo "<select value= 'RoomNo' id ='RoomID' class='form-control'>";
echo "<option>Select Room Number</option>";
while ($row = mysql_fetch_array($result))
{
echo "<option>" . $row['RoomNumber'] . "</option>";
}
echo "</select>";
?>
TIA! :))
THis is the code ofor room type with its corresponding room number
<div class="form-group">
<label for="exampleInputEmail1"> ROOMS </label>
<?php
echo "<select value= 'TRoom1' name ='TRoom1' id ='TRoom1' class='form-control'>";
echo "<option>Select Room Type</option>";
while ($row1 = mysql_fetch_array($result2))
{
echo "<option>" . $row1['Room_type'] . "</option>";
}
echo "</select>";
?>
</div>
<div class="form-group">
<?php
$select_value=$_POST['selectedValue'];
$sql = "SELECT RoomNumber FROM rooms Where Room_type = '$select_value' ";
$result = mysql_query($sql);
echo "<select value= 'RoomNo' id ='RoomID' class='form-control'>";
echo "<option>Select Room Number</option>";
while ($row = mysql_fetch_array($result))
{
echo "<option>" . $row['RoomNumber'] . "</option>";
}
echo "</select>";
?>
</div>
you need to use name attribute for your select tag if u want to fetch the value in the php part and in the option u have to pass the value attibute.that value u will get in the php part.
<html>
<head></head>
<body>
<form action="a.php" method="post">
<select name="selectname" id="someid" >
<?php
while ($row1 = mysql_fetch_array($result2))
{
?>
<option value="<?php echo $varible ?>"> <?php echo $row1['Room_type']; ?></option>
<?php } ?>
</select>
<input type="submit" value="submit">
</form>
</body>
</html>
for php part:u can fetch value like this:
filename=a.php
<?php
$select_value=$_REQUEST['selectname'];
$sql1 = "SELECT RoomNumber FROM rooms Where Room_type = '$select_Value' ";
$sql=mysql_result(mysql_query($sql1),0);
?>
Please google your doubts before posting here. There are plenty of example available. mysql_query is deprecated use mysqli_ function
<div class="form-group">
<label> ROOMS </label>
<?php
echo "<select id ='TRoom1' name ='TRoom1' class='form control'>";
echo "<option>Select Room Type</option>";
while ($row1 = mysql_fetch_array($result2))
{
echo "<option value=".$row1['Room_type'].">" . $row1['Room_type'] . "</option>";
}
echo "</select>";
?>
If you are submitting your form as post you would get values as
$sql = "SELECT Room_type, Rate, RoomNumber FROM rooms Where Room_type ='".$_POST['TRoom1']."' ";
Try like
var Sel_val = document.getElementById('TRoom1').value;
Sel_val will be the selected value of that Dropdown.Better you use ajax in your case.If it is on the same page then you use Form submit method.
For the ajax first you need the target url and the value which you want to send..So try like
$.ajax({
url : Url of the page at which the sql command will be there,
type : 'POST',
data : { Sel_val : Sel_val }
});
Then at your target file get the Sel_val via POST method.
I think you used the self action and try this below code
if($_POST){
$selectedValue = $_POST['TRoom1'];
$sql = "SELECT RoomNumber FROM rooms Where Room_type = '$selectedValue' ";
$result = mysql_query($sql);
echo "<select value= 'RoomNo' id ='RoomID' class='form-control'>";
echo "<option>Select Room Number</option>";
while ($row = mysql_fetch_array($result))
{
echo "<option>" . $row['RoomNumber'] . "</option>";
}
echo "</select>";
}