updating mysql data using php with a drop down option - php

I am trying to update my products table using a product form but it is throwing an error for an undefined variable here is my code for the productform.php
<?php
$ProductID= $_GET['ProductID'];
//Connect and select a database
mysql_connect ("localhost", "root", "");
mysql_select_db("supplierdetails");
{
$result = mysql_query("SELECT * FROM products WHERE ProductID=$ProductID");
while($row = mysql_fetch_array($result))
{
$ProductID= $_GET['ProductID'];
$ProdDesc = $row['ProdDesc'];
$SupplierID = $row['SupplierID'];
$Location = $row['Location'];
$Cost = $row['Cost'];
$Status = $row['Status'];
$MRL = $row['MRL'];
$ProductID = $row['ProductID'];
}
?>
//form
<form action="Edit_Prod.php" method="post">
<input type="hidden" name="ProductID" value="<?php echo $ProductID; ?>"/>
<label>Product Description:
<span class="small">Please enter a description for the product </span>
</label>
<input type="text" name="ProdDesc" value="<?php echo $ProdDesc ;?>" />
<label>Supplier ID
<span class="small">Please enter the Supplier ID</span>
</label>
<input type="text" name="SupplierID" value="<?php echo $SupplierID ;?>" />
<label>Bay Location:
<span class="small">Please select the bay location for this product </span>
</label>
<select name="Location" value="<?php echo $Location ;?>" />
<option>A1</option>
<option>A2</option>
<option>A3</option>
<option>A4</option>
<option>A5</option>
<option>B1</option>
<option>B2</option>
<option>B3</option>
<option>B4</option>
<option>B5</option>
<option>C1</option>
<option>C2</option>
<option>C3</option>
<option>C4</option>
<option>C5</option>
<option>D1</option>
<option>D2</option>
<option>D3</option>
<option>D4</option>
<option>D5</option>
<option>E1</option>
<option>E2</option>
<option>E3</option>
<option>E4</option>
<option>E5</option>
</select>
<label>Product Cost:
<span class="small">Please enter a new cost for the product (per roll) </span>
</label>
<input type="text" name="Cost" value="<?php echo $Cost ;?>" />
<label>Status:
<span class="small">Please select a status for the product </span>
</label>
<select name="Status" value="<?php echo $Status ;?>" />
<option>Live</option>
<option>Mature</option>
<option>Obsolete</option>
</select>
<label>MRL
<span class="small">Please enter a minimum re-order level for this product </span>
</label>
<input type="text" name="MRL" value="<?php echo $MRL ;?>" />
<input type="submit" value= "Edit Product Details"/>
Undefined variable all lines, do i have to do anything differently if using drop down areas*
*Edit_Prod.php*
<?php
$con = mysql_connect("localhost", "root", "");
mysql_select_db("supplierdetails");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
//Run a query
$ProductID = $_POST['ProductID'];
$result = mysql_query ("SELECT * FROM products WHERE Productid= '".$Productid."'") or die(mysql_error());
$row = mysql_fetch_array($result);
$ProdDesc = $_POST['ProdDesc'];
$SupplierID = $_POST['SupplierID'];
$Location = $_POST['Location'];
$Cost = $_POST['Cost'];
$Status = $_POST['Status'];
$MRL = $_POST['MRL'];
$Productid=$row['Productid'];
$query="UPDATE products SET ProdDesc='".$ProdDesc."', SupplierID='".$SupplierID."', Location='".$Location."', Cost='".$Cost."', Status='".$Status."', MRL='".$MRL."' WHERE ProductID='".$ProductID."'";
$result = mysql_query($query);
//Check whether the query was successful or not
if($result)
{
echo "Products Updated";
header ("Location: Products.php");
}
else
{
die ("Query failed");
}
?>

i think $row vairable is used to get values from database which you have missed
$productID= $_GET['productid'];
$records =mysql_query(" select * from tableName where productid=$productID");
$row=mysql_fetch_array($records);
then your code will work
$supplId= $row['supplId'];
...

Answering to your question "do i have to do anything differently if using drop down areas", of course you have to change it. Change:
<select name="Location" value="<?php echo $Location ;?>" />
<option>A1</option>
<option>A2</option>
<option>A3</option>
<option>A4</option>
<option>A5</option>
.
.
. . .. .
To:
<select name="Location" />
<option value="<?=$Location;?>" selected="selected" ></option>
<option>A1</option>
<option>A2</option>
<option>A3</option>
<option>A4</option>
<option>A5</option>
like that. I think you got the idea. You have to get value as an option under select. Try it. Good luck.

Related

Populate data from dropdown list using a load button

I am trying to populate afew fields using a load button when I select from the dropdown list. However, the load button is not working and the php codes look fine to me. So I am wondering which part went wrong.
I am wondering if I should put the load button at AuthorityCode or below the form. However, I have tried both methods and both doesn't work.
<strong>Authority Code: </strong>
<select name=authorityid value=authorityid>Select Authority</option>
<option value = "">Select</option><br/>
<?php
$connection = new mysqli("localhost", "username", "password", "dbname");
$stmt = $connection->prepare("SELECT AuthorityId FROM AuthorityList");
$stmt->execute();
$stmt->bind_result($authorityid);
while($stmt->fetch()){
echo "<option value = '$authorityid'>$authorityid</option>";
}
$stmt->close();
$connection->close();
?>
</select>
<?php
if(isset($_POST["loadbtn"])){
$authorityid = $_POST["authorityid"];
$conn = new mysqli("localhost", "username", "password", "dbname");
$stmt = $conn->prepare("SELECT AuthorityName, Address, TelephoneNo, FaxNo FROM AuthorityList WHERE AuthorityId = '$authorityid'");
$stmt->execute();
$result = mysqli_query($stmt, $conn);
$details = mysqli_fetch_array($result);
$savedName = $details["AuthorityName"];
$savedAddress = $details["Address"];
$savedTel = $details["TelephoneNo"];
$savedFax = $details["FaxNo"];
}
// while ($row = $result->fetch_array(MYSQLI_NUM)){
// $authorityname = $row[0];
// $address = $row[1];
// $telephone = $row[2];
// $fax = $row[3];
// }
?>
<form action="" method="post" >
<input type="submit" value="Load" name="loadbtn"><br/><br/>
<strong>Authority Name: </strong> <input type="text" name="authorityname" value="<?php echo $savedName; ?>"/><br/>
<strong>Address: </strong> <input type="text" name="address" value="<?php echo $savedAddress; ?>"/><br/>
<strong>Telephone No: </strong> <input type="text" name="telephone" value="<?php echo $savedTel; ?>"/><br/>
<strong>Fax No: </strong> <input type="text" name="fax" value="<?php echo $savedFax; ?>"/><br/>
Change form to:
<form action="window.location.reload()" method="post" >
Your select values are not getting submitted as they are not part of the form. Also your 2nd query was not running properly. I've cheanged it below. Please take a look
Try this-
<strong>Authority Code: </strong>
<form action="" method="post" >
<select name="authorityid">Select Authority
<option value = "">Select</option><br/>
<?php
$connection = new mysqli("localhost", "username", "password", "dbname");
$stmt = $connection->prepare("SELECT AuthorityId FROM AuthorityList");
$stmt->execute();
$stmt->bind_result($authorityid);
while($stmt->fetch()){
echo "<option value = '$authorityid'>$authorityid</option>";
}
$stmt->close();
$connection->close();
?>
</select>
<?php
if(isset($_POST["loadbtn"])){
$authorityid = $_POST["authorityid"];
$conn = new mysqli("localhost", "username", "password", "dbname");
$qry = "SELECT AuthorityName, Address, TelephoneNo, FaxNo FROM AuthorityList WHERE AuthorityId = '$authorityid'";
$result = $conn->query($qry);
$details = mysqli_fetch_array($result);
$savedName = $details["AuthorityName"];
$savedAddress = $details["Address"];
$savedTel = $details["TelephoneNo"];
$savedFax = $details["FaxNo"];
}
?>
<input type="submit" value="Load" name="loadbtn"><br/><br/>
<strong>Authority Name: </strong>
<input type="text" name="authorityname" value="<?php echo isset($savedName) ? $savedName : ''; ?>"/>
<br/>
<strong>Address: </strong>
<input type="text" name="address" value="<?php echo isset($savedAddress) ? $savedAddress : ''; ?>"/>
<br/>
<strong>Telephone No: </strong>
<input type="text" name="telephone" value="<?php echo isset($savedTel) ? $savedTel : ''; ?>"/>
<br/>
<strong>Fax No: </strong>
<input type="text" name="fax" value="<?php echo isset($savedFax) ? $savedFax : ''; ?>"/>
<br/>
</form>
Your error seems to be from mysql part of the code. Below you'll find a sample code that has mysql part removed from it and is running-
<!DOCTYPE html>
<html>
<head>
<title>Fix</title>
</head>
<body>
<strong>Authority Code: </strong>
<form action="" method="post" >
<select name="authorityid">Select Authority
<option value = "">Select</option><br/>
<?php
echo "<option value = '123'>123</option>";
echo "<option value = '234'>234</option>";
echo "<option value = '345'>345</option>";
echo "<option value = '456'>456</option>";
?>
</select>
<?php
if(isset($_POST["loadbtn"])){
$authorityid = $_POST["authorityid"];
if ($authorityid == '123') {
$savedName = 'nameTest';
$savedAddress = 'addressTest';
$savedTel = 'telTest';
$savedFax = 'faxTest';
}
}
?>
<input type="submit" value="Load" name="loadbtn"><br/><br/>
<strong>Authority Name: </strong>
<input type="text" name="authorityname" value="<?php echo isset($savedName) ? $savedName : ''; ?>"/>
<br/>
<strong>Address: </strong>
<input type="text" name="address" value="<?php echo isset($savedAddress) ? $savedAddress : ''; ?>"/>
<br/>
<strong>Telephone No: </strong>
<input type="text" name="telephone" value="<?php echo isset($savedTel) ? $savedTel : ''; ?>"/>
<br/>
<strong>Fax No: </strong>
<input type="text" name="fax" value="<?php echo isset($savedFax) ? $savedFax : ''; ?>"/>
<br/>
</form>
</body>
</html>
Edited: changed the mysqli_query function so that it works properly.

How to add to this update SQL statement and make a drop down list too for it

Okay i have this SQL statement for update which works find but I want to add locations which is in te_venue table after adding the location into sql i want for locations to be a drop down list which user can pick one location from list and when clicked on update it should update.
This is the php code that I have already (I need to add location into this)
<?php
$eventID = $_GET['id'];
if(isset($_POST['submit']))
{
$title = $_POST['title'];
$startdate = $_POST['startdate'];
$enddate = $_POST['enddate'];
$price = $_POST['price'];
$description = $_POST['description'];
$sql = "UPDATE te_events SET eventTitle='$title',eventStartDate='$startdate',eventEndDate='$enddate',eventPrice='$price',eventDescription='$description' WHERE eventID=$eventID";
if ($conn->query($sql) === TRUE)
{
header('Location:edit.php');
}
else
{
echo "Error updating record";
}
}
$sql = "SELECT * FROM te_events where eventID='$eventID'";
$result = $conn->query($sql);
while($row = $result->fetch_assoc())
{
$eventTitle = $row['eventTitle'];
$eventDescription = $row['eventDescription'];
$eventStartDate = $row['eventStartDate'];
$eventEndDate = $row['eventEndDate'];
$eventPrice = $row['eventPrice'];
}
?>
And this is where i want to have a drop down list for location.
<form method="post" action="">
<label for="title">Title</label><br/>
<input type="text" name="title" required/ value="<?php echo $eventTitle; ?>"><br/>
<label for="startdate">Start Date</label><br/>
<input type="date" name="startdate" required/ value="<?php echo $eventStartDate; ?>"><br/>
<label for="enddate">End Date</label><br/>
<input type="date" name="enddate" required/ value="<?php echo $eventEndDate; ?>"><br/>
<label for="price">Price (£)</label><br/>
<input type="number" step="any" name="price" required/ value="<?php echo $eventPrice; ?>"><br/>
<label for="description">Description</label><br/>
<textarea name="description" cols="55" rows="5"><?php echo $eventDescription; ?></textarea><br/>
<input type="submit" name="submit" value="Update" class="button">
</form>
Here is a screenshot of my tables too it might help.
https://postimg.org/image/6ui8nsi55/
Do you want something like this? Getting the location from te_venue and adding the results to a select?
<select>
<?php
$sql = "SELECT location FROM te_venue";
$result = $conn->query($sql);
while($row = $result->fetch_assoc())
}
?>
<option value="<?php echo $row['location'];?>"><?php echo $row['location'];?> </option>
<?php
}
?>
</select>
<label for="location">Choose a location: </label>
<select class="form-control" name="location">
<option value="London">London</option>
<option value="Paris">Paris</option>
<option value="Rome">Rome</option>
<option value="Berlin">Berlin</option>
<option value="Moscow">Moscow</option>
</select>

Pre-Populate Form with Dropdown Menu in PHP/MySQL?

I am creating a simple CRUD application that will be used as a blog. For the edit page, I want to have a dropdown menu with the blog titles of each post. When an option/blog post is selected, I want it to populate the "Title" and "Message" fields, so it can then be edited and saved to the database.
I got it to retrieve the titles of the blog posts, but I am struggling to make it populate the "Title" and "Message" fields so it can be edited when the option is selected.
I have 4 rows in my database: row[0] is the title, row[1] is the message, row[2] is the timestamp and row[3] is the ID.
Thanks guys. I appreciate it.
<form action="edit.php" id="myform" method="post" autocomplete=off>
<input type="hidden" name="action" value="show">
<p><label>Entry:</label><select name="blog">
<?php
$result = mysqli_query($con, "SELECT * FROM blog");
while ($row = mysqli_fetch_array($result)) {
$chosen = $row['bid'];
}
if (isset($_GET['blog'])) {
$id = $_GET['blog'];
$result = mysqli_query($con, "SELECT * FROM blog WHERE bid='$id'");
$row = mysqli_fetch_array($result);
}
$result = mysqli_query($con, "SELECT * FROM blog");
while ($row = mysqli_fetch_array($result)) {
$id = $row['bid'];
$title = $row['title'];
$selected = '';
if ($id == $chosen) {
$selected = "selected='selected'";
}
echo "<option value='$id' $selected>$title</option>\n";
}
?>
</select></p>
<p><label>Title:</label> <input type="text" id="newtitle" name="newtitle" value="<?php echo $row[0]; ?>"></p>
<p><label>Message:</label> <input type="text" id="newmessage" name="newmessage" value="<?php echo $row[1]; ?>"></p>
<p><input type="hidden" name="id" value="<?php echo $row[3]; ?>"></p>
<br><p><input type="submit" name="submit" value="Submit"></p>
</form>
I have managed to somewhat figure out the answer on my own. It's probably not the most ideal way of doing it, but for anyone else potentially stuck on this - here is my code:
The only issue I'm having now is figuring out how to keep my option selected.
<form action="edit.php" id="myform" method="post" autocomplete=off>
<input type="hidden" name="action" value="show">
<p><label>Entry:</label><select name="blog" onchange="window.location.href = blog.options[selectedIndex].value">
<option selected disabled>Select One</option>
<?php
$result = mysqli_query($con, "SELECT * FROM blog");
while ($row = mysqli_fetch_array($result)) {
$id = $row['bid'];
$title = $row['title'];
echo "<option value='edit.php?edit=$row[bid]'>$title</option>\n";
}
if (isset($_GET['edit'])) {
$id = $_GET['edit'];
$result = mysqli_query($con, "SELECT * FROM blog WHERE bid='$id'");
$row = mysqli_fetch_array($result);
}
?>
</select></p>
<p><label>Title:</label> <input type="text" id="newtitle" name="newtitle" value="<?php echo $row[0]; ?>"></p>
<p><label>Message:</label> <input type="text" id="newmessage" name="newmessage" value="<?php echo $row[1]; ?>"></p>
<p><input type="hidden" name="id" value="<?php echo $row[3]; ?>"></p>
<br><p><input type="submit" name="submit" value="Submit"></p>
</form>

How to Edit a MySQL Data base from a PHP Site

I have built a database named "artStore" and a table within that database named "inventory". I've been able to figure out how to create a new entry to the database, but I'm trying to create a page that can edit those entries.
Here is "inventory" the table I created:
$sql = "CREATE TABLE inventory (
id INT(6) AUTO_INCREMENT PRIMARY KEY,
product VARCHAR(30),
category VARCHAR(30),
seller VARCHAR(30)
)";
Here is what I'm currently trying:
<?php
$resultProduct = "product";
$resultCategory = "category";
$resultSeller = "seller";
if ($result->num_rows > 0) {
while ($row = $result->fetch_assoc()) {
$resultProduct = $row["product"];
$resultCategory = $row["category"];
$resultSeller = $row["seller"];
}
} else {
echo "No Results";
}
?>
<form action="update.php" method="POST">
<label for="product">Product</label>
<input id="product" type="text" name="product" value="<?php echo $resultProduct; ?>">
<br>
<label for="category">Category:</label>
<input id="category" type="text" name="category" value="<?php echo $resultCategory; ?>">
<br>
<label for="seller">Seller:</label>
<input id="seller" type="text" name="seller" value="<?php echo $resultSeller; ?>">
<br>
<input id="id" type="text" name="id" value="<?php echo $resultId; ?>" style="display:none;">
<input type="submit" value="Update My Record">
</form>
What I'm trying in update.php
$product = $_POST['product'];
$category = $_POST['category'];
$seller = $_POST['seller'];
$sql = "INSERT INTO inventory (product, category, seller) VALUES ('$product', '$category', '$seller')";
if ($connection->query($sql) === true) {
echo "Inserted Successfully";
} else {
echo "Error occured in the insert: " . $connection->error;
}
$connection->close();
Try the below code I added hidden field on form and changes on your sql query
<?php
$resultProduct = "product";
$resultCategory = "category";
$resultSeller = "seller";
if ($result->num_rows > 0) {
while ($row = $result->fetch_assoc()) {
$resultProduct = $row["product"];
$resultCategory = $row["category"];
$resultSeller = $row["seller"];
$resultId = $row["id"];
}
} else {
echo "No Results";
}
?>
<form action="update.php" method="POST">
<input type="text" name="update_id" value="<?php echo $resultId; ?>">
<label for="product">Product</label>
<input id="product" type="text" name="product" value="<?php echo $resultProduct; ?>">
<br>
<label for="category">Category:</label>
<input id="category" type="text" name="category" value="<?php echo $resultCategory; ?>">
<br>
<label for="seller">Seller:</label>
<input id="seller" type="text" name="seller" value="<?php echo $resultSeller; ?>">
<br>
<input id="id" type="text" name="id" value="<?php echo $resultId; ?>" style="display:none;">
<input type="submit" value="Update My Record">
</form>
In update.php
$product = $_POST['product'];
$category = $_POST['category'];
$seller = $_POST['seller'];
$updateId = $_POST['update_id'];
$sql = "UPDATE inventory set product = '$product',category='$category',seller='$seller' WHERE id = '$updateId'";
<?php
$resultProduct = "product";
$resultCategory = "category";
$resultSeller = "seller";
$resultId = "product_id";
if ($result->num_rows > 0) {
while ($row = $result->fetch_assoc()) {
$resultProduct = $row["product"];
$resultCategory = $row["category"];
$resultSeller = $row["seller"];
$resultid = $row["id"];
}
} else {
echo "No Results";
}
?>
You were using $resultId in the form but you have not declared and assigned a value to it in the loop.
<form action="update.php" method="POST">
<input id="id" type="hidden" name="id" value="<?php echo $resultId; ?>">
<label for="product">Product</label>
<input id="product" type="text" name="product" value="<?php echo $resultProduct; ?>">
<br>
<label for="category">Category:</label>
<input id="category" type="text" name="category" value="<?php echo $resultCategory; ?>">
<br>
<label for="seller">Seller:</label>
<input id="seller" type="text" name="seller" value="<?php echo $resultSeller; ?>">
<br>
<input type="submit" value="Update My Record">
</form>
<!-- Instead of passing the ID in textbox with display: none you can pass it directly in hidden -->
<?php
$product = $_POST['product'];
$category = $_POST['category'];
$seller = $_POST['seller'];
$product_id = $_POST['id'];
if($product_id!='')//MEANS WE HAVE TO UPDATE THE RECORD
{
// UPDATE QUERY WILL UPDATE THE SAME RECORD ONLY MATCHING THE UNIQUE PRODUCT ID
$sql = "UPDATE inventory SET product = '$product', category = '$category', seller = '$seller' WHERE id = '$product_id' LIMIT 1";
if ($connection->query($sql) === true) {
echo "Updated Successfully";
} else {
echo "Error occured in the insert: " . $connection->error;
}
}
else// If id is blank it means you have a new record
{
$sql = "INSERT INTO inventory (product, category, seller) VALUES ('$product', '$category', '$seller')";
if ($connection->query($sql) === true) {
echo "Inserted Successfully";
} else {
echo "Error occured in the insert: " . $connection->error;
}
}
$connection->close();
?>

check box issue during editing in php

This is advertisement query
INSERT into advertise_management(advertisement_name,adv_code) values('".$adv_name."','".$adv_code."')
post management:
<input type="text" name="post_name" value="<?php if(isset($post_value)){ echo $post_value['post_name'];} else { echo ''; }?>" required="required">
<?php $res= mysql_query("select * from advertise_management ");
while($result= mysql_fetch_array($res)){
$adv=$result['advertisement_name'];
{ ?>
<input type="checkbox" value="<?php echo $adv?>" name="post_adv[]">
<?php } ?>
Post Management insert query :
$post_name = $_POST['post_name'];
$post_adv1 = $_POST['post_adv'];
$post_adv = implode(",", $post_adv1);
$post_id = $_POST['post_id'];
$res = mysql_query("INSERT into post_managment(post_name,post_adv) values('".$post_name."','".$post_adv."')") or die(mysql_error());
Post management edit :
$result = mysql_query("SELECT * FROM post_managment where post_id='$pid'");
$post_value = mysql_fetch_array($result);
><input type="text" name="post_name" value="<?php if(isset($post_value)){ echo $post_value['post_name'];} else { echo ''; }?>" required="required">
<?php $res= mysql_query("select * from advertise_management ");
while($result= mysql_fetch_array($res)){
$adv=$result['advertisement_name'];
{ ?>
<input type="checkbox" value="<?php echo $adv?>" name="post_adv[]">
<?php } ?>
<input type="submit" value="" id="create"><input type="reset" value="" id="cancel">
I want to return the values that is checked and the values that is not checked.
The browser normally won't send anything for checkboxes that aren't checked. You can work around this by also rendering a hidden input () with the same name and a different value. Make sure it's output right before the checkbox.

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