[RESOLVED] Uable to populate passed data in my form

[cybastud]

Hi

I having a bit of trouble to print the passed data through a URL in my form...below is my code

$car_id= $_GET['car_id'];
//the URL does show the id e.g. edit.php?car_id=1
// did an echo here to test if its caught correctly and it works

$query = "SELECT manufacturer, make, year, price FROM cars WHERE car_id = '$car_id'";
$result = mysql_query($query)

while($row = mysql_fetch_array(result)){
	$manufacturer = $row['manufacturer'];
	$make = $row['make'];
            $year = $row['year'];
            $price = $row['price'];

//i tried and echo here as well for $manufacturer but it does not print anything

Then I have my form:

<form Method="POST" ACTION="edit.php">

<table border="2" align="center">
<tr>
<td>Manufacturer</td><td><input name="manufacturer" type="text" value=<?php echo "$manufacturer ?>"></input></td>
</tr>
<tr>
<td>Make</td><td><input name="make" type="text" value=<?php echo "$make?>></input></td>
</tr>

//and so on...I just need to data displayed in the form so that the user can edit it and update the records in the database, but nothing prints on the form...Any assistance will be much appreciated

Thanks🙂

[awbrowni30]

Your code...

while($row = mysql_fetch_array(result)){

Missing $ sign...

while($row = mysql_fetch_array($result)){

[cybastud]

Thanks🙂 Will test it when I get home this evening, but is everything else ok? Am I using the echo correctly in my form?

[awbrowni30]

Your code

value=<?php echo "$make?>

Try this

value="<?php echo $make; ?>"

[cybastud]

ok will do, I'll update this thread later..thanks

[cybastud]

Hey, thanks its working but now I see when change the values in the form to updates the record, it does not update it in the database. When I submit the form it does print out the echo "success" but does not updat...any idea why?

<?php
include ('connect.php');//Includes database connection details
error_reporting(0);

$car_id= $_GET['car_id'];
$Update = $_POST['update'];

$query = "SELECT manufacturer, make, year, price FROM cars WHERE car_id = '$car_id'";
$result = mysql_query($query);

while($row = mysql_fetch_array($result)){
	$manufacturer = $row['manufacturer'];
	$make = $row['make'];
    $year = $row['year'];
    $price = $row['price'];

}

if (isset($_POST['updated'])) {
	$sql = "UPDATE cars SET manufacturer = '$manufacturer', make = '$make', year = '$year', price = '$price' 
	WHERE $car_id = car_id";
	$result = mysql_query(sql);

	echo "success";

	}

?>

[cybastud]

cybastud;10987946 wrote:

Hey, thanks its working but now I see when change the values in the form to updates the record, it does not update it in the database. When I submit the form it does print out the echo "success" but does not updat...any idea why?

<?php
include ('connect.php');//Includes database connection details
error_reporting(0);

$car_id= $_GET['car_id'];
$Update = $_POST['update'];

$query = "SELECT manufacturer, make, year, price FROM cars WHERE car_id = '$car_id'";
$result = mysql_query($query);

while($row = mysql_fetch_array($result)){
	$manufacturer = $row['manufacturer'];
	$make = $row['make'];
    $year = $row['year'];
    $price = $row['price'];

}

if (isset($_POST['updated'])) {
	$sql = "UPDATE cars SET manufacturer = '$manufacturer', make = '$make', year = '$year', price = '$price' 
	WHERE $car_id = car_id";
	$result = mysql_query(sql);

	echo "success";

	}

?>

Sorry please look at revised code.....

<?php
include ('connect.php');//Includes database connection details
error_reporting(0);

$car_id= $_GET['car_id'];
$Update = $_POST['update'];

$query = "SELECT manufacturer, make, year, price FROM cars WHERE car_id = '$car_id'";
$result = mysql_query($query);

while($row = mysql_fetch_array($result)){
	$manufacturer = $row['manufacturer'];
	$make = $row['make'];
    $year = $row['year'];
    $price = $row['price'];

}

if (isset($_POST['updated'])) {
	$manufacturer = $_POST["manufacturer"];
	$make = $_POST["make"];
	$year = $_POST["year"];
	$price = $_POST["price"];

	$sql = "UPDATE cars SET manufacturer = '$manufacturer', make = '$make', year = '$year', price = '$price' 
	WHERE car_id = $car_id";
	$result = mysql_query($sql);

	echo "success";

	}

?>