entering information into mysql database from php... SEE INSIDE...

M3NT4L · May 14, 2008

ok heres the situation, i need to create a shop so that once an item is clicked... it then goes on and the user enters personal information which is then sent and added to the customer_info table i created in my mysql database...

ok so i have all that done except for the part where i get the user to enter information:

i have to files, see below

<?php
// Connect to the DB
$Sort = $_GET['Sort'];
$Submit = $_GET['submit'];
//echo "<hr><h3>Results sorted by $Sort</h3>";
//echo "\$submit is " . $submit;
//echo "\$Sort is " . $Sort;
if($Submit == "Submit Query"){
require "mysql_connect_lec.php";

switch($Sort)
{
	case "id":  $sort_var = "product_id"; break;
	case "name":  $sort_var = "product_name"; break;
	case "price": $sort_var = "product_price"; break;
}

// Create a query string
$sql_query = "SELECT * FROM product_info ORDER BY $sort_var";

// Run the query and save the results
$result = @mysql_query($sql_query) or die("Invalid query: " . mysql_error());

// Start creating a table to display the results [static part]
echo '<table border="1"><tr><th>Product ID</th><th>Team Name</th><th>Price</th></tr>';

// loop through the results - create as many table rows as there are records [dynamic part]
while ($row = mysql_fetch_array($result)) {
  // this sticks the results row by row into an array called $row.
  // rows are called via $array_name["COLUMN_NAME"]
	$out  = '<tr>';
	$out .= '<td>'.$row["product_id"].'</td>';
	$out .= '<td>'.$row["product_name"].'</td>';
	$out .= '<td><a href="mysql_lec_4.php?product_name='.$row["product_id"].'">'.$row["product_price"].'</a></td>';
	$out .= '</tr>';
	echo $out;
}

// Finish constructing the table [static part]
echo '</table>';

// Close the connection
@mysql_close($mysql_conn) or die("Error whilst closing connection: " . mysql_error());
} // End IF statement
?>

everything above seems to work fine i am just pasting so you can understand it more...

here is the bit i have a problem with:

<?php
require "mysql_connect_lec.php";
$Rank = $_GET['Rank'];
echo "the Rank I have got is ". $product_name;

if($submit == "Update"){
    $Rank = $CR['Rank'];
    echo "Now I have got Rank is " . $product_name;
$sql_query = "UPDATE product_info SET ";
$sql_query .= "Rank = '".$HTTP_POST_VARS['product_name'];
    $sql_query .= "', Gross = '".$HTTP_POST_VARS['product_id'];
    $sql_query .= "', Year = '".$HTTP_POST_VARS['product_price']."' WHERE Rank=".$product_name;
echo $sql_query;

$result = @mysql_query($sql_query) or die("Invalid query: " . mysql_error());

}
$sql_query = "SELECT * FROM product_info WHERE product_name=$product_name";
$result = @mysql_query($sql_query) or die("Invalid query: " . mysql_error());
$CR = mysql_fetch_assoc($result);
echo $CR['Name'];
@mysql_close($mysql_conn) or die("Error whilst closing connection: " . mysql_error());


?>
<H1>Record update form</H1>
<form name="form1" method="get" action="mysql_lec_4a.php">
  Rank   <input name="Rank" type="text" value="<?php echo $CR['Rank'] ?>">   <br>
  Gross sale   <input type="text" name="Gross" value="<?php echo $CR['Gross'] ?> "> <br>
  Name   <input type="text" name="Name" value=<?php echo $CR['Name'] ?>> <br>
  Year   <input type="text" name="Year" value="<?php echo $CR['Year'] ?>">
  <input name="MID" type="hidden" value="<?php echo $MID ?>">
  <input name="show" type="hidden" value="1">

  <input name="submit" type="submit" value = "Update">
<hr>
  <h2><a href="mysql_lec_3.php?Sort=Name&submit=Submit+Query">Box Office listing</a> </h2>
</form>

ok so i have tried to edit this myself with trial and error but no luck, it doesnt submit the users information into the customer_info table.

i have my customer_info table setup like this

mysql> select * from customer_info;
+-----------+--------------+----------------+-----------+-----------+----------+---------------------------+
| Name      | Address_1    | Address_2      | City      | Postcode  | Quantity | Comments                  |
+-----------+--------------+----------------+-----------+-----------+----------+---------------------------+
| name test | address test | address test 2 | city test | post test | 1        | This is working correctly |
+-----------+--------------+----------------+-----------+-----------+----------+---------------------------+

M3NT4L · May 14, 2008

ok here is my the one i edited to try and get to send info to customer_info table but wont enter it... i understand that the mysql_lec_4a.php is the file where the user will be sent after for example... THANKS FOR ENTERING INFO...

<?php
require "mysql_connect_lec.php";
$Rank = $_GET['Rank'];
echo "the Rank I have got is ". $product_name;

if($submit == "Update"){
    $Rank = $CR['Rank'];
    echo "Now I have got Rank is " . $product_name;
$sql_query = "UPDATE customer_info SET ";
$sql_query .= "Name = '".$HTTP_POST_VARS['Name'];
    $sql_query .= "', Address_1 = '".$HTTP_POST_VARS['Address_1'];
  $sql_query .= "', Address_2 = '".$HTTP_POST_VARS['Address_2'];
  $sql_query .= "', City = '".$HTTP_POST_VARS['City'];
$sql_query .= "', Postcode = '".$HTTP_POST_VARS['Postcode'];
$sql_query .= "', quantity = '".$HTTP_POST_VARS['Quantity'];
    $sql_query .= "', price = '".$HTTP_POST_VARS['Comments']."' WHERE name=".$comments;
echo $sql_query;

$result = @mysql_query($sql_query) or die("Invalid query: " . mysql_error());

}
$sql_query = "SELECT * FROM product_info WHERE product_name=$product_name";
$result = @mysql_query($sql_query) or die("Invalid query: " . mysql_error());
$CR = mysql_fetch_assoc($result);
echo $CR['Name'];
@mysql_close($mysql_conn) or die("Error whilst closing connection: " . mysql_error());


?>
<H1>Record update form</H1>
<form name="form1" method="get" action="mysql_lec_4a.php">
  id   <input name="Rank" type="text" value="<?php echo $CR['id'] ?>">   <br>
  name   <input type="text" name="Gross" value="<?php echo $CR['name'] ?> "> <br>
  price   <input type="text" name="Year" value="<?php echo $CR['price'] ?>">
  <input name="MID" type="hidden" value="<?php echo $MID ?>">
  <input name="show" type="hidden" value="1">

  <input name="submit" type="submit" value = "Update">
<hr>
  <h2><a href="mysql_lec_3.php?Sort=Name&submit=Submit+Query">Box Office listing</a> </h2>
</form>

MadTechie · May 14, 2008

whats the

echo $sql_query;

outputting?

as a guess i would say

WHERE Rank=".$product_name;

should be

WHERE Rank='$product_name';

M3NT4L · May 14, 2008

the echo is fine it tells the user what they have ordered. but again i will repeat... i am having trouble getting information from a user which enters itself into the customer_info table.

does anyone know?

Skittalz · May 14, 2008

I believe your forgetting a POST in the second file...

if($submit == "Update"){

should infact be

if($_POST['Submit'] == "Update") { //note... its a capital S as well

moselkady · May 14, 2008

What did this statement display in your browser? and did it display any errors?

echo $sql_query;

M3NT4L · May 14, 2008

My new, and its still not entering any information into the table:

mysql> select * from customer_info;
+-----------+--------------+----------------+-----------+-----------+----------+---------------------------+
| Name      | Address_1    | Address_2      | City      | Postcode  | Quantity | Comments                  |
+-----------+--------------+----------------+-----------+-----------+----------+---------------------------+
| name test | address test | address test 2 | city test | post test | 1        | This is working correctly |
+-----------+--------------+----------------+-----------+-----------+----------+---------------------------+
1 row in set (0.00 sec)

<?php
require "mysql_connect_lec.php";
$product_name = $_GET['product_name'];
echo "The product you have ordered is ". $product_name;

if($_POST['Submit'] == "Update") { //note... its a capital S as well
    $product_name = $CR['product_name'];
    echo "The product you have ordered is " . $product_name;
$sql_query = "UPDATE customer_info SET ";
$sql_query .= "Name = '".$HTTP_POST_VARS['Name'];
    $sql_query .= "', Address_1 = '".$HTTP_POST_VARS['Address_1'];
$sql_query .= "', Address_2 = '".$HTTP_POST_VARS['Address_2'];
$sql_query .= "', City = '".$HTTP_POST_VARS['City'];
$sql_query .= "', Postcode = '".$HTTP_POST_VARS['Postcode'];
$sql_query .= "', Quantity = '".$HTTP_POST_VARS['Quantity'];
    $sql_query .= "', Comments = '".$HTTP_POST_VARS['Comments']."' WHERE name=".$product_name;
echo $sql_query;

$result = @mysql_query($sql_query) or die("Invalid query: " . mysql_error());

}
$sql_query = "SELECT * FROM product_info WHERE product_name=$product_name";
$result = @mysql_query($sql_query) or die("Invalid query: " . mysql_error());
$CR = mysql_fetch_assoc($result);
echo $CR['Name'];
@mysql_close($mysql_conn) or die("Error whilst closing connection: " . mysql_error());


?>
<H1>Record update form</H1>
<form name="form1" method="get" action="mysql_lec_4a.php">
  Name   <input name="Rank" type="text" value="<?php echo $CR['Name'] ?>">   <br>
  Address Line 1   <input type="text" name="Gross" value="<?php echo $CR['Address_1'] ?> "> <br>
  Address Line 2   <input type="text" name="Gross" value="<?php echo $CR['Address_2'] ?> "> <br>
  City   <input type="text" name="Gross" value="<?php echo $CR['City'] ?> "> <br>
  Postcode   <input type="text" name="Gross" value="<?php echo $CR['Postcode'] ?> "> <br>
  Quantity   <input type="text" name="Gross" value="<?php echo $CR['Quantity'] ?> "> <br>
  Comments   <input type="text" name="Year" value="<?php echo $CR['Comments'] ?>">
  <input name="MID" type="hidden" value="<?php echo $MID ?>">
  <input name="show" type="hidden" value="1">

  <input name="submit" type="submit" value = "Update">
<hr>
  <h2><a href="mysql_lec_3.php?Sort=Name&submit=Submit+Query">LINK USED FOR LATER...</a> </h2>
</form>

moselkady · May 14, 2008

Try:

if($_POST['submit'] == "Update")

since your submit button is named "submit"

M3NT4L · May 14, 2008

Either way still not working, has anyone at least done this before with a form and the info is submitted to the table without bein in on terminal.

like there must be something like:

INSERT INTO customer_info

values('$NAME, $ADDRESS_1 ...... ETC

so that when the submit button is entered it will enter it into the row like... ya know?

<?php
require "mysql_connect_lec.php";
$product_name = $_GET['product_name'];
echo "The product you have ordered is ". $product_name;

if($_POST['submit'] == "Update") {
    $product_name = $CR['product_name'];
    echo "The product you have ordered is " . $product_name;
$sql_query = "UPDATE customer_info SET ";
$sql_query .= "Name = '".$HTTP_POST_VARS['Name'];
    $sql_query .= "', Address_1 = '".$HTTP_POST_VARS['Address_1'];
$sql_query .= "', Address_2 = '".$HTTP_POST_VARS['Address_2'];
$sql_query .= "', City = '".$HTTP_POST_VARS['City'];
$sql_query .= "', Postcode = '".$HTTP_POST_VARS['Postcode'];
$sql_query .= "', Quantity = '".$HTTP_POST_VARS['Quantity'];
    $sql_query .= "', Comments = '".$HTTP_POST_VARS['Comments']."' WHERE name=".$product_name;
echo $sql_query;

$result = @mysql_query($sql_query) or die("Invalid query: " . mysql_error());

}
$sql_query = "SELECT * FROM product_info WHERE product_name=$product_name";
$result = @mysql_query($sql_query) or die("Invalid query: " . mysql_error());
$CR = mysql_fetch_assoc($result);
echo $CR['Name'];
@mysql_close($mysql_conn) or die("Error whilst closing connection: " . mysql_error());


?>
<H1>Record update form</H1>
<form name="form1" method="get" action="mysql_lec_4a.php">
  Name   <input name="Name" type="text" value="<?php echo $CR['Name'] ?>">   <br>
  Address Line 1   <input type="text" name="Address_1" value="<?php echo $CR['Address_1'] ?> "> <br>
  Address Line 2   <input type="text" name="Address_2" value="<?php echo $CR['Address_2'] ?> "> <br>
  City   <input type="text" name="City" value="<?php echo $CR['City'] ?> "> <br>
  Postcode   <input type="text" name="Postcode" value="<?php echo $CR['Postcode'] ?> "> <br>
  Quantity   <input type="text" name="Quantity" value="<?php echo $CR['Quantity'] ?> "> <br>
  Comments   <input type="text" name="Comments" value="<?php echo $CR['Comments'] ?>">
  <input name="MID" type="hidden" value="<?php echo $MID ?>">
  <input name="show" type="hidden" value="1">

  <input name="submit" type="submit" value = "Update">
<hr>
  <h2><a href="mysql_lec_3.php?Sort=Name&submit=Submit+Query">LINK USED FOR LATER...</a> </h2>
</form>

moselkady · May 14, 2008

I see. UPDATE query is done when the record for product_name is already in your customer_info table and you are updating it. Otherwise, you should go with INSERT. Something like this:

$sql_query = "INSERT INTO customer_info (name, address_1, address_2, city, postcode, quantity, comments) VALUES('{$_POST['Name']}', '{$_POST['Address_1']}', '{$_POST['Address_2']}', '{$_POST['City']}', '{$_POST['Postcode']}', '{$_POST['Quantity']}', '{$_POST['Comments']}')";

M3NT4L · May 14, 2008

ok but just to say i have 2 tables, product_info and customer_info. so product_name isn't needed to be submitted to a column in customer_info only product_id is sent into the table... erm i hope you understand what i am trying to say.

so what your saying is that like that snippet of code you have above needs to be put say here:

<?php

require "mysql_connect_lec.php";

$product_name = $_GET['product_name'];

echo "The product you have ordered is ". $product_name;

if($_POST['submit'] == "Update") {

$product_name = $CR['product_name'];

echo "The product you have ordered is " . $product_name;

$sql_query = "UPDATE customer_info SET ";

$sql_query .= "Name = '".$HTTP_POST_VARS['Name'];

$sql_query .= "', Address_1 = '".$HTTP_POST_VARS['Address_1'];

$sql_query .= "', Address_2 = '".$HTTP_POST_VARS['Address_2'];

$sql_query .= "', City = '".$HTTP_POST_VARS['City'];

$sql_query .= "', Postcode = '".$HTTP_POST_VARS['Postcode'];

$sql_query .= "', Quantity = '".$HTTP_POST_VARS['Quantity'];

$sql_query .= "', Comments = '".$HTTP_POST_VARS['Comments']."' WHERE name=".$product_name;

echo $sql_query;

$result = @mysql_query($sql_query) or die("Invalid query: " . mysql_error());

}

$sql_query = "SELECT * FROM product_info WHERE product_name=$product_name"; <<< $sql_query = "INSERT INTO customer_info (name, address_1, address_2, city, postcode, quantity, comments) VALUES('{$_POST['Name']}', '{$_POST['Address_1']}', '{$_POST['Address_2']}', '{$_POST['City']}', '{$_POST['Postcode']}', '{$_POST['Quantity']}', '{$_POST['Comments']}')";

$result = @mysql_query($sql_query) or die("Invalid query: " . mysql_error());

$CR = mysql_fetch_assoc($result);

echo $CR['Name'];

@mysql_close($mysql_conn) or die("Error whilst closing connection: " . mysql_error());

?>

<H1>Record update form</H1>

Name <input name="Name" type="text" value="<?php echo $CR['Name'] ?>"> <br>

Address Line 1 <input type="text" name="Address_1" value="<?php echo $CR['Address_1'] ?> "> <br>

Address Line 2 <input type="text" name="Address_2" value="<?php echo $CR['Address_2'] ?> "> <br>

City <input type="text" name="City" value="<?php echo $CR['City'] ?> "> <br>

Postcode <input type="text" name="Postcode" value="<?php echo $CR['Postcode'] ?> "> <br>

Quantity <input type="text" name="Quantity" value="<?php echo $CR['Quantity'] ?> "> <br>

Comments <input type="text" name="Comments" value="<?php echo $CR['Comments'] ?>">

<hr>

<h2><a href="mysql_lec_3.php?Sort=Name&submit=Submit+Query">LINK USED FOR LATER...</a> </h2>

</form>

moselkady · May 14, 2008

Ok. When user hits Update button, we have product_name and we want to get product_id from product_info. Then insert this product_id along with customer information into customer_info, right? If so you can do this:

if($_POST['submit'] == "Update") {
  $sql_query = "SELECT * FROM product_info WHERE product_name=$product_name";
  $result = @mysql_query($sql_query) or die("Invalid query: " . mysql_error());
  $CR = mysql_fetch_assoc($result);
  product_id = $CR['product_id'];
  $sql_query = "INSERT INTO customer_info (product_id , name, address_1, address_2, city, postcode, quantity, comments) VALUES('{$_POST['Name']}', '{$product_id , $_POST['Address_1']}', '{$_POST['Address_2']}', '{$_POST['City']}', '{$_POST['Postcode']}', '{$_POST['Quantity']}', '{$_POST['Comments']}')";
  $result = @mysql_query($sql_query) or die("Invalid query: " . mysql_error());
}

I hope I got right this time.

M3NT4L · May 14, 2008

afraid not.

what is needed is.

forget about product_name and info etc that is not need anymore.

when the user selects a product, the only thing that is sent to the customer_info table is the product_id.

the titles under the customer_info table are

name address_1 address2 etc etc.... so the information that the user enters into the textbox goes into the columns in the customer_info table... you understand what i am saying??? i dont know how else to put it...?

just forget the product info i have done that.... just now onto that form and the information put it needs to be sent to database... yes?

EDIT: for that code you said... nothing works its a blank screen but from what you think i say its wrong either way... read up.. ^

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