Php and SQl help please

bribren2001

hey all, the following php/sql code will not bring back the required information

im trying to get back just one field-order number

it keeps giving me back the table with the label but no order number value

any help greatly appreciated

<?php
session_start();
header("cache-control: private");
$Student_Username = $_POST;
// Make a MySQL Connection
mysql_connect("xx", "xx", "xx") or die(mysql_error());
mysql_select_db("gamesemp_cart") or die(mysql_error());
$sql = mysql_query("SELECT * FROM orders WHERE Student_UserName = 'Smith");

echo "<table border='1'>";
echo "<tr> <th>Order Number</th></tr>";

{
echo "<table border='1'>";
echo "<tr><td>";
echo $row;
echo "</td></tr>";
}
echo "</table>";
?>

seamus

OK.

$sql = mysql_query("SELECT * FROM orders WHERE Student_UserName = 'Smith");

Missing a closing single quote after Smith.

bribren2001

cheers seamus
i put that in but still the same unfortunately

Ph3n0m

first of all edit your code, you left in your db name and password

second of all - make sure the information is there in the database, otherwise nothing will be returned

bribren2001

Oops,
ye there is loads of information in the database

seamus

Do us a favour. Run the following queries and tell us the results you get.

SELECT COUNT(*) FROM orders WHERE Student_UserName = 'Smith'
SELECT COUNT(*) FROM orders WHERE Student_UserName LIKE 'Smith'

Ph3n0m

<?php
session_start();
header("cache-control: private");
$Student_Username = $_POST['Student_Username'];
// Make a MySQL Connection
mysql_connect("xx", "xx", "xx") or die(mysql_error());
mysql_select_db("gamesemp_cart") or die(mysql_error());
$sql = mysql_query("SELECT * FROM orders WHERE Student_UserName = 'Smith");

echo "<table border='1'>";
echo "<tr> <th>Order Number</th></tr>";
while ($row = mysql_fetch_array($sql)) {

echo "<table border='1'>";
echo "<tr><td>";
echo $row['OrderNumber'];
echo "</td></tr>";
}
echo "</table>";
?>

that should do it

bribren2001

seamus-it give me errors on line 12 similar to the one below.

ph3nom- it gave me the following error-Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/gamesemp/public_html/getordernum.php on line 12

seamus

bribren2001 wrote:

seamus-it give me errors on line 12 similar to the one below.

Can you run them through a MySQL interface, and not through PHP? Just to ensure your DB is returning the correct info.

Ph3n0m

oops, try this;

<?php
session_start();
header("cache-control: private");
$Student_Username = $_POST['Student_Username'];
// Make a MySQL Connection
mysql_connect("xx", "xx", "xx") or die(mysql_error());
mysql_select_db("gamesemp_cart") or die(mysql_error());
$sql = mysql_query("SELECT OrderNumber FROM orders WHERE Student_UserName = 'Smith'");

echo "<table border='1'>";
echo "<tr> <th>Order Number</th></tr>";
while ($row = mysql_fetch_array($sql)) {

echo "<table border='1'>";
echo "<tr><td>";
echo $row['OrderNumber'];
echo "</td></tr>";
}
echo "</table>";
?>

bribren2001

Worked a treat Ph3n0m-thanks a million for that!!!

big thanks to you also seamus

seamus

D'oh!

Should have read Ph3nom's posts. The simple ones are sometimes the hardest to spot .