[john_zakaria]

Now i have this page and in the image part i want to passs the id in the selected row to the page delete.php and after that deleteing this row


<a href='delete.php?id='$id'><img src='untitled.jpg' width='10' height='11' border='0' /></a>

i want here to set the id to be like the id in the database to use it in the where condition.

can anyone help me on how to set the vaiable$id to be like the id in the selected row?



but till now i can not pass the id to the next page

PHP Code:

            
<?
    $DBhost = "localhost";
    $DBuser = "root";
    //$DBpass = "123456";
    $DBname = "gateway";

    /// 1. create a database connection
    $connection = mysql_connect($DBhost, $DBuser)
    or die ("database connect error: " . mysql_error());
    //// 2. select a database to use
    $db_connect = mysql_select_db($DBname, $connection)
    or die ("database select error: " . mysql_error());    

    ///// 3. Perform database query
    $qryp = "SELECT * FROM hotels";
    print "qryp: " . $qryp . "<br>";
    $result = mysql_query($qryp)
    or die ("table error: " . mysql_error());
    
    
    
    ?>
    
    

        <table border="1">
    <tr>
    <td> ID </td>
        <td>Hotel </td>
        <td>Tel</td>
        <td>Tel2</td>    
         <td>Fax</td>
        <td>Contact Person</td>
        <td>E-mail</td>
        <td>Notes</td>                
    <td></td>
    </tr>
    <p>
      <?
    
    
$hotel_name=$_POST["hotel_name"]; 
$tel=$_POST["tel"]; 
$tel2=$_POST["tel2"]; 
$fax=$_POST["fax"]; 
$contact_person=$_POST["contact_person"]; 
$emal=$_POST["emal"]; 
$notes=$_POST["notes"]; 


?>

<?

 $qryp2="INSERT INTO `hotels` (
`Hotel_Name` ,
`Tel` ,
`Tel2` ,
`Fax` ,
`Contact_Person` ,
`E-mail` ,
`Notes` 
)
VALUES (
'$hotel_name', '$tel', '$tel2', '$fax', '$contact_person', '$emal', '$notes'
)";
    $result2 = mysql_query($qryp2);
    ///// 3. Perform database query
    $qryp = "SELECT * FROM hotels";
    print "qryp: " . $qryp . "<br>";
    $result = mysql_query($qryp)
    or die ("table error: " . mysql_error());
    
    //// 4. Use returned data
    while ($tr_res = mysql_fetch_array($result)){
    
        echo "<tr>";
        
        echo 

        "<td>" . $tr_res['id']  . "</td>

        <td>" . $tr_res["Hotel_Name"] . "</td>
        <td>" . $tr_res["Tel"] . "</td>
        <td>" . $tr_res["Tel2"] . "</td>
        <td>" . $tr_res["Fax"] . "</td>
        <td>" . $tr_res["Contact_Person"] . "</td>
        <td>" . $tr_res["E-mail"] . "</td>
        <td>" . $tr_res["Notes"] . "</td>

        <td> <a href='delete.php?id='$id'><img src='untitled.jpg' width='10' height='11' border='0' /></a></td>
        ";
    
        echo "</tr>";
    }
    
    echo("</table>");






?>

[landing]

Change this

PHP Code:

<a href='delete.php?id='$id'><img src='untitled.jpg' width='10' height='11' border='0' /></a> 

To this

PHP Code:

<a href='delete.php?id='" . $tr_res[id] . "'><img src='untitled.jpg' width='10' height='11' border='0' /></a> 

Providing 'id' is the name of your ID column

[foxed]

Response above got there before I finished writing

[john_zakaria]

till now everythin gworks well but i have a problem in update syntax


i have a syntax error in update and i can not modify it
can u help me?
$qryp = "UPDATE hotels SET
Hotel_Name='$hotel_name',
Tel='$tel',
Tel2='$tel2',
Fax= '$fax',
Contact_Person='$contact_person',
E-mail='$emal',
Notes='$notes'
WHERE id=$id";

1- i made a form and i post its value in this page
2- i echoed these values in my page so it works well
3- but i have a syntax error in mysql statment and i can not modify it
can u help me?

my code is:
<?

$id = $_POST["id1"];
echo ("$id");
$hotel_name=$_POST["hotel_name"];
echo ("$hotel_name");
$tel=$_POST["tel"];
$tel2=$_POST["tel2"];
$fax=$_POST["fax"];
$contact_person=$_POST["contact_person"];
$emal=$_POST["emal"];
$notes=$_POST["notes"];


?>
<?


///// 3. Perform database query
$qryp = "UPDATE hotels SET
Hotel_Name='$hotel_name',
Tel='$tel',
Tel2='$tel2',
Fax= '$fax',
Contact_Person='$contact_person',
E-mail='$emal',
Notes='$notes'
WHERE id=$id";
print "qryp: " . $qryp . "<br>";
$result = mysql_query($qryp)
or die ("table error: " . mysql_error());
while ($tr_res = mysql_fetch_array($result)){
//echo $tr_res["id"] ;
}

?>

[landing]

PHP Code:

///// 3. Perform database query

$qryp = "UPDATE `hotels` SET 

                `Hotel_Name` = '$hotel_name',

                `Tel` = '$tel',

                `Tel2` = '$tel2',

                `Fax` = '$fax',

                `Contact_Person` = '$contact_person',

                `E-mail`= '$emal',

                `Notes` = '$notes'

                 WHERE `id` = '$id' LIMIT 1";



print "qryp: " . $qryp . "<br>";

$result = mysql_query($qryp)

or die ("table error: " . mysql_error());

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

//echo $tr_res["id"] ;

}