Generating a random result & updating it using buttons...

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ladyath
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Posts: 1
Joined: Thu Mar 03, 2011 7:54 am

Generating a random result & updating it using buttons...

Post by ladyath »

Hi
I generate a random result from a MySQL database, but I need a button that updates the results so it is not displayed again. My problem lies with the UPDATE command. How do I update the id that was generated to change 'used' to 'yes'?

Table structure is a simple 'id, riskid, used' structure

Unfortunately I know incredibly little about programming and I am not sure how to complete the php code to work. Help would be very much appreciated!

Code snippets below:

Code: Select all

<?php 
include 'includes/db_write.php'; 
$conn = @mysql_connect($db_host,$db_usr,$db_pass); 
    if (!$conn) { 
    echo( "<P>Unable to connect to the database server (".$db_host.") at this time.</P>" ); 
    exit(); 
    } 

    @mysql_select_db($db_name, $conn); 
    if (! @mysql_select_db($db_name) ) { 
    echo( "<P>Unable to locate the ".$db_name." database at this time.</P>" ); 
    exit(); 
    } 
     
       $sql = "SELECT * FROM idgen where used='no' ORDER BY RAND() limit 1"; 
    $result = mysql_query($sql); 
    if (!$result) { 
    echo("<P>Error performing query: " . 
    mysql_error() . "</P>"); 
    exit(); 
    } 
?> 


<table align="center" class="table2" style="width:300px"> 
<tr><td class="td1">Risk ID</td></tr> 
<tr>     


<?php 
while ( $row = mysql_fetch_array($result) ) { 
$msg .= "<td class='td2'>".$row['riskid']."</td>\r\n"; 
$msg .= "</tr>\r\n"; 
} 
$msg .= "</table>\r\n"; 
echo "<p>". $msg ."</p>"; 
?> 

<br> 
   
 <table width="800" border="0" align="center"> 
    <tr><td align="center"><form action="index.php" method="post"><input type="submit" value="Use This ID" /> 
     
    <?php 
include 'includes/db_write.php'; 
$conn = @mysql_connect($db_host,$db_usr,$db_pass); 
    if (!$conn) { 
    echo( "<P>Unable to connect to the database server (".$db_host.") at this time.</P>" ); 
    exit(); 
    } 

    @mysql_select_db($db_name, $conn); 
    if (! @mysql_select_db($db_name) ) { 
    echo( "<P>Unable to locate the ".$db_name." database at this time.</P>" ); 
    exit(); 
    } 

$sql = "UPDATE `idgen` SET `used` = 'yes' WHERE `riskid` = ??????????";    // <--------  How do I complete this?  And how do I make sure it only executes the update if the button is pressed?
$result = mysql_query($sql,$conn) or die("Could not execute sql!");
?>

         
    </form></td></tr> 
</table> 

  <table width="800" border="0" align="center"> 
           <tr><td align="center"><form><input type=button value="Generate New ID" onClick="window.location.reload()"></form></td></tr> 
        
</table> 
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