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To display two records from a table based on the selection

Posted: Tue Feb 18, 2003 3:59 pm
by minds_gifts
Hello,

I'm displaying some topic names on a list box, based on the selection and when i hit a button, it displays the sub-topic names in another list box.fine, upto here.

Now, when i select a sub-topic, and hit this button, I want to show the article names and article author related to this sub-topic.

This articles table has topic_id, subtopic_id, article_id and so on....
Also, could you please tell me where should i place the form fields to show the article names and article content??

Do i need to loop them through array???I commented my code well, so somebody can easily point out where exactly i can modify it.

Many thanks

Code: Select all

<?php 
//DB connection details 
include "../dbconnection.php";

?> 
<html> 
<head> 
<title>Delete articles</title> 
</head> 
<body>     

<?php 

// If the Change button is pressed. 
if (isset($_POST["send"])) 
{ 
// Check to see that the user selected an item from the topics. 
    if(isset($_POST["TOPIC_ID"])) 
    { 
    
    $TOPIC_ID = (int)$_POST["TOPIC_ID"]; 
    $q = mysql_query("select * from topic where TOPIC_ID=$TOPIC_ID"); 
         
        if (!mysql_num_rows($q)) 
        $TOPIC_ID = 0;       
    } 
    else {   
            $TOPIC_ID = 0;   
         }    
     
// If the user has selected a topic and a subtopic. 
   if((isset($TOPIC_ID)) && (isset($_POST["SUBTOPIC_ID"]))) 
   { 
   $SUBTOPIC_ID = (int)$_POST["SUBTOPIC_ID"]; 
   $q = mysql_query("select * from subtopic where TOPIC_ID=$TOPIC_ID and SUBTOPIC_ID=$SUBTOPIC_ID"); 
      
       if (!mysql_num_rows($q)) 
    $SUBTOPIC_ID = 0; 
   } 
   else { 
            $SUBTOPIC_ID = 0; 
 
         }       
     
//If the user has selected the topic, subtopic, print the article names and article content    
//I'm not sure if this part of the code is correct...
    if ((isset($TOPIC_ID)) && (isset($SUBTOPIC_ID)) && (isset($_POST["ARTICLE_ID"])))  
    $select_sql = "select ARTICLE_NAME, ARTICLE_CONTENT from articles where SUBTOPIC_ID=$SUBTOPIC_ID";
    $result_id = mysql_query($select_sql) or die('Error retrieving records from articles.<br />Mysql Reported: '.mysql_error());
// Till here........... 
}
// If we got here, we need to display the form... 

echo "<form method=post action="$PHP_SELF">"; 
echo "<select name=TOPIC_ID>"; 

$q = mysql_query("SELECT * FROM topic ORDER BY TOPIC_NAME"); 
while ($l = mysql_fetch_array($q)) 
{ 
     $selected=""; 
     if ($l["TOPIC_ID"] == $TOPIC_ID) 
         $selected = "selected=1"; 
     echo "<option value="".$l["TOPIC_ID"]."" $selected>".$l["TOPIC_NAME"]; 
} 

echo "</select><br><br>"; 

echo "<select name=SUBTOPIC_ID>"; 
    if ($TOPIC_ID) 
    { 
    $q=mysql_query("SELECT * FROM subtopic WHERE TOPIC_ID=$TOPIC_ID"); 
    while ($l=mysql_fetch_array($q)) 
        { 
        $selected=""; 
        if ($l["SUBTOPIC_ID"]==$SUBTOPIC_ID) 
            $selected="selected=1"; 
        echo "<option value="".$l["SUBTOPIC_ID"]."" $selected>".$l["SUBTOPIC_NAME"]; 
        } 
    } else { 
     echo "<option value=0>Please select topic first"; 
           } 
echo "</select><br><br>"; 

echo " <input type=submit name=send value="Change"> "; 
echo "</form>"; 

?>
</body> 
</html>