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View Full Version : Retrieving the other two records from a table based on the selection in a list box



chinni
02-18-2003, 08:56 PM
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




<?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>



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