How to display n -n relationship in php : Many to many relationship

If I have following tables


Many to many relationship

Category A
ID , CategoryName

CategoryB


ID_CategorB , FK_CategoryA, CategoryName


Third Table Relationship Table

Relation
ID ,ID_CategoryB

How can I show the values in "select "  box
Puneet AroraFounderAsked:
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Mike EghtebasDatabase and Application DeveloperCommented:
Table_CategoryA
ID_A     -- PK
CategoryName

Table_CategoryB
ID_B     -- PK
CategoryName

Table_A_B      <-- this is known as bridge table
ID_A     -- FK
ID_B     -- FK


You can make write SQLs you need from these table like:

Select a.CategoryName as Cat_A, b.CategoryName As Cat_B
From Table_CategoryA a inner join Table_A_B ab
on a.ID_A = ab.ID_A inner join Table_CategoryB b
on ab.ID_B = b.ID_B

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0
Puneet AroraFounderAuthor Commented:
Thanks for the reply :

I want   to show it in select box , then uses these multiple values
to select some more in the Category  C table , In php do I need to simply use for loop
to get all the values  and show in "select option box " and use java script to transfer
it to input box with multiple values ? Question is How to ?
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hieloCommented:
In php you need to query the db and iterate over the records using a while loop. Since you seem to need just the <option> tags, echo them directly within the while -construct:
	$host="localhost";
	$username="XXX";
	$password="YYY";
	$db_name="yourDBName";

	$mysqli = new mysqli("$host", "$username", "$password", "$db_name");

	/* check connection */
	if (mysqli_connect_errno()) {
	    printf("Connect failed: %s\n", mysqli_connect_error());
	    exit();
	}

	// The way this is setup, on the client-side you need to send two fields:
	// category: This should have the value that the user selected on the first list
	// ajax-request: you can set this to anything.  It is just meant to allow the
	// if clause below to execute and give you only the <option>s for the second dropdown
	if( array_key_exists('ajax-request',$_POST) )
	{
		$catcode = $_POST['categoryA'];
		
		$sql ="SELECT B.ID_CategoryB as `B_ID`, B.CategoryName as `B_CategoryName`
				FROM `CategoryB` B INNER JOIN `ThirdTable` C ON B.ID_CategoryB=C.ID_CategoryB
				WHERE B.FK_CategoryA=? ORDER BY B.CategoryName ASC";
				
		$stmt = $mysqli->prepare($sql);
		if($stmt)
		{
			// assuming that the value you are passing from the first select 
			// is an integer, then use "i" as the first argumen to bind_param()
			// If it is a string, use "s" instead
			$stmt->bind_param("i", $_POST['category']);
			
			$stmt->execute();
			
			$stmt->bind_result($B_ID, $B_CategoryName);
			
			while ( $stmt->fetch() ) {
				echo sprintf('<option value="%s">%s</option>', $B_ID, htmlentities($B_CategoryName, ENT_QUOTES, "UTF-8"));
			}
			$stmt->close();	
		}
		$smt=null
		exit;
	}

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The ajax callback function will "see" just the options and append them to the second drop down (like the example I referred you in the other question).
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