「テーブルに値が挿入されました」というメッセージをどのように出力するのですか？

Question

可能性の重複：
​​3210

<?php 
    error_reporting(0); 
    $con=mysql_connect("localhost","root",""); 

    if (!$con) { 
    die('could not connect:'.mysql_error()); 
    } 

    mysql_select_db("final?orgdocs",$con); 

    $org_name = $_POST["org_name"]; 
    $org_type = $_POST["org_type"]; 
    $org_code = $_POST["org_code"]; 
    $description = $_POST["description"]; 

    $stmt = "INSERT INTO organization 
      VALUES('".$org_name."','".$org_type."','".$org_code."','".$description."')"; 
    $result = mysql_query("SELECT * FROM organization WHERE org_name = '$org_name' "); 

    echo '<TABLE BORDER = "1">'; 
    $result1 = $result; 

    while ($row = mysql_fetch_array($result1)){ 
    echo '<TR>'.'<TD>'.'Organization Name'.'</TD>'.'<TD>'.'Organization Type'.'</TD>'.'<TD>'.'Organization Code'.'</TD>'.'<TD>'.'Description'.'</TD>'.'<TD>'.'Constitution'.'</TD>'; 
    echo '</TR>'; 
    echo '<TR>'.'<TD>'.$row['org_name'].'</TD>'.'<TD>'.$row['org_type'].'</TD>'; 
    echo '<TD>'.$row['org_code'].'</TD>'.'<TD>'.$row['description'].'</TD>'.'<TD>'; 
    echo '</TR>'; 
    }  
    echo '</TABLE>'; 
?> 

Answer 1

$count = mysql_affected_rows(); 
    echo "$count values inserted to table”; 

Answer 2

Shortest answer：echo 'values inserted to table';