このプログラムストリップは検索されたディレクトリを出力しますが、見つからない場合はメッセージを出力する必要があります

Question

this program must output a directory that you searched for how ever if its not found a message must appear that the org_name is not found,i don't know how to do that, i keep trying on some if-else but it just won't output it. 

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

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

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

    $org_name = $_POST["org_name"]; 
    $position = $_POST["position"]; 

    $result = mysql_query("SELECT * FROM directory WHERE org_name = '$org_name' OR position = '$position' ORDER BY org_name"); 



    echo '<TABLE BORDER = "1">'; 
    $result1 = $result; 
    echo '<TR>'.'<TD>'.'Name'.'</TD>'.'<TD>'.'Organization Name'.'</TD>'.'<TD>'.'Position'.'</TD>'.'<TD>'.'Cell Number'.'</TD>'.'<TD>'.'Email-Add'.'</TD>'; 
    echo '</TR>'; 

    while ($row = mysql_fetch_array($result1)){ 

    echo '<TR>'.'<TD>'.$row['name'].'</TD>'.'<TD>'.$row['org_name'].'</TD>'; 

    echo '<TD>'.$row['position'].'</TD>'.'<TD>'.$row['cell_num'].'</TD>'.'<TD>'.$row['email_add'].'</TD>'; 
    echo '</TR>'; 
    } 

    echo '</TABLE>'; 

    ?> 

Answer 1

mysql_num_rowsは、結果の行数をクエリで確認します。

http://de2.php.net/manual/en/function.mysql-num-rows.php

if (mysql_num_rows($result)>0) { 
    // your thing above with mysql_fetch_array($result1) etc 
} else { 
    echo 'nor found'; 
}