-1

このスニペットのどこにエラーがありますか?

<?php
include('db.php');
switch ($_REQUEST['tab']) {
  case 'categorie':
    $sql = mysql_query("SELECT * FROM categorie");
    echo "INSERT INTO categorie ( id, nome, free, pos ) VALUES " ;
    $a=0;
    while($tmp = mysql_fetch_array($sql))
    {
      if($a!=0)
        echo ",";
      echo "(".$tmp['id'].", '".$tmp['nome']."', ".$tmp['free'].", ".$tmp['pos'].")";
      $a=1;
    }
    break;
  case 'articoli':
    $sql = mysql_query("SELECT * FROM articoli");
    echo "INSERT INTO articoli ( id, titolo, testo, sottocategoria, free, pos ) VALUES " ;
    $a=0;
    while($tmp = mysql_fetch_array($sql))
    {
      if($a!=0)
        echo ",";
      echo "(".$tmp['id'].", '".$tmp['titolo']."', '".$tmp['testo']."', ".$tmp['sottocategoria'].", ".$tmp['free'].", ".$tmp['pos'].")";
      $a=1;
    }     
    break;
  case 'sottocategorie':
    $sql = mysql_query("SELECT * FROM sottocategorie");
    echo "INSERT INTO sottocategorie ( id, nome, categoria, free, pos ) VALUES " ;
    $a=0;
    while($tmp = mysql_fetch_array($sql))
    {
      if($a!=0)
        echo ",";
      echo "(".$tmp['id'].", '".$tmp['nome']."', ".$tmp['categoria'].", ".$tmp['free'].", ".$tmp['pos'].")";
      $a=1;
    }
    break;
}
?>

の近くに構文エラーがあると言います。

コード全体を投稿します.................................................................. ...................................................

4

1 に答える 1