php的問題pg_query

Question

我遇到了PHP的問題。我想用PHP進行表格更新。

$name = pg_escape_string($_POST['NAME']); 
    $place = pg_escape_string($_POST['PLACE']); 
    $zip = pg_escape_string($_POST['ZIP']); 
    $nation = pg_escape_string($_POST['NATION']); 

    $name = "'" . $name . "'"; 
    $place = "'" . $place . "'"; 
    $zip = "'" . $zip . "'"; 
    $nation = "'" . $nation . "'"; 
    $club_id = "'" . $club_id . "'";   



    $result = pg_query($db_connect, "UPDATE club SET name_c = $name, place_c = $place, zip_c = $zip, nation_c = $nation WHERE id_c = $club_id;"); 

爲什麼不起作用？

謝謝！

Answer 1

您的代碼中沒有定義club_id。並且爲了避免任何問題和清理代碼，我會這樣做：

$club_id = 1; 
$dbconn = pg_connect("connectionstring"); 
$sql = 'UPDATE club SET name_c = $1, place_c = $2, zip_c = $3, nation_c = $4 WHERE id_c = $5;'; 
$result = pg_query_params($dbconn, $sql, array(
    $_POST['NAME'], 
    $_POST['PLACE'], 
    $_POST['ZIP'], 
    $_POST['NATION'], 
    $club_id 
)); 

// Do what you need 

它會爲您跳過值，因此無需處理奇怪的情況。

http://php.net/manual/en/function.pg-query-params.php