PHP：准备好的语句，需要 IF 语句帮助

admin

PHP：准备好的语句，需要 IF 语句帮助

sql

我有以下代码：

$sql = "SELECT name, address, city FROM tableA, tableB WHERE tableA.id = tableB.id";

if (isset($price) ) {
    $sql = $sql . ' AND price = :price ';
}
if (isset($sqft) ) {
    $sql = $sql . ' AND sqft >= :sqft ';
}
if (isset($bedrooms) ) {
    $sql = $sql . ' AND bedrooms >= :bedrooms ';
}


$stmt = $dbh->prepare($sql);


if (isset($price) ) {
    $stmt->bindParam(':price', $price);
}
if (isset($sqft) ) {
    $stmt->bindParam(':price', $price);
}
if (isset($bedrooms) ) {
    $stmt->bindParam(':bedrooms', $bedrooms);
}


$stmt->execute();
$result_set = $stmt->fetchAll(PDO::FETCH_ASSOC);

我注意到的是我拥有的冗余的多个 IF 语句。

问题：有什么方法可以清理我的代码，以便我没有这些多个 IF 语句用于准备好的语句？

阅读 241

2021-07-01

共1个答案

admin

这与用户最近在我的书 SQL Antipatterns 的论坛上问我的问题非常相似。我给了他类似的答案：

$sql = "SELECT name, address, city FROM tableA JOIN tableB ON tableA.id = tableB.id";

$params = array();
$where = array();

if (isset($price) ) {
    $where[] = '(price = :price)';
    $params[':price'] = $price;
}
if (isset($sqft) ) {
    $where[] = '(sqft >= :sqft)';
    $params[':sqft'] = $sqft;
}
if (isset($bedrooms) ) {
    $where[] = '(bedrooms >= :bedrooms)';
    $params[':bedrooms'] = $bedrooms;
}

if ($where) {
  $sql .= ' WHERE ' . implode(' AND ', $where);
}

$stmt = $dbh->prepare($sql);

$stmt->execute($params);
$result_set = $stmt->fetchAll(PDO::FETCH_ASSOC);

2021-07-01