MySQL计数匹配行数总是返回1

Question

我在mySQL中检查了下面的SQL，它返回用户拥有的产品的正确计数。我试图将该语句移入PHP以基于产品计数来执行条件代码。但是，回声总是返回1.我错过了什么？

$sql = "select count(product_id) from dap_users_products_jn where product_id not in(13,34) and user_id = 1"; 
$stmt = $dap_dbh->prepare($sql); 
$stmt->execute(); 
$stmt = $stmt->rowCount(); 
echo "the product count is: ".$stmt; 

Answer 1

你总是会得到使用COUNT()这样，当你有得到一个结果集即使COUNT()结果是零一行返回。或者你怎么知道结果是什么？如果您需要COUNT()的结果，则需要正常查看查询结果，然后检查COUNT()操作的值。

$sql = "select count(product_id) AS prod_count from dap_users_products_jn where product_id not in(13,34) and user_id = 1"; 
$stmt = $dap_dbh->prepare($sql); 
$stmt->execute(); 
$result = $stmt->fetch(PDO::FETCH_OBJ); 
echo "the product count is: ".$result->prod_count; 

Answer 2

你应该做的是，拿到专栏。 PDO有获取列的函数fetchColumn函数。你应该用它代替$stmt->rowCount()

$sql = "select count(product_id) from dap_users_products_jn where product_id not in(13,34) and user_id = 1"; 
$stmt = $dap_dbh->prepare($sql); 
$stmt->execute(); 
$stmt = $stmt->fetchColumn(); 
echo "the product count is: ".$stmt; 

由于SQL取出已行数为什么使用PDO的rowCount时（）？