Php 为什么PDO fetchColumn()在这里不起作用
我试图计算查询返回的行数,我是这样做的:Php 为什么PDO fetchColumn()在这里不起作用,php,pdo,Php,Pdo,我试图计算查询返回的行数,我是这样做的: $what = 'Norman'; $stmt = $conn->prepare('select names as names from names where names = :what'); $stmt->bindParam('what', $what); $stmt->execute(); $rows = $stmt->fetchColumn(); echo 'Rows found '.$rows; $stmt->
$what = 'Norman';
$stmt = $conn->prepare('select names as names from names where names = :what');
$stmt->bindParam('what', $what);
$stmt->execute();
$rows = $stmt->fetchColumn();
echo 'Rows found '.$rows;
$stmt->setFetchMode(PDO::FETCH_ASSOC);
while($row = $stmt->fetch())
{
echo $row['names'] . "<br>";
}
$what='Norman';
$stmt=$conn->prepare('select names as names from names where name=:what');
$stmt->bindParam('what',$what);
$stmt->execute();
$rows=$stmt->fetchColumn();
回显“已找到行”。$行;
$stmt->setFetchMode(PDO::FETCH_ASSOC);
而($row=$stmt->fetch())
{
echo$row['names']。“
”;
}
但我一直一无所获。一片空白。正确的方法是什么?如果要获取返回的行数,请使用
看起来您在这里使用了不正确的函数
$what = 'Norman';
$stmt = $conn->prepare('select names from names where names = ?');
$stmt->execute(array($what));
$rows = $stmt->fetchAll(); // it will actually return all the rows
echo 'Rows found '.count($rows);
foreach ($rows as $row)
{
echo $row['names'] . "<br>";
}
$what='Norman';
$stmt=$conn->prepare('从名称=?'的名称中选择名称');
$stmt->execute(数组($what));
$rows=$stmt->fetchAll();//它实际上将返回所有行
回显“找到行”。计数($Rows);
foreach($行作为$行)
{
echo$row['names']。“
”;
}
或者,您可以通过使用一维数组而不是二维数组,使其更加整洁
$rows = $stmt->fetchAll(PDO::FETCH_COLUMN, 0);
echo 'Rows found '.count($rows);
foreach ($rows as $name)
{
echo $name . "<br>";
}
$rows=$stmt->fetchAll(PDO::FETCH_列,0);
回显“找到行”。计数($Rows);
foreach($行作为$name)
{
echo$name。“
”;
}
但是您必须首先检查PDO错误,是的,这个fetchColumn()在这里看起来很不一致。您是否收到
“找到的行数”的任何输出。$Rows代码>?@YourCommonSense实际上,我刚刚意识到fetchColumn()正在返回一个名称。我原以为它会返回一个数字(select中的行数),我以为这对mysql不起作用,但它实际上是返回表中的行数。我通篇都读到rowCount()不适用于selects{mazed}。@Norman它不适用于某些数据库系统。它没有列出检测到的工作项。phpdelusions.com:)
$rows = $stmt->fetchAll(PDO::FETCH_COLUMN, 0);
echo 'Rows found '.count($rows);
foreach ($rows as $name)
{
echo $name . "<br>";
}