这是我尝试的代码：
2
1775$query = $database->prepare('SELECT * FROM table WHERE column LIKE "?%"');
$query->execute(array('value'));
while ($results = $query->fetch())
{
echo $results['column'];
} 
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发布后就找到答案了：
$query = $database->prepare('SELECT * FROM table WHERE column LIKE ?'); $query->execute(array('value%')); while ($results = $query->fetch()) { echo $results['column']; }