Question

我正在使用MySQL和PHP，我需要做这样的事情（伪代码）：

if (sql row exists where username='bob')
{
    // do this stuff
}

Answer 1

如果您使用的是mysql数据库，请使用以下 -

$query = "SELECT username from my_table where username='bob'";
$result = mysql_query($query);

if(mysql_num_rows($result) > 0)
{
    // row exists. do whatever you would like to do.
}

如果您想像alex建议的那样使用PDO (PHP Data Object)，请使用以下代码 -

$dbh = new PDO("mysql:host=your_host_name;dbname=your_db_name", $user, $pass);
$stmt = $dbh->prepare("SELECT username from my_table where username = ':name'");
$stmt->bindParam(":name", "bob");
$stmt->execute();

if($stmt->rowCount() > 0)
{
    // row exists. do whatever you want to do.
}

Answer 2

Sayem的答案有最多的赞成，但我认为这对于PDO是不正确的。

来自PHP docs：

对于大多数数据库，PDOStatement :: rowCount（）不会通过SELECT语句返回受影响的行数。相反，使用PDO :: query（）发出SELECT COUNT（*）语句，其语句与预期的SELECT语句相同，然后使用PDOStatement :: fetchColumn（）来检索将返回的行数。

$sql = "SELECT COUNT(*) FROM fruit WHERE calories > 100";
if ($res = $conn->query($sql)) {

  /* Check the number of rows that match the SELECT statement */
  if ($res->fetchColumn() > 0) {

       /* Issue the real SELECT statement and work with the results */
       $sql = "SELECT name FROM fruit WHERE calories > 100";
       foreach ($conn->query($sql) as $row) {
           print "Name: " .  $row['NAME'] . "\n";
       }
  }
  /* No rows matched -- do something else */
  else {
      print "No rows matched the query.";
  }
}

$res = null;
$conn = null;

Answer 3

另一种方法

$user = "bob";
$user = mysql_real_escape_string($user);
$result = mysql_query("SELECT COUNT(*) AS num_rows FROM my_table WHERE username='{$user}' LIMIT 1;");
$row = mysql_fetch_array($result);
if($row["num_rows"] > 0){
   //user exists
}

检查PHP是否存在SQL行

3 个答案: