php - 在 PHP 中询问 mysql_num_rows 的正确方法

Question

因为如果没有返回行，mysql_num_rows 返回 false，最好这样做：

$query = mysql_query("SELECT id FROM table WHERE something = 'this'"); 
$result = mysql_num_rows($query);

if ($result) { }

或者我应该这样做：

if ($result >= 1) { }

Answer 1

正确的方法是使用PDO而不是古老的mysql_*函数：

$stmt = $dbh->prepare('SELECT item_id FROM Items WHERE name = :param');
$stmt->bindParam( ':param', $some_name, PDO::PARAM_STR, 127 );
if ( $stmt->execute() )
{
   echo $stmt->rowCount();
   var_dump( $stmt->fetchAll( PDO::FETCH_ASSOC ));
}

Answer 2

合适的

$result = mysql_query("SELECT id FROM table WHERE something = 'this'"); 
if (mysql_num_rows($result)){
    //there are results
}

但是，您可以更轻松地执行此操作，而无需检查

$result = mysql_query("SELECT id FROM table WHERE something = 'this'"); 
while($row = mysql_fetch_assoc($result))
    //there are results
}

请。给你的变量正确的名字

Answer 3

false如果没有返回行，它不会返回，它会在false错误的情况下返回。你可以这样处理：

 if ($result === false) {
    /* An error occurred - do something */
 } else {
    /* $result is set to some number >= 0 */
 }

Answer 4

老实说，我认为

$query = mysql_query("SELECT id FROM table WHERE something = 'this'"); 
if (mysql_num_rows($query)!==FALSE){
    //there are results
}

更合适。

Answer 5

count会返回一个值，不能count再调用mysql_num_rows。它是其中之一。

你可以做

$isExist = mysql_query("Select count(id) from ..."); 
$r = mysql_fetch_array($isExist);
if($r['COUNT(id)'] > 0){
//item exists
}else{
//item doesnt exist
}

如果或者您可以执行以下查询：

$isexist = mysql_query("select * from wcddl_filehosts where downloadid = '".$download[id]."'");
if(mysql_num_rows($isExists)>0){
//we have items
}else{
//we dont have items
}