Question

此代码的主要问题是什么？测试时 SELECT count（获胜者）AS num FROM subscribe_user WHERE获胜者=＆＃39;是＆＃39; AND hitdate = CURDATE（）这个代码在数据库sql中的sql中它显示num为3 但是下面显示的代码只显示了应该为3的代码。

$link = mysqli_connect("localhost", "root", "", "bot");
$count_winner=mysqli_query($link,"SELECT count(winner) AS num FROM 
subscribe_user WHERE winner ='yes' AND hitdate=CURDATE()")or 
 die(mysql_error());
$w=mysqli_num_rows($count_winner);
echo $w;

Answer 1

您所看到的是预期的。 mysqli_num_rows为您提供结果中的行数 - 1。结果中的num字段会显示count(winner) - 3的值。这些是不同的指标。

但我想在DB SQL中打印3。我怎样才能做到这一点？

而不是抓住mysqli_num_rows，实际上是读取结果：

$result = mysqli_query('...') or die('...');    
$w = $result->fetch_assoc()['num'];

此外，mysql_error()不会与mysqli_ api混在一起。它应该是mysqli_error()并将连接作为其（单个）参数传递。

参考：

http://php.net/manual/en/mysqli.error.php

Answer 2

$link = mysqli_connect("localhost", "root", "", "bot");
$count_winner = mysqli_query($link,"SELECT count(winner) AS num FROM subscribe_user WHERE winner ='yes' AND hitdate=CURDATE()") or mysqli_error($link));
$row = mysqli_fetch_array($count_winner);
echo $row['num']; //Will output the count of number of winners

mysqli_num_rows()函数返回结果中的行数集。

mysqli_num_rows无效

2 个答案: