Question

我正在尝试检索一个整数表元素（max），取决于课程的3或4，并将其与已输入表格的元素数量进行比较。

以下是我用来获取课程最高入学率和注册课程总数的代码：

//Store the value of the course enrolment limit
$course1 = mysql_real_escape_string($_POST['coursename']);
$qw2 = "SELECT max FROM course WHERE name = '$course1'";
$courselimit = mysql_query($qw2) or die (mysql_error());


//check how many students are enrolled in the course
$qw3 = "SELECT COUNT(*) FROM enrolled WHERE course = '$course1' LIMIT 4";
$totalenrolled = mysql_query($qw3) or die (mysql_error());

不幸的是，即使表条目数超过了课程限制，我的if语句似乎也没有工作，因为它注册为true。

    if ($totalenrolled >= $courselimit ){
        die('That course has reached the enrolment limit, please select another');
    }

EDIT1：这是课程表的转储，我已经尝试了Alexey的解决方案，但遗憾的是该声明似乎尚未运行。我已将问题隔离到$ courselimit变量

--
-- Table structure for table `course`
--

CREATE TABLE `course` (
`name` varchar(50) DEFAULT NULL,
`code` varchar(10) DEFAULT NULL,
`max` int(2) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

--
-- Dumping data for table `course`
--

INSERT INTO `course` (`name`, `code`, `max`) VALUES
('HTML', 'PROG-1288', 4),
('Javascript', 'PROG-2283', 3),
('Dreamweaver', 'MEDA-1380', 4),
('Photoshop', 'PHOT-1382', 3);

这是我将值插入并尝试读取的表的转储：

--
-- Table structure for table `enrolled`
--

CREATE TABLE `enrolled` (
`student` varchar(50) DEFAULT NULL,
`studentnum` varchar(50) DEFAULT NULL,
`course` varchar(50) DEFAULT NULL,
FULLTEXT KEY `student` (`student`,`studentnum`,`course`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1;

--
-- Dumping data for table `enrolled`
--

INSERT INTO `enrolled` (`student`, `studentnum`, `course`) VALUES
('graham', '987654', 'HTML'),
('bruce', '123456', 'HTML'),
('jane', '111222', 'HTML');

Answer 1

您需要使用mysql_num_rows（）。

//Store the value of the course enrolment limit
$course1 = mysql_real_escape_string($_POST['coursename']);
$qw2 = "SELECT max as `maxitem` FROM course WHERE name = '$course1'";
$res = mysql_query($qw2) or die (mysql_error());
$result = mysql_fetch_assoc($res);
$courselimit = $result['maxitem'];


//check how many students are enrolled in the course
$qw3 = "SELECT COUNT(*) FROM enrolled WHERE course = '$course1'";
$res = mysql_query($qw3) or die (mysql_error());
$totalenrolled = mysql_num_rows($res);

Answer 2

您是否回复了$totalenrolled和$courselimit这两个值，并根据数据表中的数据进行了检查，以确保您获得所需内容？

此外，LIMIT 4中的$qw3无法做任何事情 - 因为您使用COUNT(*)只会有一个结果。

Answer 3

首先，忘记使用mysql，而不是mysqli_或PDO进行所有操作，更加强大和安全！

You can get basic understanding of mysqli_ from here

现在，问题：

在if条件下，当您必须在objects上执行此操作时，您将在value上进行隔离：

<强>解决方案

在比较之前获取值：

$row = mysqli_fetch_array($courselimit,MYSQLI_ASSOC); /* associative array */
$row2 = mysqli_fetch_array($totalenrolled,MYSQLI_ASSOC); /* associative array */

然后比较它们

if ($row['max'] >= $row2['count'] ) //assuming count is column name for 2nd resukt

Answer 4

您没有获取Mysql对象。试试这个我希望能有效。

//Store the value of the course enrolment limit
$course1 = mysql_real_escape_string($_POST['coursename']);
$qw2 = "SELECT  COUNT(*) as count1 FROM course WHERE name = '$course1'";
$courselimit = mysql_fetch_assoc((mysql_query($qw2)) or die (mysql_error());


//check how many students are enrolled in the course
$qw3 = "SELECT COUNT(*) as count2 FROM enrolled WHERE course = '$course1' LIMIT 4";
$totalenrolled = mysql_fetch_assoc(mysql_query($qw3)) or die (mysql_error());


 if ($totalenrolled['count2'] >= $courselimit['count1 '] ){
    die('That course has reached the enrolment limit, please select another');
}

将表条目的数量与最大数量进行比较

4 个答案: