Question

这是我的表结构

ID名称值
1个问题50
2持续时间120

现在，我想根据名称显示值。我无法正确执行sql语句。这就是我所做的：

$qry = "SELECT * FROM (
SELECT (SELECT value FROM pq_examsettings) as duration,
       (SELECT value FROM pq_examsettings) as questions
 ) ";
$result= mysql_query($qry);
$row = mysql_fetch_array($result);
$duration = $row['duration'];
$questions = $row['questions'];

有人帮助我如何进行查询，以便我的$ duration和$ questions变量显示120和50。

Answer 1

尝试一下，

SELECT  MAX(CASE WHEN name = 'questions' THEN value END) questions,
        MAX(CASE WHEN name = 'duration' THEN value END) duration 
FROM    pq_examsettings

Answer 2

使用mysqli或pdo函数mysql_ *函数已被正式弃用：

$mysqli = new mysqli($dbserver, $dbmanager, $dbpass, $dbname);
if($mysqli->connect_errno) {
    echo "Error connecting to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}

$query = $mysqli->query("SELECT * FROM pq_examsettings");
if($query->num_rows){
$new_row = array();
while($row = $query->fetch_assoc()){
    foreach($row as $key => $value){
        if($key == 'name'){
            $array_key = $value;
        }
        if($key == 'value'){
            $new_row[$array_key] = $value;
        }
    }
}
$duration = $new_row['duration'];
$questions = $new_row['questions'];
}

Answer 3

我对MySQL不熟悉，知道一种简单的方法将行转换为列并将结果返回到一个选择中 - 看起来像JW给出的答案就是这样做。

另一种选择是只进行两次数据库调用：

$qry = "SELECT value FROM PQ_EXAMSETTINGS WHERE NAME = 'duration'";
$result= mysql_query($qry);
$row = mysql_fetch_array($result);
$duration = $row['value'];

$qry = "SELECT value FROM PQ_EXAMSETTINGS WHERE NAME = 'questions'";
$result= mysql_query($qry);
$row = mysql_fetch_array($result);
$questions = $row['value'];

从一个表中选择多行

3 个答案: