绑定准备语句的多个参数

时间:2015-05-29 13:30:10

标签: php mysql

我的PHP看起来像这样:

$sql1="SELECT @rownum := @rownum + 1 Rank, q.* FROM (SELECT @rownum:=0) r,(SELECT  * ,sum(`number of cases`) as tot, sum(`number of cases`) * 100 / t.s AS `% of total` FROM `myTable` CROSS JOIN (SELECT SUM(`number of cases`) AS s FROM `myTable` where `type`=:criteria and `condition`=:diagnosis) t where `type`=:criteria and `condition`=:diagnosis group by `name` order by `% of total` desc) q"";
$stmt = $dbh->prepare($sql1);
$stmt->bindParam(':criteria', $search_crit, PDO::PARAM_STR);
$stmt->bindParam(':diagnosis', $diagnosis, PDO::PARAM_STR);
$stmt->execute();
$result1 = $stmt->fetchAll(PDO::FETCH_ASSOC);
header('Content-type: application/json');
echo json_encode($result1);

我在这一行收到错误:$stmt->execute();

错误说:

  

PHP致命错误:未捕获的异常' PDOException'消息' SQLSTATE [HY093]:参数号无效'在php / rankings.php:39

堆栈追踪:

  

"#" 0 php / rankings.php(39):PDOStatement-> execute()

     

"#" 1 {main}     在第39行的php / rankings.php中抛出

我该如何解决这个问题?我知道我可以用准备好的声明传递多个变量,但我不太清楚如何去做。

2 个答案:

答案 0 :(得分:3)

您只能在查询中使用一次参数

$sql1="SELECT @rownum := @rownum + 1 Rank, q.* FROM (SELECT @rownum:=0) r,(SELECT  * ,sum(`number of cases`) as tot, sum(`number of cases`) * 100 / t.s AS `% of total` FROM `myTable` CROSS JOIN (SELECT SUM(`number of cases`) AS s FROM `myTable` where `type`=:criteria and `condition`=:diagnosis) t where `type`=:criteria2 and `condition`=:diagnosis2 group by `name` order by `% of total` desc) q";
$stmt = $dbh->prepare($sql1);       
$stmt->execute(array(':criteria' => $search_crit, ':diagnosis' => $diagnosis, ':criteria2' => $search_crit, ':diagnosis2' => $diagnosis));

答案 1 :(得分:1)

您可以像下面这样在execute语句中添加一个数组:

curl: (28) Operation timed out
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