php - when in 从句与 mysql pdo

Question

我想在 mysql pdo 中使用“where in”子句运行一个简单的选择查询。

$开始= 12；$end = 14;

for($i=$start; $i<=$end;$i++)
{
$limitin[] = array($qstep,$i);
}

数组是

Array
(
    [0] => Array
        (
            [0] => 3
            [1] => 12
        )

    [1] => Array
        (
            [0] => 3
            [1] => 13
        )

    [2] => Array
        (
            [0] => 3
            [1] => 14
        )

) 

然后

$questionmarks = str_repeat("?,", count($limitin)-1) . "?";

查询是

$getans = $this->db->prepare("SELECT * 
                              FROM answers
                              WHERE qstep = ? 
                              and ansid in ($questionmarks) ");


$getans->execute($limitin); //$limitin is the array.

我得到一个空白的结果集和一个通知，它是“通知:数组到字符串的转换”

Answer 1

$limitin 是一个多维数组，不是在 $getans->execute($limitin); 中使用的有效数组。您需要创建一个数组。尝试类似 ->

$inarray=array();

foreach($limitin as $key=>$val){
       if($key==0) {$inarray[]=$val[0];} // set the qstep as the 1st array value
       $inarray[] = $val[1]; // add each value
}

$questionmarks = str_repeat("?,", count($limitin)-1) . "?";

$getans = $this->db->prepare("SELECT * 
                              FROM answers
                              WHERE qstep = ? 
                              and ansid in ($questionmarks) ");


$getans->execute($inarray);

编辑

正如@Fred-ii- 所指出的，bindParam() 是另一种方法，例如

$questionmarks = str_repeat("?,", count($limitin)-1) . "?";

$getans = $this->db->prepare("SELECT * 
                              FROM answers
                              WHERE qstep = ? 
                              and ansid in ($questionmarks) ");


foreach($limitin as $key=>$val){
       if($key==0) {$getans->bindParam(1,$val[0]);} // set the qstep as the 1st array value
       $getans->bindParam($key+2,$val[1]); // add each value
}

$getans->execute();