子查询返回多于1行的错误

时间:2016-03-08 17:25:43

标签: php html mysql

我有两个想要运行的查询。结果应该是列出的困境/问题行以及问题下面的答案。答案是通过javascript隐藏的,直到我选择按下图像按钮,然后应该显示属于所选问题的答案。

代码:

$result = mysqli_query($mysqli,"SELECT rid, pid, qid, aid, points FROM result WHERE rid=$val");
$answertext = mysqli_query($mysqli, "SELECT answer FROM answer_det WHERE pid=(SELECT pid FROM result WHERE rid=$val) AND qid=(SELECT qid FROM result WHERE rid=$val) AND aid=(SELECT aid FROM result WHERE rid=$val)");

while($row = mysqli_fetch_array($result) AND $row2 = mysqli_fetch_array($answertext))
{
    $resultp = $row['points'];
    $color = "#000000";

if (($resultp >= 1) && ($resultp <= 3))
   $color = "#FF0000";
else if (($resultp >= 3) && ($resultp <= 6))
   $color = "#FF9900";
else if (($resultp >= 6) && ($resultp <= 10))
   $color = "#07d407";

    echo "<div class='question-wrap'>
    <b><small>Dilemma ".$row['qid']." - Answer ". $row['aid'].": </small><span style=\"color: $color\">". $resultp."</span></b> of <b>10  <small>Points</small></b>
    <input type='image' class='answer-toggle' title='Information' src='img/down.png' width='13' height='10'>
    <p class='answer'>". $row2['answertext']."</p></div>";  }

我无法弄清楚出了什么问题。这是我得到的信息:

Warning: mysqli_query(): (21000/1242): Subquery returns more than 1 row in D:\home\site\wwwroot\devlopment\respondent2.php on line 122 Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in D:\home\site\wwwroot\devlopment\respondent2.php on line 125

这是第122行:

$answertext = mysqli_query($mysqli, "SELECT answer FROM answer_det WHERE pid=(SELECT pid FROM result WHERE rid=$val) AND qid=(SELECT qid FROM result WHERE rid=$val) AND aid=(SELECT aid FROM result WHERE rid=$val)");

这是第125行:

while($row = mysqli_fetch_array($result) AND $row2 = mysqli_fetch_array($answertext))

2 个答案:

答案 0 :(得分:2)

使用

之类的查询时
WHERE your_column = (SELECT ... WHERE ...)

子选择必须只返回一行;如果没有,那么你会得到你所看到的错误。

快速解决方案可能是将其更改为

WHERE your_column = (SELECT ... WHERE ... LIMIT 1)

但我会改用联接:

SELECT answer
FROM answer_det
JOIN result USING (pid, qid, aid)
WHERE result.rid = $val

答案 1 :(得分:0)

删除子查询并对表结果执行左连接 或者这样做(你只会看到出现的第一行):

$answertext = mysqli_query($mysqli, "SELECT answer FROM answer_det WHERE pid=(SELECT top 1 pid FROM result WHERE rid=$val) AND qid=(SELECT top 1 qid FROM result WHERE rid=$val) AND aid=(SELECT top 1 aid FROM result WHERE rid=$val)");

您的子查询不能返回多个要比较的值或其他任何值。