我问了一个类似的问题,但是无法完全让查询给我想要的结果,它输出了一个问题,但不是正确的问题。我将在这里更好地澄清这个问题,并对表格进行一些修改。我有一个程序,如果一个人键入“让我们谈谈”,该程序会从表格中向用户询问一些问题(例如 20 qns)
我有 3 张桌子。
(peopleiknow) (questions) (questionsasked)
NID FName QID Question QID PID Answer
1 Mark 1 Are you human 1 1 'Definately yes.'
2 Sue 2 Do you like soup? 1 2 'Most of the time.'
2 2 'Yes especially tomato.'
(*问题表一开始可以为空)
基本上,该程序会为不在 questionsasked 表中的用户“$name”获取第一个问题。然后提出问题并将答案添加到问题表中。(起初我试图得到正确的问题)
与 php 交互的 javascript/jQuery 片段如下:
var uname='Lindsay'; //will be assigned dynamically after I get this fixed.
$.post('ajax/getQuestions.php', { name: uname },function(data2) {
var qn = data2.value1; //question
var pid = data2.value2; // PID
var qid = data2.value3; //QID
alert(qn);
alert(pid);
alert(qid);
var answer1 = prompt(qn, 'Please answer here');
$.post('ajax/addQuestionAsked.php', {answer: answer1, PID: pid, QID: qid},
function(data3) {alert('added question');} );
//$('div#PlaceOfResponse').html(data3);}
});
PHP文件的重要部分如下:
*连接部分省略 * $link 是连接
if (isset($_POST['name']) && !empty($_POST['name']))
{
$name = trim($_POST['name']); //could do better protection..
$query1 = "SELECT * //(q.Question, q.QID, p.person)?
FROM questions q
INNER JOIN questionsasked qa
ON q.QID = qa.QID
INNER JOIN peopleiknow p
ON qa.PID = p.NID
WHERE qa.Asked='N' OR qa.Asked IS NULL
AND p.Person = '$name'";
上面的查询不太奏效,也尝试了下面评论中的那个..但没有奏效..
/*
$query1 = "SELECT FIRST(Question)
FROM questions q
WHERE NOT EXISTS
(SELECT *
FROM questionsasked qa
RIGHT JOIN peopleiknow p
ON p.NID = qa.PID
INNER JOIN questions q
ON qa.QID = q.QID
WHERE q.QID = qa.QID
AND p.FName = '$name')";
*/
if ($result = mysqli_query($link, $query1))
{
if (0 == mysqli_num_rows($result))
{
echo('all asked');
}
else
{
/* fetch associative array */
$row = mysqli_fetch_assoc($result);
{header('Content-Type: application/json');
$question = $row['Question'];
$PID = $row['PID'];
$QID = $row['QID'];
$response = array('value1' => $question, 'value2' => $QID, 'value3' => $PID);
echo json_encode($response);
}
}
}
}
else {
?><script>alert('$name');</script>
<?php
echo 'variable \'name\' not set properly';}
/* close connection */
mysqli_close($link);
?>
我几乎可以正常工作,但是当我将答案添加到问题表中,并刷新并说“让我们谈谈”时,它又问了同样的问题。然后我稍微更改了代码,现在 js 文件中的警报现在正在为所有 3 个发回的变量发出警报(未定义)。(在我稍微修改表之前工作)。现在看不出它为什么这样做。
这是我对查询的最新尝试:
$query1 = "SELECT FIRST(q.question,p.NID,q.QID)
FROM questions q
LEFT JOIN questionsasked qa
ON q.QID = qa.QID
LEFT JOIN peopleiknow p
ON qa.PID = p.NID
WHERE qa.QID
NOT IN (SELECT * FROM questionsasked WHERE q.QID = qa.QID)
AND p.FName = '$name'";