我是新来的，不是很方便PHP和MySql。虽然我设法得到了一些工作（在一些朋友的帮助下;-)并且它可能有点凌乱），但我现在已经陷入困境。
这就是情况; 我有三张桌子; 参与者，项目和面试。
参与者表格在访谈开始时填写了我面试的人员的详细信息。
项目表由我在访谈中使用的一组主题组成。
在面试中，面试表将填满数据。
我的工作是，我让系统在一组复选框旁边的弹出页面中显示项目，文本字段和选择框。 输入表单字段的数据将添加到访谈表中。
我仍然需要将参与者的ID以及项目和项目ID添加到面试表中。有人可以帮助我吗？
这是弹出页面调用的代码;
<div class="communication">
<a href="popups/DeveloperCommunication.php" onclick="window.open('popups/DeveloperCommunication.php','communication', 'width=800,height=350,scrollbars=no,toolbar=no,location=no'); return false">
<?php
// get the records from the database
if ($result = $conn->query('SELECT Categories FROM categories WHERE ID="2"'))
{
// display records if there are records to display
if ($result->num_rows > 0)
{
while ($row = $result->fetch_object())
{
// set up a row for each record
echo $row->Categories;
}
}
// if there are no records in the database, display an alert message
else
{
echo "No results to display!";
}
}
// show an error if there is an issue with the database query
else
{
echo "Error: " . $conn->error;
}
?>
</a></a>
</div>
这是弹出页面的代码;
<?php
include '../../include/dbh.inc.php';
date_default_timezone_set('Europe/Amsterdam');
?>
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Developer Communication</title>
<link href="../../css/chapterDeveloper.css" rel="stylesheet" type="text/css" media="screen">
</head>
<body>
<div class="popupTitle">
<p>Developer Communication</p>
</div>
<?php
$comment = "";
// connect to the database
$result = $conn->query('SELECT participant.ID, items.items, participant.yes, Participant.no, Participant.question, Participant.comment
FROM participant
RIGHT JOIN items ON participant.itemID=items.ID
WHERE items.role="DE" AND items.categorie="Communication"
ORDER BY items.items')
?>
<form method="post" action="../../include/func.participant.inc.php" onsubmit="refreshAndClose()">
<table>
<tr>
<!--<th></th>--><th>Items</th><th>Y</th><th>N</th><th>?</th><th>Comment</th><th>levels</th>
<?php
while ($row = mysqli_fetch_assoc($result)) {?>
</tr>
<tr>
<td>
<?php echo $row['items'] ?>
</td>
<td>
<input name="items[<?php echo $row['ID']; ?>][yes]" type="checkbox" value="yes" <?php if ($row['yes'] == "yes") echo "checked"; ?>/>
</td>
<td>
<input name="items[<?php echo $row['ID']; ?>][no]" type="checkbox" value="no" <?php if ($row['no'] == "yes") echo "checked"; ?>/>
</td>
<td>
<input name="items[<?php echo $row['ID']; ?>][question]" type="checkbox" value="question" <?php if ($row['question'] == "yes") echo "checked"; ?>/>
</td>
<td>
<textarea name="items[<?php echo $row['ID']; ?>][comment]" rows="1" cols="25" placeholder="comment"><?php echo $comment;?></textarea>
</td>
<td>
<select name="items[<?php echo $row['ID']; ?>][level]">
<option value="">Select...</option>
<option value="1">1. Starter</option>
<option value="2">2. Junior</option>
<option value="3">3. Intermediate</option>
<option value="4">4. Senior</option>
<option value="5">5. Expert</option>
<option value="6">6. Un Known</option>
<option value="7">7. Future</option>
<option value="8">8. Not relefant</option>
</select>
</td>
</tr>
<?php } ?>
</table>
<input type="submit" name="submit" value="Submit">
</form>
<?php
function refreshAndClose() {
window.opener.location.reload(true);
window.close();
} ?>
</body>
</html>
这是弹出页面中使用的功能;
<?php include 'dbh.inc.php'; ?>
<?php
var_dump($_POST); //met $_POST wordt de array met de verschillende items getoond
/*
* Je lust nu door deze array met items, en ieder item sla je op in de database obv de id
*/
foreach($_POST['items'] as $item) {
$query = "INSERT INTO interview (participantID, itemID, item, role, yes, no, question, comment, level)
VALUES ('DE', '11', '12', '13', '".mysqli_real_escape_string($conn, $item['yes'])."',
'".mysqli_real_escape_string($conn, $item['no'])."',
'".mysqli_real_escape_string($conn, $item['question'])."',
'".mysqli_real_escape_string($conn, $item['comment'])."',
'".mysqli_real_escape_string($conn, $item['level'])."',
)";
mysqli_query($conn, $query);
echo "<br>Record toegevoegd! ($query)<hr>";
}
#
职位11需要是参与者ID，职位12需要是项目ID，职位13需要是自己的项目。
我希望有人可以帮助我。
格尔茨。
答案 0 :(得分：0)
FWIW，我发现这更易于阅读和理解，但是查看这些表的CREATE和INSERT语句以及期望的结果会很有用。
SELECT p.ID
, i.items
, p.yes
, p.no
, p.question
, p.comment
FROM items i
LEFT
JOIN participant p
ON p.itemID = i.ID
WHERE i.role = "DE"
AND i.categorie = "Communication"
ORDER
BY i.items
顺便说一句，同时具有yes和no列是奇数
答案 1 :(得分：-1)
感谢您帮助我！
这读起来确实比较容易，虽然我不习惯别名喷射......
为我正在使用的三个表创建语句;
CREATE TABLE `participants` (
`ID` int(11) NOT NULL,
`first_name` text NOT NULL,
`insertion` text NOT NULL,
`last_name` text NOT NULL,
`position` text NOT NULL,
`email` text NOT NULL,
`datetime` datetime NOT NULL DEFAULT CURRENT_TIMESTAMP,
`role` varchar(2) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
CREATE TABLE `items` (
`ID` int(3) NOT NULL,
`items` varchar(75) DEFAULT NULL,
`categorie` varchar(16) DEFAULT NULL,
`role` varchar(2) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
CREATE TABLE `interview` (
`ID` int(11) NOT NULL,
`participantID` int(11) NOT NULL,
`ItemID` int(11) NOT NULL,
`item` varchar(255) NOT NULL,
`Categorie` varchar(255) NOT NULL,
`role` varchar(2) NOT NULL,
`yes` varchar(1) NOT NULL,
`no` varchar(1) NOT NULL,
`question` varchar(1) NOT NULL,
`comment` text NOT NULL,
`level` int(11) NOT NULL,
`datetime` datetime NOT NULL DEFAULT CURRENT_TIMESTAMP
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
插入语句已经在所提供的函数中。
参与者的插入声明是;
<?php include 'dbh.inc.php'; ?>
<?php
// create a variable
$first_name=$_POST['first_name'];
$insertion=$_POST['insertion'];
$last_name=$_POST['last_name'];
$position=$_POST['position'];
$role=$_POST['role'];
$email=$_POST['email'];
//Execute the query
mysqli_query($conn,"INSERT INTO participant (first_name, insertion, last_name,position,email,role) VALUES ('$first_name','$insertion','$last_name','$position','$email','$role')");
if(mysqli_affected_rows($conn) > 0){
echo "<p>Participant Added</p>";
echo "<a href='../indexnl.php'>Select role to scan</a>";
} else {
echo "Participant NOT Added<br />";
echo mysqli_error ($conn);
}