问题是我有一个html表,其中包含有关从MySQL数据库获取的员工的简短信息。我正在尝试创建一个带有一些ID的按钮,该按钮将打开一个模态窗口,其中包含有关员工的完整信息,该信息在其名为“ButtonID”的字段中具有相同的值。我尝试进行所需的查询并将其设置为php中的某个变量,然后在javascript中访问它,但它并没有很好地结束。
<!--
in <input id = "1" type="button" value="Full Info" input id equals to sql
'ButtonID' fields value
-->
Edit:
<form action="">
<input id = "1" type="button" value="Full Info"
onclick="fullInfo()">
</form>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<title>Список сотрудников</title>
<meta name="viewport" content="width=device-width, initial-scale=1">
</head>
<body>
<div class = "header">
<img src = "img/header.png"></img>
</div>
<div class = "container">
<?php
echo "<table id='table_id' class = 'table table-striped'>";
echo "<thead>
<tr>
<th>№</th>
<th>ФИО</th>
<th>Имя транслитом</th>
<th>Дата рождения</th>
<th>Должность</th>
<th>Дата приёма</th>
<th>№ удостоверения</th>
<th>Полная информация</th>
</tr>
</thead>
";
class TableRows extends RecursiveIteratorIterator {
function __construct($it) {
parent::__construct($it, self::LEAVES_ONLY);
}
function current() {
return "<td>" . parent::current(). "</td>";
}
function beginChildren() {
echo "<tr>";
}
function endChildren() {
echo "</tr>" . "\n";
}
}
$servername = "localhost";
$username = "user";
$password = "password";
$dbname = "test";
try {
$conn = new PDO("mysql:host=$servername;dbname=$dbname;charset=utf8", $username, $password);
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$stmt = $conn->prepare("SELECT number, fullname, engname,birthdaydate, position, recruitmentDate,id,buttonid FROM employees");
$ids = $conn->prepare("SELECT * FROM employees");
$stmt->execute();
// set the resulting array to associative
$result = $stmt->setFetchMode(PDO::FETCH_ASSOC);
foreach(new TableRows(new RecursiveArrayIterator($stmt->fetchAll())) as $k=>$v) {
echo $v;
}
}
catch(PDOException $e) {
echo "Error: " . $e->getMessage();
}
$conn = null;
echo "</table>";
?>
</div>
<div class = "footer">
<img src = "img/footer.jpg"></img>
</div>
<!--<link rel="stylesheet" type="text/css" href="css/styles.css">-->
<link rel="stylesheet" type="text/css" href="css/bootstrap.css">
<link rel="stylesheet" type="text/css" href="css/normalize.css">
<link rel="stylesheet" type="text/css" href="https://cdn.datatables.net/1.10.15/css/jquery.dataTables.min.css">
<link rel="stylesheet" type="text/css" href="https://cdn.datatables.net/1.10.15/js/jquery.dataTables.min.js">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<script src="https://code.jquery.com/jquery-1.10.2.js"></script>
<script src="http://code.jquery.com/ui/1.11.1/jquery-ui.js"></script>
<link rel="stylesheet" type="text/css" href="//cdn.datatables.net/1.10.15/css/jquery.dataTables.css">
<script type="text/javascript" charset="utf8" src="//cdn.datatables.net/1.10.15/js/jquery.dataTables.js"></script>
<script>
$(document).ready(function() {
$('#table_id').DataTable( {
"language": {
"url": "//cdn.datatables.net/plug-ins/1.10.13/i18n/Russian.json"
}
} );
} );
</script>
<!--
<script>
function fullInfo(){
var ids = <?php echo json_encode($ids); ?>;
for (int i = 0;i<ids.length;i++){
if (ids['buttonID'] == this.id){
alert ("good");
}
}
};
</script>
-->
</body>
</html>
答案 0 :(得分:0)
您创建了一个处理对您网站的调用的功能,如下所示:
http://my-site.com/employee.php?id=12
employee.php(想法)
$stmt = $conn->prepare("SELECT * FROM employees WHERE id=:id");
$stmt->execute(array(':id' => $_GET['id']));
$employee = $stmt->fetch(PDO::FETCH_ASSOC);
print "<ul>";
//list the employee data
foreach($employee as $key => $value)
{
$key = htmlspecialchars($key, ENT_QUOTES, 'UTF-8');
$value = htmlspecialchars($value, ENT_QUOTES, 'UTF-8')
print "<li>". $key .': ' . $value . '</li>';
}
print "</ul>";
我建议您使用<a>
代码而不是<button>
,因为此处有链接,而不是表单。只需stylize it with CSS。在您的表格的“Полнаяинформация”栏中创建一个链接:
<a target="_blank" href=\"/employee.php?id=$row['id']\">Полная информация</a>
这将打开一个新页面。使用PHP处理请求,我给了你示例代码。
然后添加装饰品,如制作窗户模态,改变它的大小等。
答案 1 :(得分:0)
对于每一行,回显一个id为员工ID的按钮。 在按钮上的id之前添加一个字母,以确保它不会重复。
echo "<button id='emp*$employeeid*'>view</button>