这是我第一次使用stackoverflow.com,我几乎放弃了代码问题,我希望有人可以帮我解决我的代码问题。
这是我的问题,我有2个表,我将数据从数据库加载到第一个表(onload),我想在数据行上有一个onclick事件来加载第二个表,我想使用单击值作为加载第二个表的参数。
<!DOCTYPE html>
<html lang="en">
<head>
</head>
<body>
<table>
<thead>
<tr>
<th style="text-align:center">empid</th>
<th style="text-align:center">empname</th>
<th style="text-align:center">age</th>
</tr>
</thead>
<tbody>
<?php
include ("dbconn.php");
$sql = "select * from employee";
$result = mysqli_query($conn, $sql);
while($row = mysqli_fetch_assoc($result)) {
?>
<tr>
<td><a href="#"><u><?php echo $row["empid"]; ?></u></a></td>
<td><?php echo $row["empname"]; ?></td>
<td><?php echo $row["age"]; ?></td>
</tr>
<?php
}
mysqli_close($conn);
?>
</tbody>
</table>
<! -- additional table -->
<table>
<thead>
<tr>
<th>employee name</th>
<th>position</th>
<th>salary</th>
</tr>
</thead>
<tbody id="tbl">
<?php
$sql = "select * from employee_salary where empid = $empid";
$result = mysqli_query($conn, $sql);
while($row = mysqli_fetch_assoc($result)) {
?>
<tr>
<th><?php echo $row["empname"]; ?></th>
<td><?php echo $row["position"]; ?></td>
<td><?php echo $row["salary"]; ?></td>
</tr>
<?php
}
mysqli_close($conn);
?>
</tbody>
</table>
</body>
</html>
任何人都应该帮助我......谢谢你。
答案 0 :(得分:0)
完成以下步骤:
1)这里是您的初始页码:
<body>
<table>
<thead>
<tr>
<th style="text-align:center">empid</th>
<th style="text-align:center">empname</th>
<th style="text-align:center">age</th>
</tr>
</thead>
<tbody>
<?php
include ("dbconn.php");
$sql = "select * from employee";
$result = mysqli_query($conn, $sql);
while($row = mysqli_fetch_assoc($result)) {
?>
<tr onclick="getDetail('<?php echo $row["empid"]; ?>')">
<td><a href="#"><u><?php echo $row["empid"]; ?></u></a></td>
<td><?php echo $row["empname"]; ?></td>
<td><?php echo $row["age"]; ?></td>
</tr>
<?php
}
mysqli_close($conn);
?>
</tbody>
</table>
<! -- additional table -->
<table id="secondTable" style="display:none;">
<thead>
<tr>
<th>employee name</th>
<th>position</th>
<th>salary</th>
</tr>
</thead>
<tbody id="tbl">
</tbody>
</table>
</body>
</html>
2)现在点击员工ID进行一次ajax调用,如:
<script>
function getDetail(id){
$.post( "../ajax/getData.php", {emp_id:id},function(data){
$('#secondTable').show();
$('#secondTable tbody').html(data);
});
}
</script>
3)这里是你的getData.php文件:
<?php
include ("dbconn.php");
$emp_id = $_POST['emp_id'];
$sql = "select * from employee where empid= $emp_id";
$result = mysqli_query($conn, $sql);
$data = '';
while($row = mysqli_fetch_assoc($result)) {
$data.='<tr>
<td><a href="#"><u>'.$row["empid"].'</u></a></td>
<td>'. $row["empname"].'</td>
<td>'. $row["age"].'</td>
</tr>';
}
mysqli_close($conn);
echo $data; exit;
?>
如果有任何疑问,请告诉我,希望对此有所帮助!