基本上，此代码由studentid从id.php中获取信息，并返回地址、文档成本等。单击studentid按钮时会返回代码，但仅返回第一行。第二行没有返回任何数据，我似乎无法解决问题。请帮忙。
<div id="briefinfo" style="display:inline-block;text-align:center;margin-left:20px;">
<?php require_once("db.php");
if ($result = $mysqli->query("SELECT * FROM requests WHERE status = 1 ORDER BY id"))
{
if ($result->num_rows > 0)
{
while ($row = $result->fetch_object())
{
echo "<div id='thumbnail'>";
echo " ". $row->lastname;
echo " ". $row->firstname;
echo "</div>";
echo "document id:" . $row->id;
echo "<br>";
echo "requested: " . $row->document;
echo "<br>";
if ($row->paidstatus == 1){
echo "payment status: paid";
}
else{
echo "payment status: not paid";
}
echo "<br>";
echo "<input type='button' value='$row->student_id' id='studentid'/>";
/*echo "<td>" . $row->document . "</td>";
echo "<td><a href='records.php?id=" . $row->id . "'>Edit</a></td>";
echo "<td><a href='delete.php?id=" . $row->id . "'>Delete</a></td>";
echo "<td><a href='unverify.php?id=" . $row->id . "'>unverify</a></td>";
echo "<td><a href='comments.php?id=" . $row->id . "'>comment</a></td>";
echo "<td>" . $row->paymentamount . " pesos";"</td>";
echo "<td><a href='paymentamount.php?id=" . $row->id . "'>set amount</a></td>";*/
}
}
else
{
echo "No results to display!";
}
}
else
{
echo "Error: " . $mysqli->error;
} ?>
</div>
这是JS
<script type='text/javascript' src='http://ajax.googleapis.com/ajax/libs/jquery/1.6.4/jquery.min.js'></script>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.2/jquery.min.js"></script>
<script>
//in your Javascript, put
$(document).ready ( function () {
$("#studentid").click (function(){
var studentid = $(this).val();
$.ajax({
type: 'POST',
url: 'id.php',
data: {
id: studentid
},
success: function(response) {
$("#moredata").html(response);
}
});
});
});
</script>
这个id.php
<?php
// connect to the database
include('connect-db.php');
// confirm that the 'id' variable has been set
require_once("db.php");
$id = $_POST['id'];
if ($result = $mysqli->query("SELECT * FROM requests WHERE student_id=$id"))
{
if ($result->num_rows > 0)
{
while ($row = $result->fetch_object())
{
echo $row->paymentamount . " pesos";
echo "<br>";
echo $row->address;
echo "<br>";
echo $row->address2;
echo "<br>";
echo $row->country;
}
}
else
{
echo "No results to display!";
}
}
else
{
echo "Error: " . $mysqli->error;
}
?>
当单击按钮时，输入按钮的第一行显示中的更多数据。第二行不再显示任何内容。帮助
对于while循环中的以下语句，应该使用class而不是id。您不应该对多个元素使用相同的ID。
echo "<input type='button' value='$row->student_id' class='studentid'/>";
在jquery脚本中，您应该这样称呼它：
<script>
//in your Javascript, put
$(document).ready ( function () {
$(".studentid").click (function(){
var studentid = $(this).val();
$.ajax({
type: 'POST',
url: 'id.php',
data: {
id: studentid
},
success: function(response) {
$("#moredata").html(response);
}
});
});
});
</script>