我目前有以下数据表来显示数据库中的联系人列表。每行,我还插入一个“编辑”&amp; “删除”按钮并传递data-id = "<?=$row['id']?>"
,以便jQuery在单击按钮时进行处理。当单击按钮时,通过ajax,我将从getcontact.php获取数据,并将其设置为模式形式的相应输入,但数据似乎不显示在表单值中。我可以知道哪里出错了吗?
tables.php
<table width="100%" class="display table table-striped table-bordered table-hover" id="contactsTable">
<thead>
<tr>
<th>ID</th>
<th>Name</th>
<th>Company</th>
<th>Position</th>
<th>Phone</th>
<th>Email</th>
<th>Gender</th>
<th>Date Registered</th>
<th>Edit</th>
<th>Delete</th>
</tr>
</thead>
<tbody>
<?php
while($row = mysqli_fetch_array($result)){
?>
<tr>
<td><?=$row['id']?></td>
<td><?=$row['name']?></td>
<td><?=$row['company']?></td>
<td><?=$row['position']?></td>
<td><?=$row['phone']?></td>
<td><?=$row['email']?></td>
<td><?=$row['gender']?></td>
<td><?=$row['dateregistered']?></td>
<td>
<!-- Edit Button trigger modal -->
<button id="editButton" type="button" class="btn btn-primary" datatoggle="modal" data-target="#editModal" data-id="<?=$row['id']?>">
Edit
</button>
</td>
<td>
<!-- Delete Button trigger modal -->
<button type="button" class="btn btn-primary" data-toggle="modal" data-target="#deleteModal" data-id="<?=$row['id']?>">
Delete
</button>
</td>
</tr>
<?php
}
?>
</tbody>
</table>
模态(与tables.php在同一页面中)
<!-- Edit Contact Modal -->
<div class="modal fade" id="editModal" tabindex="-1" role="dialog" aria labelledby="myModalLabel">
<div class="modal-dialog" role="document">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-label="Close"><span aria-hidden="true">×</span></button>
<h4 class="modal-title" id="myModalLabel">Edit Contact</h4>
</div>
<div class="modal-body">
<form class="form-horizontal" method="POST" id="editForm" role="form">
<div class="form-group animated fadeIn">
<label for="nameInput" class="col-sm-2 control-label">Name</label>
<div class="col-sm-10">
<input type="name" name="name" class="form-control" id="nameInput" placeholder="Name" required>
</div>
</div>
<div class="form-group animated fadeIn">
<label for="companyInput" class="col-sm-2 control-label">Company</label>
<div class="col-sm-10">
<input type="company" name="company" class="form-control" id="companyInput" placeholder="Company" required>
</div>
</div>
<div class="form-group animated fadeIn">
<label for="posInput" class="col-sm-2 control-label">Position</label>
<div class="col-sm-10">
<input type="position" name="position" class="form-control" id="posInput" placeholder="Position/Job Title">
</div>
</div>
<div class="form-group animated fadeIn">
<label for="contactInput" class="col-sm-2 control-label">Contact Number</label>
<div class="col-sm-10">
<input type="number" name="contact" class="form-control" id="contactInput" placeholder="Office/Mobile Number" data-error="Please enter a valid mobile number" required>
</div>
</div>
<div class="form-group animated fadeIn">
<label for="emailInput" class="col-sm-2 control-label">Email</label>
<div class="col-sm-10">
<input type="email" name="email" class="form-control" id="emailInput" placeholder="Email Address">
</div>
</div>
<div class="form-group animated fadeIn">
<label for="genderInput" class="col-sm-2 control-label">Gender</label>
<div class="col-sm-10">
<input type="gender" name="gender" class="form-control" id="genderInput" placeholder="Male/Female">
</div>
</div>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
<button id="editContact" type="button" class="btn btn-primary">Save</button>
</div>
</form>
</div>
的jQuery
$("#editButton").click(function(e){
e.preventDefault();
var uid = $(this).data('id'); // get id of clicked row
$.ajax({
url: 'getcontact.php',
type: 'POST',
data: 'id='+uid,
dataType: 'json',
success: function(response){
$("#nameInput").val(result[0]);
$("#companyInput").val(result[1]);
$("#posInput").val(result[2]);
$("#contactInput").val(result[3]);
$("#emailInput").val(result[4]);
$("#genderInput").val(result[5]);
}
});
});
getcontact.php
<?php
include("dbconnect.php");
$id = $_POST['id'];
$result = mysqli_query($link, "SELECT * FROM businesscontact WHERE id=$id");
$rows = array();
while($row = mysqli_fetch_array($result)){
$rows[] = $row['*'];
}
echo json_encode($rows);
?>
答案 0 :(得分:1)
1 - 您正在从您的php文件发送编码的json并直接从您的javascript代码中使用它,这是无效的,可以这么说,
2 - 您正在使用ajax错误地将数据对象发送到php,它应该如下data: {id: uid},
3 - 您声明了错误的数据ID,它应该如下:var uid = $(this).attr('data-id');
你需要decode
你的json回复如下:
var uid = $(this).attr('data-id');
$.ajax({
url: 'getcontact.php',
method: 'POST',
data: {id: uid},
success: function(response){
var result = JSON && JSON.parse(response) || $.parseJSON(response);
...
// rest of your code
...
你在这一部分遇到了问题:
while($row = mysqli_fetch_array($result)){
$rows[] = $row['*'];
}
添加您需要执行的整个数组,如下所示:
while($row = mysqli_fetch_array($result)){
$rows[] = $row;
}