[英]update the database without page refresh
Whenever the hyperlink with the yes id is clicked i dont want the page to refresh and then show the status, i want the status to change instant without page refreshing.每当点击带有 yes id 的超链接时,我不希望页面刷新然后显示状态,我希望状态在不刷新页面的情况下立即更改。 I know Ajax deals with this, but can anyone provide me with a working example with my code please?我知道 Ajax 会处理这个问题,但是谁能给我提供一个包含我的代码的工作示例? As it melting my head :/因为它融化了我的头:/
<h3 class="page-header"> Enquiries </h3>
<form id="enquiry" method="post" action="enquiry_csv.php">
<table class="table table-bordered table-hover">
<thead>
<tr>
<th> First Name</th>
<th> Last Name</th>
<th>Email</th>
<th>Message</th>
<th>Date</th>
<th>Responded to Enquiry?</th>
<th>Status</th>
<th></th>
<th><input class='btn-success' name='export' id='btnExport' type='submit' value='Export to CSV'/></th>
</tr>
</thead>
<tbody>
<?php
$query = "SELECT * FROM enquiries";
$select_enquiries = mysqli_query($connection,$query);
while($row = mysqli_fetch_assoc($select_enquiries)) {
$Enquiry_ID = $row['Enquiry_ID'];
$FirstName = $row['First_Name'];
$LastName =$row['Last_Name'];
$Email = $row['Email'];
$Message = $row['Message'];
$Date =$row['Date'];
$Responded =$row['Responded'];
echo "<tr>";
echo "<td>$FirstName </td>";
echo "<td>$LastName </td>";
echo "<td>$Email </td>";
echo "<td>$Message </td>";
echo "<td>$Date </td>";
echo "<td> <a id='yes' class='success' style='' href='enquiries.php?Yes=$Enquiry_ID'>Yes</a> | <a class='success' href='enquiries.php?No=$Enquiry_ID'>No</a> </td>";
echo "<td> $Responded</td>";
echo "<td> <a class='btn btn-danger' href ='enquiries.php?delete=$Enquiry_ID'>Delete</a> </td>";
echo "</tr>";
}
?>
<?php
if(isset($_GET['Yes'])){
$enquiry_id = $_GET['Yes'];
$query = "UPDATE enquiries SET Responded = 'Yes' WHERE Enquiry_ID = {$Enquiry_ID}";
$query = mysqli_query($connection, $query);
}
if(isset($_GET['No'])){
$enquiry_id = $_GET['No'];
$query = "UPDATE enquiries SET Responded = 'No' WHERE Enquiry_ID = {$Enquiry_ID}";
$query = mysqli_query($connection, $query);
}
if(isset($_GET['delete'])){
$review_id = $_GET['delete'];
$query = "DELETE FROM enquiries WHERE Enquiry_ID = {$Enquiry_ID} ";
$delete_query = mysqli_query($connection, $query);
}
?>
<tr>
<td></td>
<td> </td>
<td> </td>
<td> </td>
<td> </td>
</tr>
</tbody>
</table>
</form>
You can do it with jquery ajax api it performs an asynchronous HTTP (Ajax) request您可以使用 jquery ajax api 执行异步 HTTP (Ajax) 请求
http://api.jquery.com/jquery.ajax/ http://api.jquery.com/jquery.ajax/
Refer the above link which @Sanya Zahid posted and try to do some thing like this:请参阅@Sanya Zahid 发布的上述链接并尝试执行以下操作:
<script type="text/javascript" src="jquery.min.js"></script>
<script type="text/javascript" >
$(function() {
$(".submit").click(function() {
var name = $("#name").val();
var age = $("#age").val();
var dataString = 'name='+ name +'&age='+ age;
if(name=='' || age =='')
{
$('.success').fadeOut(200).hide();
$('.error').fadeOut(200).show();
}
else
{
$.ajax({
type: "POST",
url: "submit.php",
data: dataString,
success: function(){
$('.success').fadeIn(200).show();
$('.error').fadeOut(200).hide();
}
});
}
return false;
});
});
</script>
<form method="post" name="form">
<input id="name" name="name" type="text" /><br>
<input id="age" name="age" type="text"/>
<div>
<input type="submit" value="Submit" class="submit"/>
<span class="error" style="display:none"> Please Enter Data</span>
<span class="success" style="display:none"> Data Saved!!</span>
</div>
</form>
Here submit.php
contains database related stuff( insert/update/delete
).这里submit.php
包含数据库相关的东西( insert/update/delete
)。 I just added name and age fileds here in code.我只是在代码中添加了姓名和年龄文件。 Add fields as per your form.根据您的表单添加字段。
submit.php :提交.php :
<?php
$conn = mysqli_connect("localhost","root","","test");
/* Insert form data with out page refresh */
if($_POST)
{
$name=$_POST['name'];
$age = $_POST['age'];
mysqli_query($conn,"Query will come here");
}else{
echo "Please try again!!";
}
/* Insert form data with out page refresh */
?>
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