我想从数据库中删除数据而不刷新页面。我的代码可以正常工作,但删除产品后需要刷新页面。我想要类似this的东西
这是我的js代码:
<script>
$(document).on('click', '.delete-it', function() {
var id = $(this).attr('delete-id');
bootbox.confirm("Are you sure?", function(result) {
if (result) {
$.ajax({
type: "POST",
async: false,
url: "delete_product.php",
data: {
object_id: id
},
dataType: 'json',
success: function(data) {
location.reload();
}
});
}
});
return false;
});
</script>
和delete_product php代码:
<?php
// check if value was posted
if($_POST){
// get database connection
$database = new Database();
$db = $database->getConnection();
// prepare product object
$product = new Product($db);
// set product id to be deleted
$product->id = $_POST['object_id'];
// delete the product
if($product->delete()){
echo "Object was deleted.";
}
// if unable to delete the product
else{
echo "Unable to delete object.";
}
}
?>
请告诉我一种制作方法!
最佳答案
我没有看到您在页面中定位某些内容的地方,但这就是我会使用的:
<script>
function swapContent(href, url_data, target) {
$.ajax({
type: 'GET',
cache: false,
url: href+'?' + url_data, //add a variable to the URL that will carry the value in your i counter through to the PHP page so it know's if this is new or additional data
success: function (data) { // this param name was confusing, I have changed it to the "normal" name to make it clear that it contains the data returned from the request
//load more data to "target" value div
target.innerHTML = (data); // as above, data holds the result of the request, so all the data returned from your results.php file are in this param but please see below
}
})
}
$(document).on('click', '.delete-it', function() {
var id = $(this).attr('delete-id');
bootbox.confirm("Are you sure?", function(result) {
if (result) {
swapContent(base_url, url_data, target) //set variables
}
});
return false;
});
</script>
注意:由于我使用了很多ajax,我自己保留了一个ajax函数
关于javascript - 从数据库中删除数据而不重新加载页面,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35759221/