[英]How to delete record using ajax?
我試圖使用AJAX從數據庫中刪除記錄。 確認窗口不會出現,因此可以刪除記錄。 這是代碼..
<?php
$q = $_GET['q'];
$p = $_GET['p'];
$sql="SELECT * FROM course_details WHERE sem='" . $q . "' AND branch='" . $p . "' ORDER BY course_codes ASC";
$result = mysql_query($sql);
while($row = mysql_fetch_assoc($result)){
echo '<tr class="record">';
echo "<td>" . $row['course_codes'] . "</td>";
echo "<td>" . $row['course_names'] . "</td>";
echo "<td>" . $row['course_instructors'] . "</td>";
echo "<td>" . $row['course_credits'] . "</td>";
echo '<td><div align="center"><a href="#" id="' . $row['course_id'] . '" class="delbutton" title="Click To Delete">delete</a></div></td>';
echo '</tr>';
}
echo "</table>";
mysql_close($bd);
?>
這里$ p和$ q由另一個頁面的AJAX腳本發送。 它工作正常。 記錄按預期顯示。 如果我不使用AJAX顯示記錄,刪除工作使用AJAX。我用來刪除的腳本是:
<script src="jquery.js"></script>
<script type="text/javascript">
$(function() {
$(".delbutton").click(function(){
var element = $(this);
var del_id = element.attr("id");
var info = 'id=' + del_id;
if(confirm("Are you sure you want to delete this Record?")){
$.ajax({
type: "GET",
url: "deleteCourse.php",
data: info,
success: function(){
}
});
$(this).parents(".record").animate({ backgroundColor: "#fbc7c7" }, "fast")
.animate({ opacity: "hide" }, "slow");
}
return false;
});
});
</script>
deleteCourse.php
if($_GET['id']){
$id=$_GET['id'];
$id = mysql_escape_string($id);
}
$del = "DELETE from course_details where course_id = '$id'";
$result = mysql_query($del);
問題是因為您正在創建動態元素,因此您必須使用delagate $(document).on()
才能將click事件綁定到元素。 這是更正后的代碼
<script type="text/javascript">
$(function() {
$(document).on('click','.delbutton',function(){
var element = $(this);
var del_id = element.attr("id");
var info = 'id=' + del_id;
if(confirm("Are you sure you want to delete this Record?")){
$.ajax({
type: "GET",
url: "deleteCourse.php",
data: info,
success: function(){ }
});
}
return false;
});
});
</script>
和你的deletCourse.php
if($_GET['id']){
$id=$_GET['id'];
$id = mysql_escape_string($id);
}
$del = "DELETE from course_details where course_id = ".$id."";
$result = mysql_query($del);
希望這有幫助,謝謝
嘗試這個:
<script type="text/javascript" src="jquery.js"></script>
試試這個
<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
<a href="#" id="1" onclick="del(this.id);return false;">delete</a>
<a href="#" id="2" onclick="del(this.id);return false;">delete</a>
<a href="#" id="3" onclick="del(this.id);return false;">delete</a>
<a href="#" id="4" onclick="del(this.id);return false;">delete</a>
<a href="#" id="5" onclick="del(this.id);return false;">delete</a>
JavaScript的
function del(id)
{
var info = 'id=' + id;
if(confirm("Are you sure you want to delete this Record?")){
var html = $.ajax({
type: "POST",
url: "delete.php",
data: info,
async: false
}).responseText;
if(html == "success")
{
$("#delete").html("delete success.");
return true;
}
else
{
$("#captchaStatus").html("incorrect. Please try again");
return false;
}
}
}
ajax文件
if($_GET['id']){
$id=$_GET['id'];
$id = mysql_escape_string($id);
}
$del = "DELETE from course_details where course_id = '$id'";
$result = mysql_query($del);
if($result)
{
echo "success";
}
<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
<script>
function del(id)
{
var info = 'id=' + id;
if(confirm("Are you sure you want to delete this Record?")){
var html = $.ajax({
type: "GET",
url: "deletCourse.php",
data: info,
async: false ,
success: function() {
window.location.reload(true);}
}).responseText;
}
}
</script>
<?php
$link=mysql_connect("localhost","root","") or die(mysql_error());
mysql_select_db("cart");
$sql=mysql_query("SELECT * FROM `details`");
echo "<table>";
echo "<tr><th>Name</th><th>NO of Items</th></tr>";
while($row = mysql_fetch_assoc($sql)){
echo '<tr class="record">';
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['num'] . "</td>";
echo '<td><div align="center"><a href="#" id="' . $row['id'] . '" onclick="del('.$row['id'].')">delete</a></div></td>';
echo '</tr>';
}
echo "</table>";
mysql_close($link);
?>
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