使用Ajax的每个按钮的不同动作 [英] Different actions for each button using Ajax
问题描述
嗯,我的问题是我试图插入,更新和删除通过使用Ajax的表单发送数据,但我正在努力提交按钮,因为我需要每个人执行一个特定的操作,每次我点击它们调用PHP的指令,但你可以在Ajax中看到的动作只执行提交,而不是取决于我点击的按钮,我真的需要解决这个问题...
FORM
< form method =postid =form_shirt ENCTYPE = 多部分/格式数据 >
ID:< br>
< input type =hiddenname =id_shirtid =id_shirtclass =form-control>
名称:< br>
< input type =textname =name_shirtid =name_shirtclass =form-controlrequired =required>
价格:< br>
< input type =textname =price_shirtid =price_shirtclass =form-controlrequired =required>
< input type =submitname =btninsertid =btninsertvalue =Insertclass =btn btn-success/>
< input type =submitname =btnupdateid =btnupdatevalue =Updateclass =btn btn-warning>
< input type =submitname =btndeleteid =btndeletevalue =Deleteclass =btn btn-danger>
< / form>
AJAX
< $($#$ $ $ $''$ $ $ $ $' event.preventDefault();
$ .ajax({
url:insert_shirts.php,
method:POST,
data:new FormData(this),
contentType:false,
cache:false,
processData:false,
success:function(data){
$('#form_shirt')[0] .reset ();
$('#table_shirt')。html(data);
}
});
});
$ b $ (submit,function(event){
event.preventDefault();
$ .ajax({
url:update_shirts.php ,
method:POST,
data:new FormData(this),
contentType:false,
cache:false,
processData:false,
成功:函数(数据){
$('#form_shirt')[0] .reset();
$('#tab le_shirt)的html(数据)。
}
});
}); $ {
$ b $('#form_shirts')。on(submit,function(event){
event.preventDefault();
$ .ajax({
url:delete_shirts.php,
方法:POST,
data:new FormData(this),
contentType:false,
cache:false,
processData:false,
success:function(data){
$('#form_shirt')[0] .reset();
$('#table_shirt')。 html(data);
}
});
});
});
PHP
<?php
$ connect = mysqli_connect(localhost,root,,shirts);
$ output ='';
$ name_shirt = mysqli_real_escape_string($ connect,$ _POST [name_shirt]);
$ price_shirt = mysqli_real_escape_string($ connect,$ _POST [price_shirt]);
$ query =插入衬衫(名称,价格)
VALUES('$ name_shirt','$ price_shirt');
if(mysqli_query($ connect,$ query))
{
$ output。='< label class =text-success>插入数据< / label>';
$ select_query =SELECT id_shirt,name,price FROM shirts;
$ result = mysqli_query($ connect,$ select_query);
$ output。='
< table id =shirtsclass =table table-bordered>
< thead>
< tr>
< th> ID< / th>
< th> NAME< / th>
PRICE< / th>
< / tr>
< / thead>
';
while($ row = mysqli_fetch_array($ result))
{
$ output。='
< tr>
< tbody>
< td>'。 $ row [id_shirt]。 < / TD>
< td>'。 $ row [name]。 < / TD>
< td>'。 $ row [price]。 < / TD>
< / tr>
< / tbody>
';
}
$ output。='< / table>';
}
echo $ output;
?>
。除了可以在单击按钮时引用的同义类之外,只需创建三个具有唯一类名称的按钮。然后引用该同义类的单击处理程序(而不是表单提交): p>
< form method =postid =form_shirtenctype =multipart / form-data>
ID:
< br>
< input type =hiddenname =id_shirtid =id_shirtclass =form-control>名称:
< br>
< input type =textname =name_shirtid =name_shirtclass =form-controlrequired =required>价格:
< br>
< input type =textname =price_shirtid =price_shirtclass =form-controlrequired =required>
< button name =btninsertid =btninsertvalue =Insertclass =submit insert_shirts btn btn-success/>
< button name =btnupdateid =btnupdatevalue =Updateclass =submit update_shirts btn btn-warning/>
< button name =btndeleteid =btndeletevalue =Deleteclass =submit delete_shirts btn btn-danger/>
< / form>
JavaScript :
函数(){
$(document).ready(function(){
$('。submit')。 .ajax({
url:this.classList [1] +.php,
method:POST,
data:new FormData(this),
contentType: false,
cache:false,
processData:false,
success:function(data){
$('#form_playera')[0] .reset();
$('#table_playeras')。html(data);
}
});
});
});
请注意,您可以在一个 $(document) .ready()
,这三个函数都可以合并为一个,通过使用 url 参数发送给AJAX =https://developer.mozilla.org/en/docs/Web/API/Element/classList =nofollow noreferrer> this.classList : url :this.classList [1] +.php
。 1
表示每个元素的第二个类( insert_shirts
, update_shirts
和 delete_shirts
)。
另外请注意,您不需要事件.preventDefault()
,因为你不再针对表单提交。
希望这有助于! :)
Well, my problem is that I am trying to insert, update and delete sending the data through a form using Ajax but I'm struggling with the submit buttons because I need each one to perform one specific action everytime I click them calling the url where are the php instructions, but as you can see in the Ajax the action only performs on submit and not depending on the button I click, I really need to solve this...
FORM
<form method="post" id="form_shirt" enctype="multipart/form-data">
ID:<br>
<input type="hidden" name="id_shirt" id="id_shirt" class="form-control" >
Name:<br>
<input type="text" name="name_shirt" id="name_shirt" class="form-control" required="required">
Price:<br>
<input type="text" name="price_shirt" id="price_shirt" class="form-control" required="required">
<input type="submit" name="btninsert" id="btninsert" value="Insert" class="btn btn-success"/>
<input type="submit" name="btnupdate" id="btnupdate" value="Update" class="btn btn-warning">
<input type="submit" name="btndelete" id="btndelete" value="Delete" class="btn btn-danger">
</form>
AJAX
$(document).ready(function(event){
$('#form_shirts').on("submit", function(event){
event.preventDefault();
$.ajax({
url:"insert_shirts.php",
method:"POST",
data:new FormData(this),
contentType: false,
cache: false,
processData:false,
success:function(data){
$('#form_shirt')[0].reset();
$('#table_shirt').html(data);
}
});
});
$('#form_shirts').on("submit", function(event){
event.preventDefault();
$.ajax({
url:"update_shirts.php",
method:"POST",
data:new FormData(this),
contentType: false,
cache: false,
processData:false,
success:function(data){
$('#form_shirt')[0].reset();
$('#table_shirt').html(data);
}
});
});
$('#form_shirts').on("submit", function(event){
event.preventDefault();
$.ajax({
url:"delete_shirts.php",
method:"POST",
data:new FormData(this),
contentType: false,
cache: false,
processData:false,
success:function(data){
$('#form_shirt')[0].reset();
$('#table_shirt').html(data);
}
});
});
});
PHP
<?php
$connect = mysqli_connect("localhost", "root", "", "shirts");
$output = '';
$name_shirt = mysqli_real_escape_string($connect, $_POST["name_shirt"]);
$price_shirt = mysqli_real_escape_string($connect, $_POST["price_shirt"]);
$query = "INSERT into shirts ( name, price)
VALUES ('$name_shirt','$price_shirt') ";
if(mysqli_query($connect, $query))
{
$output .= '<label class="text-success">Data Inserted</label>';
$select_query = "SELECT id_shirt, name, price FROM shirts";
$result = mysqli_query($connect, $select_query);
$output .= '
<table id="shirts" class="table table-bordered">
<thead>
<tr>
<th>ID</th>
<th>NAME</th>
<th>PRICE</th>
</tr>
</thead>
';
while($row = mysqli_fetch_array($result))
{
$output .= '
<tr>
<tbody>
<td>' . $row["id_shirt"] . '</td>
<td>' . $row["name"] . '</td>
<td>' . $row["price"] . '</td>
</tr>
</tbody>
';
}
$output .= '</table>';
}
echo $output;
?>
You can't have three buttons for a single form. Simply create three buttons that will have unique class names, in addition to a synonymous class that you can reference on button click. Then reference the click handler for that synonymous class (rather than the form submission):
HTML:
<form method="post" id="form_shirt" enctype="multipart/form-data">
ID:
<br>
<input type="hidden" name="id_shirt" id="id_shirt" class="form-control"> Name:
<br>
<input type="text" name="name_shirt" id="name_shirt" class="form-control" required="required"> Price:
<br>
<input type="text" name="price_shirt" id="price_shirt" class="form-control" required="required">
<button name="btninsert" id="btninsert" value="Insert" class="submit insert_shirts btn btn-success" />
<button name="btnupdate" id="btnupdate" value="Update" class="submit update_shirts btn btn-warning" />
<button name="btndelete" id="btndelete" value="Delete" class="submit delete_shirts btn btn-danger" />
</form>
JavaScript:
$(document).ready(function() {
$('.submit').on("click", function() {
$.ajax({
url: this.classList[1] + ".php",
method: "POST",
data: new FormData(this),
contentType: false,
cache: false,
processData: false,
success: function(data) {
$('#form_playera')[0].reset();
$('#table_playeras').html(data);
}
});
});
});
Note that you can have the three separate functions in one $(document).ready()
, and the three functions can all be combined into one, sending different url
parameters to AJAX by using this.classList: url: this.classList[1] + ".php"
. The 1
denotes the second class of each element (insert_shirts
, update_shirts
and delete_shirts
respectively).
Also note that you won't need event.preventDefault()
, as you're no longer targeting the form submission.
Hope this helps! :)
这篇关于使用Ajax的每个按钮的不同动作的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!