[英]how to get ajax form submit return value
I use ajax to submit a form, however when i finished the whole process, it stopped at the api.php and did get back to the ajax success. 我使用ajax提交表单,但是当我完成整个过程时,它停止在api.php上,并确实恢复了ajax的成功。 Here the code:
这里的代码:
<form method="POST" id="form_1" action="../api.php" enctype="multipart/form-data" novalidate="novalidate">
<input type="text" name="name" id="name" value="Amy" readonly="readonly">
<input id="fileBox" class="fileUpload" type="file" name="img" accept="image/*" required>
<input type="hidden" name="isSubmit" value="1"/>
</form>
<a id="btnSubmit" href="javascript:void(0)">Submit</a>
$("#form_1").submit(function(){
var formData = new FormData(($this)[0]);
$ajax({
type: "POST",
url: "../../api.php",
data: { action:"formSubmit", formData:formData},
processData: false,
contentType: false,
success: function(){
console.log('success return');
}
})
});
$("#btnSubmit").click(function(){
if($("#form_1").valid()){
// inside click event handler you trigger form submission
$("#form_1").trigger('submit');
}
});
in my api.php 在我的api.php中
if($action=='formSubmit'){
$name = $_POST['name'];
if(createUser($name)){
return true;
}else{
return false;
}
}
I checked the database , record is successfully created. 我检查了数据库,记录创建成功。 But I can't get the console in success.
但是我无法成功获得控制台。 The url show on the browser there is the path of the api.php.
在浏览器上显示的url是api.php的路径。 Why I stopped there can't get the return?
为什么我停在那里无法获得回报?
Thanks for your help!! 谢谢你的帮助!!
I finallly did it !I am missing "return false" after the ajax; 我最终做到了!阿贾克斯过后,我错过了“ return false”; But actually I dont really know the function of return false;
但是实际上我并不真正知道return false的功能。
$("#form_1").submit(function(){
var formData = new FormData((this)[0]);
$.ajax({
type: "POST",
url: "../../api.php",
data: { action: "formSubmit", formData: formData},
processData:false,
contentType:false,
success: function(data){
$("#thankyou").show();
}
})
return false;
});
Redirecting to other page will have the header()
to use: 重定向到其他页面将具有
header()
来使用:
if ($action == 'formSubmit') {
$name = $_POST['name'];
if (createUser($name)) {
// Redirect to right.htm
header("Location: right.htm");
} else {
// Redirect to wrong.htm
header("Location: wrong.htm");
}
die();
}
javascript doesn't recognize return value from php. javascript无法识别php的返回值。 Use
echo
instead. 请改用
echo
。
if($action=='formSubmit'){
$name = $_POST['name'];
if(createUser($name)){
echo 'true';
}else{
echo 'false';
}
}
If you want to see actual output coming from api.php
then use below code in your ajax
. 如果要查看来自
api.php
实际输出,请在ajax
使用以下代码。
success: function(res){
console.log(res);
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.