Question

我使用的是另一个stackoverflow文章所使用的代码，但有一部分对我不起作用。

<script>

// variable to hold request
var request;

// bind to the submit event of our form
$("#notification").submit(function(event){

// prevent default posting of form
event.preventDefault();

// abort any pending request
if (request) {
    request.abort();
}

// let's select and cache all the fields
var $inputs = $form.find("input, select, button, textarea");

// let's disable the inputs for the duration of the ajax request
$inputs.prop("disabled", true);

//this code was added by me cause i saw that when i pass as data {notTitle:"smth"} 
//it works, so i wanted to remake that structure and include all my elements.
// Apparently it does not work, maybe because i create a string and by 
//{notTitle:"smth"}, notTitle is not passed as string.
 x=$("form").serializeArray();
 var str="{";
 var size = x.length;
 $.each(x, function(i, field){
    if (field.value!="")
     str = str.concat(field.name + ":" + field.value);
   else
     str = str.concat(field.name + ":" + "k");
   if(i!=size-1)
     str=str.concat(",");
 });

str=str.concat("}");

// fire off the request to /form.php
request = $.ajax({
    url: "test.php",
    type: "POST",
    dataType: 'json',
    data: str//{notTitle:"s",notType:''}
});


// callback handler that will be called on success
request.done(function (response, textStatus, jqXHR){
    alert(response.status);
});

// callback handler that will be called on failure
request.fail(function (jqXHR, textStatus, errorThrown){
    // log the error to the console
    alert(
        "The following error occured: "+
        textStatus, errorThrown
    );
});

// callback handler that will be called regardless
// if the request failed or succeeded
request.always(function () {
    // reenable the inputs
    $inputs.prop("disabled", false);
});


});

</script>

这是我的php

<?php $result = array("status" => "1",  "a" => "2");
$notificationTitle = htmlspecialchars($_POST["notTitle"],ENT_QUOTES,'ISO-8859-1');
echo json_encode($result); ?>

在您建议使用序列化表单之前，我尝试了它并且它不起作用。我注意到的是，当我尝试将数据传递给我的php文件时，我无法通过$ _POST ['name of element']读取它们。所以，如果我评论php文件的第二行并作为数据传递{notTitle：'s'}它可以工作，我看到了成功的消息。

无法通过post将jquery ajax传递给php

0 个答案: