I've checked out various other posts on SO, but I don't seem to see the problem, and was hoping if you could help me shed some light on this issue. Basically, I'm doing a microblogging appliation and inserting a tweet when a button is clicked, which calls the jQuery ajax function. Here's the respective code:
我已经检查了其他各种关于SO的帖子,但我似乎没有看到问题,并希望你能帮助我解决这个问题。基本上,我正在进行微博应用并在点击按钮时插入推文,该按钮调用jQuery ajax函数。这是相应的代码:
home.js
This is the ajax jquery call
这是ajax jquery调用
function sendTweet(single_tweet) {
var tweet_text = $("#compose").val();
tweet_text = tweet_text.replace(/'/g, "'");
tweet_text = tweet_text.replace(/"/g, """);
var postData = {
author : $("#username").text().split("@")[1], // be careful of the @! - @username
tweet : tweet_text,
date : getTimeNow()
};
$.ajax({
type : 'POST',
url : '../php/tweet.php',
data : postData,
dataType : 'JSON',
success : function(data) {
alert(data.status);
}
})
}
The ajax call works successfully, and the tweet is inserted, but I can't get the alert call to fireback under the success parameter. I tried something basic like alert('abc');
but it didn't work either.
ajax调用成功,并且插入了推文,但我无法在success参数下获得对fireback的警报调用。我尝试了像alert('abc')这样基本的东西;但它也没有用。
tweet.php
This is just a wrapper, looks like this:
这只是一个包装器,看起来像这样:
<?php
include 'db_functions.php';
$author = $_POST['author'];
$tweet = $_POST['tweet'];
$date = $_POST['date'];
insert_tweet($author, $tweet, $date);
$data = array();
$data['status'] = 'success';
echo json_encode($data);
?>
This just inserts the tweet into the database, and I wanted to try sending simple JSON formatted data back, but data.status
didn't work on the success callback.
这只是将推文插入数据库,我想尝试发回简单的JSON格式数据,但data.status对成功回调不起作用。
db_functions.php
This is where the insert_tweet function is in, and it looks like this:
这是insert_tweet函数所在的位置,它看起来像这样:
function insert_tweet($author, $tweet, $date) {
global $link;
$author_ID = get_user_ID($author);
$query = "INSERT INTO tweets (`Author ID`, `Tweet`, `Date`)
VALUES ('{$author_ID}', '{$tweet}', '{$date}')";
$result = mysqli_query($link, $query);
}
I've tested it, and I'm pretty sure this runs fine. I doubt this is the cause of the problem, but if it is, I'm all ears. I've tested the $link
, which is defined in another file included in the top of the db_functions.php
file, and this works.
我已经测试过了,我很确定它运行良好。我怀疑这是问题的原因,但如果是,我全都听见了。我已经测试了$ link,它是在db_functions.php文件顶部包含的另一个文件中定义的,这是有效的。
Would appreciate some advice regarding this, thanks!
非常感谢有关此的一些建议,谢谢!
UPDATE
Changed success
to complete
, and it works. However, the data
object seems a bit odd:
改变成功完成,它的工作原理。但是,数据对象似乎有点奇怪:
data.status
pops up 200 in the alert
data.status在警报中弹出200
I tried changing the JSON array element name to data['success']
in PHP, and accessed it in the front end with data.success
, and it outputted this in the alert box:
我尝试在PHP中将JSON数组元素名称更改为data ['success'],并使用data.success在前端访问它,并在警告框中输出:
function () {
if ( list ) {
// First, we save the current length
var start = list.length;
(function add( args ) {
jQuery.each( args, function( _, arg ) {
var type = jQuery.type( arg );
if ( type === "function" ) {
if ( !options.unique || !self.has( arg ) ) {
list.push( arg );
}
} else if ( arg && arg.length && type !== "string" ) {
// Inspect recursively
add( arg );
}
});
})( arguments );
// Do we need to add the callbacks to the
// current firing batch?
if ( firing ) {
firingLength = list.length;
// With memory, if we're not firing then
// we should call right away
} else if ( memory ) {
firingStart = start;…
What does this mean??
这是什么意思??
UPDATE 2
Okay, I don't know if this helps, but I've printed the console log from Chrome's inspector, and if I'm not mistaken, the JSON data is sent back just fine. Here's the entire log:
好吧,我不知道这是否有帮助,但是我已经从Chrome的检查员打印了控制台日志,如果我没有弄错的话,JSON数据会被发回。这是整个日志:
Object {readyState: 4, getResponseHeader: function, getAllResponseHeaders: function, setRequestHeader: function, overrideMimeType: function…}
abort: function ( statusText ) {
always: function () {
complete: function () {
arguments: null
caller: null
length: 0
name: ""
prototype: Object
__proto__: function Empty() {}
<function scope>
done: function () {
error: function () {
fail: function () {
getAllResponseHeaders: function () {
getResponseHeader: function ( key ) {
overrideMimeType: function ( type ) {
pipe: function ( /* fnDone, fnFail, fnProgress */ ) {
progress: function () {
promise: function ( obj ) {
readyState: 4
responseJSON: Object
status_success: "success"
__proto__: Object
responseText: "{"status_success":"success"}"
status_success: "success"
__proto__: Object
responseText: "{"status_success":"success"}"
setRequestHeader: function ( name, value ) {
state: function () {
status: 200
statusCode: function ( map ) {
statusText: "OK"
success: function () {
then: function ( /* fnDone, fnFail, fnProgress */ ) {
__proto__: Object
__defineGetter__: function __defineGetter__() { [native code] }
__defineSetter__: function __defineSetter__() { [native code] }
__lookupGetter__: function __lookupGetter__() { [native code] }
__lookupSetter__: function __lookupSetter__() { [native code] }
constructor: function Object() { [native code] }
hasOwnProperty: function hasOwnProperty() { [native code] }
isPrototypeOf: function isPrototypeOf() { [native code] }
propertyIsEnumerable: function propertyIsEnumerable() { [native code] }
toLocaleString: function toLocaleString() { [native code] }
toString: function toString() { [native code] }
valueOf: function valueOf() { [native code] }
get __proto__: function __proto__() { [native code] }
set __proto__: function __proto__() { [native code] }
UPDATE 3
Console error scrn shot
控制台错误scrn拍摄
3 个解决方案
#1
1
Try this:
$.ajax({
type : 'POST',
url : '../php/tweet.php',
data : postData,
dataType : 'json',
complete : function(data) {
alert(data.status);
}
})
#2
0
Try the below code. I've added content type while sending data to server, parseJson method in success method,
请尝试以下代码。我在向服务器发送数据时添加了内容类型,在成功方法中使用了parseJson方法,
$.ajax({
url: "../php/tweet.php",
data: '{"author": "' + $("#username").text().split("@")[1] + '","tweet": "' + tweet_text + '","date": "' + getTimeNow() + '"}',
dataType: "json",
contentType: "application/json; charset=utf-8",
type: "POST",
success: function (data) {
var myResult = $.parseJSON(data);
},
error: function (err) {
alert(err)
}
});
if you have any clarification you can go through jquery ajax document
如果您有任何说明,可以查看jquery ajax文档
Note:it's helpful for you if you add callback for error method
注意:如果为error方法添加回调,它会对您有所帮助
#3
0
Try to set content type in your PHP file:
尝试在PHP文件中设置内容类型:
header( 'Content-Type: application/json' );
Then echo your data:
然后回显你的数据:
echo json_encode( $data );
And stop the script:
并停止脚本:
exit;
Hope it helps!
希望能帮助到你!
#1
1
Try this:
$.ajax({
type : 'POST',
url : '../php/tweet.php',
data : postData,
dataType : 'json',
complete : function(data) {
alert(data.status);
}
})
#2
0
Try the below code. I've added content type while sending data to server, parseJson method in success method,
请尝试以下代码。我在向服务器发送数据时添加了内容类型,在成功方法中使用了parseJson方法,
$.ajax({
url: "../php/tweet.php",
data: '{"author": "' + $("#username").text().split("@")[1] + '","tweet": "' + tweet_text + '","date": "' + getTimeNow() + '"}',
dataType: "json",
contentType: "application/json; charset=utf-8",
type: "POST",
success: function (data) {
var myResult = $.parseJSON(data);
},
error: function (err) {
alert(err)
}
});
if you have any clarification you can go through jquery ajax document
如果您有任何说明,可以查看jquery ajax文档
Note:it's helpful for you if you add callback for error method
注意:如果为error方法添加回调,它会对您有所帮助
#3
0
Try to set content type in your PHP file:
尝试在PHP文件中设置内容类型:
header( 'Content-Type: application/json' );
Then echo your data:
然后回显你的数据:
echo json_encode( $data );
And stop the script:
并停止脚本:
exit;
Hope it helps!
希望能帮助到你!