当按下ID为#share的按钮时,我使用以下代码发送用户ID。但我不知道如何获取数据,一个变量,当为真时将运行FB.ui。
<script type="text/javascript">
$(document).ready(function(){
$("#share").click(function(){
$.ajax({
type: 'POST',
url: 'upload.php',
data: 'user_id=$user_id',
success: // I don't know what to put here, maybe "function(result){ if (result == true) {"?
FB.ui({
method: 'feed',
name: '$app_name',
link: '$app_url',
picture: '$upload_picture',
caption: 'Aluxian Apps',
description: '$app_description'
});
});
});
});
</script>
答案 0 :(得分:3)
是的,您可以json_encode您要发送回客户端的变量。
echo json_encode( array('result' => true ) );
然后在你的js
$(document).ready(function(){
$("#share").click(function(){
$.ajax({
type: 'POST',
url: 'upload.php',
dataType: 'json',
data: 'user_id=$user_id',
success: function( data ) {
if( data.result ) // you can access the result variable here
FB.ui({
method: 'feed',
name: '$app_name',
link: '$app_url',
picture: '$upload_picture',
caption: 'Aluxian Apps',
description: '$app_description'
});
}
});
});
});
答案 1 :(得分:1)
<script type="text/javascript">
$(document).ready(function(){
$("#share").click(function(){
$.ajax({
type: 'POST',
url: 'upload.php',
data: 'user_id=$user_id',
success: function(data){
if (data.result == true) {
FB.ui({
method: 'feed',
name: '$app_name',
link: '$app_url',
picture: '$upload_picture',
caption: 'Aluxian Apps',
description: '$app_description'
});
}
});
});
});
</script>
答案 2 :(得分:1)
你的假设是正确的。
$.ajax({
type: 'POST',
url: 'upload.php',
data: 'user_id=$user_id',
success: function(result) {
if (result == true) {
FB.ui({
method: 'feed',
name: '$app_name',
link: '$app_url',
picture: '$upload_picture',
caption: 'Aluxian Apps',
description: '$app_description'
});
}
}
});
答案 3 :(得分:0)
请务必在PHP文件中添加Content Type标头,以便将信息正确地传回客户端。在php文件的顶部:header('Content type: application/json);