我们如何在jQuery/php/mySQL中显示最近的通知?更新 [英] How can we show recent notifications in jQuery/php/mySQL? update
问题描述
So as stated below in the answers. I added a neew column to DB called seen. Default value is 0. If I understand correctly, then I would need to change the seen=0 to 1 as soon the notification is displayed, so it wont loop anymore and show me infinite amount of the same notification.
这就是我现在所拥有的:
That's what I have at the moment:
function fetch_notification(){
setInterval(function(){
//GET ALL DATA WHERE SEEN=0
$.ajax({
url: "fetchResults.php",
success: function(data){
$.each(data.vormid, function(i, vormid) {
$("#noti-box").append('<div class="alert alert-info "><button data-dismiss="alert" class="close close-sm" type="button"><i class="fa fa-times"></i></button>New form filled out by Dr. '+data.vormid[i].arsti_eesnimi+' '+data.vormid[i].arsti_perekonnanimi+'</div>');
});
update_notification();
}, dataType: "json"});
}, 5000);
}
fetch_notification();
//UPDATE SEEN=0 to 1
function update_notification(){
console.log("updating");
}
我的两个PHP文件是fetchResults.php和updateResults.php
My two PHP files are fetchResults.php and updateResults.php
fetchResults.php:
fetchResults.php:
<?php
header('Content-Type: application/json');
include_once '../dbconfig.php';
$stmt4 = $DB_con->prepare("SELECT * FROM ravim WHERE seen =0 ORDER BY date_created DESC");
$stmt4->execute();
$vormid = $stmt4->fetchAll(PDO::FETCH_ASSOC);
echo json_encode(array("vormid" => $vormid));
?>
updateResults.php:
updateResults.php:
<?php
header('Content-Type: application/json');
include_once '../dbconfig.php';
$ravim_id = $_POST['ravim_id'] ;
$stmt4 = $DB_con->prepare("UPDATE ravim SET seen=1 WHERE ravim_id=:ravim_id");
$stmt4->execute();
?>
编辑 所以我设法只显示一次通知.现在，我的问题与音频有关，通常来说，我编写的代码是否正常"，还是应该更改某些内容. 音频:每次通知到时，我都想播放声音，但从一开始它就一直播放一次.我尝试添加一个循环计数器，并检查对象是否为空，但是它不起作用.有什么建议或好的做法，如何处理音频?
EDIT So i managed to show the notification only once. My question now comes with the audio and in general if the code I wrote is "okay" or should I change something. Audio: I would like to play a sound every time the notification comes, but it keeps playing once in the beginning. I tried adding a loopcounter, and checking if the object is empty or not, but it doesn't work. Any advice or good practice how to deal with audio?
代码:
$.ajaxSetup ({
cache: false
});
var loopLimit = 1;
var loopCounter = 0;
setInterval(function(){
$.getJSON('fetchResults.php', function(data) {
$("#loadingDiv").show();
$('#noti-box').empty();
$("#notificationTitle").empty();
if(jQuery.isEmptyObject(data)){
console.log("there is no data");
}else{
console.log(data);
if (loopCounter < loopLimit){
var sound = $("#notification")[0];
sound.currentTime = 0;
sound.load();
sound.play();
loopCounter++;
}
$.each(data.vormid, function(i, vormid) {
$("#noti-box").append('<div class="alert alert-info"><button id='+data.vormid[i].ravim_id+' data-dismiss="alert" class="close close-sm" type="button"><i class="fa fa-times"></i></button>New form filled out by Dr. <b>'+data.vormid[i].arsti_eesnimi+' '+data.vormid[i].arsti_perekonnanimi+' </b> at '+data.vormid[i].date_created+'</div>');
var id= $(".alert.alert-info").val();
$("#notificationTitle").append('<li style="font-size: 14px; padding: 0; margin: 0 0 10px 10px; color: #666666;" >New form filled out by Dr. '+data.vormid[i].arsti_eesnimi+' '+data.vormid[i].arsti_perekonnanimi+'</li>');
$(".close").on('click', function () {
var ravim_id = $(this).attr('id');
$.ajax({
type: "POST",
url:'updateResults.php',
data:"ravim_id=" + ravim_id,
success:function(data){
console.log("success");
},error: function(data){
console.log("not saved");
}
});
});
});
}
$("#loadingDiv").fadeOut("slow");
});
}, 5000);
推荐答案
目前尚不清楚问题出在哪里，但是这里您的代码有错误:
It is not entirely clear what the problem is, but here you have an error in your code:
$stmt4 = $DB_con->prepare("UPDATE ravim SET seen=1 WHERE ravim_id=:ravim_id");
$stmt4->execute();
您正在使用绑定参数，但没有绑定它.您可以通过将数组添加到execute()方法来解决该问题，或者在单独的语句中手动将其绑定.
You are using a bound parameter but you are not binding it. You can solve that by adding an array to the execute() method or you bind it manually in a separate statement.
使用第一个选项:
$stmt4 = $DB_con->prepare("UPDATE ravim SET seen=1 WHERE ravim_id=:ravim_id");
$stmt4->execute(array(
':ravim_id' => $ravim_id
));
请注意，您还可以设置PDO在遇到问题时引发异常，因此您应该这样做，以便在有问题时得到通知.
Note that you can also setup PDO to throw exceptions when it runs into problems so you should probably do that so that you get notified when there is one.
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