我使用ajax进程修改index.php
文件上的用户状态。
它有效,但我想为用户状态的div函数着色
我的代码:
function recupstatut() {
$.post('recup.php', function(data) {
$('.cont2').html(data);
var content = document.querySelector('#cont2');
var status2 = content.innerHTML;
if (status2 == "En-ligne") {
content.style.backgroundColor = "#4CAF50";
} else {
content.style.backgroundColor = "#f44336";
}
});
}
setInterval(recupstatut, 1000);
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.0/jquery.min.js"></script>
<div class="cont2" id="cont2">
</div>
条件始终应用else状态:
content.style.backgroundColor = "#f44336";
我认为问题来自var status2 =
我该如何解决这个问题?
答案 0 :(得分:5)
<强> HTML 强>
<div class="cont2" id="cont2"></div>
<强> SCRIPT 强>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.0/jquery.min.js"></script>
<script>
function recupstatut() {
$.post('recup.php', function(data) {
console.log(data);
var status2 = data.trim();
console.log(status2);
$('.cont2').html(status2);
if (status2 == "En-ligne") {
content.style.backgroundColor = "#4CAF50";
} else {
content.style.backgroundColor = "#f44336";
}
});
}
setInterval(recupstatut, 1000);
</script>
出了什么问题,你在调用函数
后导入了jquery文件所以在调用你的函数之前进行导入
你的错误是你在调用函数后进行了导入,这就是你得到未定义错误的原因。
答案 1 :(得分:0)
正如您在页面中提到的echo
字符串,您可以直接从data
按照以下代码检查此字符串。
<强>脚本:强>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.0/jquery.min.js"></script>
<script>
$(function(){
function recupstatut() {
$.post('recup.php', function(data) {
$('#cont2').html(data); // If the data return from the php page as a string then you can compare it directly.
if (data == "En-ligne") {
$('#cont2').css("backgroundColor","#4CAF50");
} else {
$('#cont2').css("backgroundColor","#f44336");
}
});
}
setInterval(recupstatut, 1000);
});
</script>
HTML:
<div class="cont2" id="cont2"></div>
答案 2 :(得分:0)
function recupstatut(){
$.post('recup.php',function(data){
console.log(data);
$('.cont2').html(data);
var status2 = data;
if (status2 == "En-ligne") {
$('#cont2').css("backgroundColor","#4CAF50");
} else {
$('#cont2').css("backgroundColor","#f44336");
}
});
}
setInterval(recupstatut,1000);
现在我的div中没有任何内容出现在console.log中......
答案 3 :(得分:-1)
有很多方法可以完成这个。您可以通过将 $。post 作为变量发送来使用 $ .post()功能。例如:
// Fire off the request to /form.php
request = $.post({
url: "recup.php",
});
// Callback handler that will be called on success
request.done(function (response, textStatus, jqXHR){
// Log a message to the console
console.log("Hooray, it worked!");
});
// Callback handler that will be called on failure
request.fail(function (jqXHR, textStatus, errorThrown){
// Log the error to the console
console.error(
"The following error occurred: "+
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);
});
或(我推荐)以这种方式使用 $。ajax({})功能:
// Fire off the request to /form.php
$.ajax({
url: "recup.php",
type: "post",
data: { //specify data to be sent },
beforeSend:function(){
/* before sending the data to the other page
may be a loader to show waiting animation
*/
},
success:function(status){
/* this will check the response received from the previous page
and the determine the below conditions
*/
if (status == "En-ligne") {
content.style.backgroundColor = "#4CAF50";
} else {
content.style.backgroundColor = "#f44336";
}
}
});