【发布时间】：2021-12-25 01:43:12
【问题描述】：
我必须创建一个网页，允许收件人预订或删除可用餐点（如果之前已预订），如果餐点已预订，则其样式必须更改。
但我一直面临这个问题，除非我刷新页面，否则（餐点名称）的样式不会改变。另外，当我单击保留时，数据库会更新并显示删除按钮，但 它被禁用，除非我刷新页面，否则我无法再单击它。
这是收件人主页的 Ajax Javascript 部分：
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.6.0/jquery.min.js" > </script>
<script>
$(document).ready(function(){
//when remove button is clicked
$(".rmvJQ").click(function(){
var classIdValue = this.value;
var toResvBtn = "#" + classIdValue +"toResvBtn";
var ToRmv = "#" + classIdValue +"ToRmv";
// var styleM = "#" + classIdValue +"toStyle";
$.get("rmov.php", {classID : classIdValue},
function(response){
var opeartion = "success";
if(response == opeartion){
alert("The remove operation was failed");
}else{
// alert(classIdValue);
}
$("#RemovedSt").css("background-color");
$(toResvBtn).html("<button class ='rsvJQ' value ="+classIdValue+" >Reserve</button>");
$(ToRmv).html("");
});
});
//when reserve button is clicked
$(".rsvJQ").click(function(){
var classIdValue = this.value;
var toResevededString = "#" +classIdValue + "toResevededString";
var tormovBtn = "#" + classIdValue + "tormovBtn";
var ReservedSt = "#" + classIdValue + "ReservedSt";
$.get("rsv.php", {classID : classIdValue},
function(response){
var opeartion = "success";
if(response == opeartion){
alert("The reseve operation was failed");
}else{
alert("The reseve operation was successful!");
}
$(ReservedSt).css({"background-color": "yellow", "font-size": "200%"});
$(toResevededString).html("Reserved");
$(tormovBtn).html("<button class ='rmvJQ' value = '"+classIdValue+"' >Remove</button>");
});
});
});
</script>
这是同一页面的 php/html 部分
<?php
$error= mysqli_connect_errno();
if($error!=null){
echo "<p>unable to connect with db</p>";
exit($output);
}
else
{
$sql="SELECT * FROM restaurant";
$res=mysqli_query($con,$sql);
while($row=mysqli_fetch_assoc($res)){
$restaurantID=$row['id'];
$sql="SELECT * FROM meal WHERE restaurant_id=".$restaurantID;
$meals=mysqli_query($con,$sql);
$check = mysqli_num_rows($meals);
if($check>0){
//list of meals for each restaurant
echo '<table id= "mealsList">';
echo '<thead>';
echo '<tr>';
echo "<caption>".$row['name']."</caption>" ;
echo "<tr ><th > Meals </th>"
. "<th colspan='2' > Status </th>"
."<br><br></tr>";
echo'</thead>';
//each meal
echo'<tbody>';
$resv=array(); //reserved meals
$remv=array(); //removed meals
while($row=mysqli_fetch_assoc($meals)){
$sql4="SELECT * FROM reservation WHERE meal_id=".$row['id']." AND recipient_id=".$_SESSION['recipient_id'];
$res4=mysqli_query($con,$sql4);
$row4=mysqli_fetch_assoc($res4);
if($row4){
$resv[]=$row;
}
else{
$remv[]= $row;
}
}
$i=0;
while($i<count($resv) ){
echo '<tr >';
echo '<td id="'.$resv[$i]['id'].'ReservedSt" ><a style="background-color:#00ff00" href="MealInfo.php?mealID='.$resv[$i]['id'].'">'.$resv[$i]['name'].''.'</a> </td>';
echo '<td id="'.$resv[$i]['id'].'toResvBtn">Reserved</td>';
echo '<td id="'.$resv[$i]['id'].'ToRmv"><button class = "rmvJQ" value = "'.$resv[$i]['id'].'" >Remove</button></td>';
echo '</tr>';
$i++;
}
$i=0;
while($i<count($remv)){
$checkZero = mysqli_query($con,"SELECT qty FROM meal WHERE id =".$remv[$i]['id']);
$checkZ=mysqli_fetch_assoc($checkZero);
if($checkZ['qty']>0){
echo '<tr id="RemovedSt" style="background-color:white" >'
. '<td ><a href="MealInfo.php?mealID='.$remv[$i]['id'].'">'.$remv[$i]['name'].'</a></td>' //<a href="enroll.php?classID='.$row['id'].'">Enroll</a>
. '<td id= "'.$remv[$i]['id'].'toResevededString"> <button class = "rsvJQ" value = "'.$remv[$i]['id'].'" >Reseve</button> </td>'
. '<td id = "'.$remv[$i]['id'].'tormovBtn"></td>'
. '</tr>';
}else{
echo '<tr>';
echo '<th ><a style="color:black;" href="MealInfo.php?mealID='.$remv[$i]['id'].'">'.$remv[$i]['name'].'</a> </th>';
echo '<td colspan="2" > meal is not available </td >';
echo '</tr>';
}
$i++;
}//end of while loop
echo '</tbody>';
echo '</table>';
}//end of if($check>0)
}//end of big while
}//end of big if-else
?>
an image of Recipient Home page with each restaurant and their available meals
请问有没有人遇到过这个问题或者知道如何解决呢？
【问题讨论】：
-
欢迎来到 Stack Overflow。有点不清楚为什么要为元素分配 CSS 样式而不是分配/删除特定的类名称。这使您更轻松地更改样式，并且通常一起更快。我们还需要查看生成的 HTML 而不是 PHP 才能提供适当的帮助。请提供一个最小的、可重现的示例：stackoverflow.com/help/minimal-reproducible-example
-
您说的是同一页面的 html/php？，因为您使用的是 AJAX，所以最好将 php 代码放在单独的文件中，甚至在您需要控制执行流程的单个文件中。跨度>
标签： javascript php html jquery ajax