不知道如何最好地描述这一点,但我会尝试放入一些上下文。我对Javascript都很陌生,在PHP方面仍然是初学者,所以请指出您看到的任何问题 - 总是愿意学习!
话虽如此,这就是我正在做的事情
- 我有一个基本的 php 页面,它使用 AJAX 在单击div(.todo 或 .complete)时更新表格
- AJAX 返回已更新项的数组,我使用它来拉取 itemText 和 itemNo,并在div 后面附加或附加一个包含其中的信息的数组。
- 单击的 .todo 项目将向上滑动,然后添加为 .complete,单击时 .已完成的项目将恢复为 .todo。
问题是,如果我单击 .todo 项目,它会使用 slideUp 删除,然后作为 .complete 前缀,没有问题。但是,如果我随后单击新的 .done 项目以将其恢复为 .todo,则会附加一个 COPY 并保留原始内容。
我不太确定该怎么做,非常感谢任何帮助。
我的简斯
$('#needToDo').on("click", ".todo", function() {
var itemNo = $(this).attr("id");
var updateQuery = "UPDATE items SET done=1 WHERE itemNo = " + itemNo;
$.ajax({
type: "POST",
url: "toDoProcess.php",
dataType: "json",
data: { query: updateQuery, itemNo: itemNo },
success: function(resultArray){
if(resultArray != ""){
var itemNo = resultArray[0]['itemNo'];
var itemText = resultArray[0]['itemText'];
$('#' + itemNo).slideUp(function(){
$('#completedItems').prepend('<div class=''item completed'' id='''+itemNo+'''>'+itemText+'</div>');
});
} else {
console.log("Could not complete that at this time");
}
}
});
});
$('#completedItems').on("click", ".completed", function() {
var itemNo = $(this).attr("id");
console.log(itemNo);
var updateQuery = "UPDATE items SET done=0 WHERE itemNo = " + itemNo;
$.ajax({
type: "POST",
url: "toDoProcess.php",
dataType: "json",
data: { query: updateQuery, itemNo: itemNo },
success: function(resultArray){
if(resultArray != ""){
var itemNo = resultArray[0]['itemNo'];
var itemText = resultArray[0]['itemText'];
$('#' + itemNo).slideUp(function(){
$('#needToDo').append('<div class=''item todo'' id='''+itemNo+'''><input type="checkbox">'+itemText+'</div>');
});
} else {
console.log("Could not complete that at this time");
}
}
});
});
我的网页
<div class="container">
<h1 class="sectionTitle">To Do App</h1>
<h2 class="sectionTitle">Need to complete</h2>
<div id="needToDo">
<?php
// connect to DB
try {
$db = new PDO("mysql:host=localhost;dbname=toDoApp;port=3306","root","root");
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
} catch (Exception $e) {
echo $e;
}
// Query for to-do items
// (where items are not marked as done & TO BE ADDED date = selected date)
// REMEMBER TO ADD WHERE DATE = TODAY etc.
try {
$query = $db->query("SELECT itemNo, itemText, needShouldWant FROM items WHERE done != 1 ORDER BY itemNo DESC");
} catch (Exception $e) {
echo $e;
}
// array returned by the query
// (assigned to result and told to use assoc keys)
$result = $query->fetchAll(PDO::FETCH_ASSOC);
// Each variable in the result array is assigned as item while looping through, each item returns html
foreach($result as $item){
$itemNo = $item['itemNo']; ?>
<div class="item todo" id="<?php echo $itemNo ?>"><input type="checkbox" id="checkbox<?php echo $itemNo ?>"><?php echo $item['itemText'];?></div>
<?}?>
</div>
<h2 class="sectionTitle">Completed Items</h2>
<div id="completedItems">
<?php
// Query for to-do items
// (where items are not marked as done & TO BE ADDED date = selected date)
// REMEMBER TO ADD WHERE DATE = TODAY etc.
try {
$query = $db->query("SELECT * FROM items WHERE done = 1 ORDER BY itemNo DESC");
} catch (Exception $e) {
echo $e;
}
// array returned by the query
// (assigned to result and told to use assoc keys)
$result = $query->fetchAll(PDO::FETCH_ASSOC);
// Each variable in the result array is assigned as item while looping through, each item returns html
foreach($result as $item){
$itemNo = $item['itemNo']; ?>
<div class="item completed" id="<?php echo $itemNo ?>"><?php echo $item['itemText'];?></div>
<?}?>
</div>
</div>
</div>
问题是您正在创建具有相同 id 的多个元素。作为解决方案,尝试使用slideUp
隐藏元素后将其删除
$('#' + itemNo).slideUp(function () {
$(this).remove()
$('#completedItems').prepend('<div class=''item completed'' id=''' + itemNo + '''>' + itemText + '</div>');
});
.....
$('#' + itemNo).slideUp(function () {
$(this).remove()
$('#needToDo').append('<div class=''item todo'' id=''' + itemNo + '''><input type="checkbox">' + itemText + '</div>');
});