[英]event.target in AJAX
希望我的問題不太復雜。
function showSkills(event,str) {
xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
event.target.parentElement.parentElement.parentElement.nextElementSibling.innerHTML = this.responseText;
}
};
xmlhttp.open("POST","skills.php",true);
xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded");
xmlhttp.send("q=" + str);
event.target.parentElement.parentElement.parentElement.nextElementSibling.style.display = "block";
}
該塊已顯示,因此最后一行代碼很好,但是我無法發布$q
。 我知道問題出在哪里,但我不知道如何解決:如果我將document.getElementById("skills").innerHTML = this.responseText;
而不是event.target.parentElement.parentElement.parentElement.nextElementSibling.innerHTML = this.responseText;
一切正常,但僅適用於<div id="skills'>
,當然不適用於skills2
,我必須分別將此腳本用於更多的div ID (skills, skills2, skills3)
。
我的HTML:
<div class="bgimg" style="background-image:url('pic/a3.jpeg');">
<h3 onclick="displayGrow(event)">bla bla</h3>
<div id="sport" class="hide resetInput"><?php include 'sport.php'; ?></div>
<div id="skills" class="hide resetInput"><?php include 'skills.php'; ?></div>
</div><!-- end of the 1st Parallax -->
<div class="bgimg" style="background-image: url('pic/a5.jpg');">
<h3 onclick="displayGrow(event)">bla bla</h3>
<div id="sport2" class="hide resetInput"><?php include 'sport.php'; ?></div>
<div id="skills2" class="hide resetInput"><?php include 'skills.php'; ?></div>
</div><!-- end of the 2nd Parallax -->
和sport.php
代碼:
<?php
include 'cookies.php';
if ($_SERVER["REQUEST_METHOD"] == "POST") {
if (isset($_POST['q'])) {
$q = $_POST['q'];
}}
?>
<div class="dropdown marginTop">
<button><!-- select sport -->
<?php // "select your sport"
if (isset($q)) {
$result = mysqli_query($con, 'SELECT ' .$language. ' FROM sports WHERE name1="' .$q. '" LIMIT 1');
} else {
$result = mysqli_query($con, 'SELECT ' .$language. ' FROM page WHERE title="selSport1" LIMIT 1');
}
$row = mysqli_fetch_row($result);
print_r($row[0]);
?>
</button>
<div class="dropdown-content">
<?php // sports list
$sportlist = mysqli_query($con, 'SELECT * FROM sports WHERE title = "sports" ORDER BY ' .$language. ' ASC');
while ($row = mysqli_fetch_array($sportlist)) {
echo '
<button class = "buttonList" value="' . $row[1] . '"';?>
onclick="showSport(event, this.value); showSkills(event, this.value);"
<?php echo ' style="border:0;">' . $row[$language] . '</button>
';
}
?>
</div>
</div>
因此,在調用showSport()
之后,在sport.php
顯示具有$q
值的按鈕。 這對於兩個div ID都適用: sport
和sport2
。 另一項功能showSkills()
應該打開skill.php
在<div id="skills">
或<div id="skills2">
和交$q
那里。 該部分已打開,但內部沒有$q
。 有任何想法嗎? 對我有很大幫助。
event.target.parentElement.parentElement.parentElement.nextElementSibling
對於我的口味來說,爬上DOM太多了。 當我理解正確時,您可以將相同的功能用作多個元素的事件處理程序。 為什么不只將特定的關聯div的ID作為參數傳遞呢? 我在下面添加一個示例。
// Note, I do not know what your 'str' is, so here it is just 'foo' document.getElementById('event-div-1').addEventListener('click', showSkills.bind(null, 'skills1', 'foo')); document.getElementById('event-div-2').addEventListener('click', showSkills.bind(null, 'skills2', 'foo')); document.getElementById('event-div-3').addEventListener('click', showSkills.bind(null, 'skills3', 'foo')); function showSkills(skills, event, str) { // Ajax stuff document.getElementById(skills).style.display = 'block'; }
.skills { display: none; }
<div class="skills" id="skills1">1</div> <div class="skills" id="skills2">2</div> <div class="skills" id="skills3">3</div> <div id="event-div-1">Click me for skills 1</div> <div id="event-div-2">Click me for skills 2</div> <div id="event-div-3">Click me for skills 3</div>
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.