希望我的问题不会太复杂......
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";
}
“block”是输出了的,所以最后一行代码没问题,但是我不能发布$q。我知道问题出在哪里,但我不知道如何解决它:如果我用document.getElementById("skills").innerHTML = this.responseText;而不用event.target.parentElement.parentElement.parentElement.nextElementSibling.innerHTML = this.responseText;,一切都运行良好,但这只适用于<div id="skills'>,不适用于skills2。我必须分别使用这个脚本来获得更多的div ID(skills, skills2, skills3)。
HTML:
<div class="bgimg" style="background-image:url('pic/a3.jpeg');">
<h3 onclick="displayGrow(event)">bla bla</h3>
<div id="sport" class="hide resetInput"><?php include 'sport.php'; ?></div>
<div id="skills" class="hide resetInput"><?php include 'skills.php'; ?></div>
</div><!-- end of the 1st Parallax -->
<div class="bgimg" style="background-image: url('pic/a5.jpg');">
<h3 onclick="displayGrow(event)">bla bla</h3>
<div id="sport2" class="hide resetInput"><?php include 'sport.php'; ?></div>
<div id="skills2" class="hide resetInput"><?php include 'skills.php'; ?></div>
</div><!-- end of the 2nd Parallax -->
sport.php代码:
<?php
include 'cookies.php';
if ($_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,在调用showSport()之后,在sport.php中将显示具有$ q值的按钮,div ID:sport和sport2都可以正常工作。 另一个函数showSkills()应该在<div id =“ skills”>或 <div id =“ skills2”>中打开skill.php并在其中发布$ q。该部分已打开,但内部没有$q。你们有任何建议吗?