不确定这是怎么回事。我将首先解释文件的组织，然后进入代码。
应用程序：允许用户选择不同的属性（即年份或名称），并从数据库中查看匹配结果的图片。
文件-我有一个gallery.php文件，其中包含一系列HTML表单元素，这些元素用作选择器以从数据库中获取过滤结果。每当设置或更改选择器时，文件都会向get.php文件发送新请求，该文件使用Ajax刷新结果而不加载新页面。一切都很好。我的下一个任务是实现一个模式部分，在其中可以单击图像并查看打算使用JavaScript和CSS进行放大的版本，但是我的中间目标是仅更改{{ 1}}再次使用JavaScript，只是第一步。但是我似乎无法启动任何用get.php编写的JavaScript。
代码-
这是get.php：
gallery.php
这是<?php
include_once("./../php/navbar.php");
?>
<html>
<head>
<link href="/css/siteTheme.css" rel="stylesheet">
<style>
.attributes span {
margin-right: 1rem;
}
</style>
<script>
function changeParams() {
if (window.XMLHttpRequest) {
xmlhttp = new XMLHttpRequest();
} else {
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("test").innerHTML = this.responseText;
}
};
year = document.getElementById("years_select").value;
nameFirst= document.getElementById("nameFirst_select").value;
/* Ton of other getElementById statements here that I'm excluding to keep things shorter */
url = "get.php?year="+year+......;
xmlhttp.open("GET", url, true);
xmlhttp.send();
} /* End function */
</script>
</head>
<body>
<div class="content">
<h1>Gallery</h1>
<form>Details:
<!-- -------------------- Year -------------------- -->
<select onchange="changeParams()" name="years" id="years_select">
<option value="All">Year</option>
<?php
include("/var/www/admin.php");
$conn = mysqli_connect($dbServername, $publicdbUsername, $publicdbPass, $dbName);
if (!$conn) {
die('Could not connect: ' . mysqli_error($conn));
}
$sql = "select year from db group by year order by year desc";
$result = mysqli_query($conn, $sql);
while ($row = mysqli_fetch_array($result)) {
echo "<option value=" . $row['year'] . ">" . $row['year'] . "</option>";
}
mysqli_close($conn);
?>
</select>
<!-- A bunch of other selectors are created here following the same pattern as above of getting results from a DB -->
<!-- -------------------- Searchbar -------------------- -->
<input type="text" size="50" onkeyup="changeParams()" name="search" id="search" placeholder="Search"></input>
</form>
<div id="test">Choose some filters to see some results!</div>
</form>
</div> <!-- Ends content div -->
</body>
</html>：
get.php
很抱歉，如果那是一团糟，我砍掉了一堆只获取/选择/过滤一些数据的东西。所有有效的方法和其中保留的所有变量都在我的完整代码中设置。
但是我遇到的问题是在<html>
<head>
<style>
/* Bunch of CSS that's not relevant here */
</style>
<script type="text/javascript">
console.log(".....");
var parts = document.querySelectorAll("div.imgContainer");
console.log("Found something:", parts);
parts.addEventListener("click", function(){
document.getElementById("anID").innerHTML = "Test...";
});
</script>
</head>
<body>
<?php
include("/var/www/admin.php");
$year = $_GET['year'];
//Bunch of other variables set here following the same logic, getting data from gallery.php
$conn = mysqli_connect($dbServername, $publicdbUsername, $publicdbPass, $dbName);
if (!$conn) {
die('Could not connect: ' . mysqli_error($conn));
echo '$conn';
}
$sql = 'select * db';
/* ---------- Creating SQL statement ---------- */
$clauses = 0;
if ($year != "All") {
if ($clauses == 0) {
$sql = $sql . ' where year = "' . $year . '" and';
$clauses = $clauses + 1;
} else {
$sql = $sql . ' year = "' . $year . '" and';
}
} /* Bunch of other if statements to get set information and add to sql statement as such */
// Need to chop of the last ' and' from the sql statement
$sql = substr($sql, 0, -4);
$sql = $sql . ' order by year desc';
$result = mysqli_query($conn, $sql);
$num_results = mysqli_num_rows($result);
if ($num_results == 0 or $clauses == 0) {
echo "<p>No matches to your query. Try refining your search terms to get some results.</p>";
} else {
echo "<p>" . $num_results . " results matched your query.</p>";
echo "<div class=results>";
//echo "<div>";
echo '<script type="text/javascript">
function modalFunction() {
document.getElementById("anID").innerHTML = "test";
}
</script>';
while ($row = mysqli_fetch_array($result)) {
$pic = $row['pathToPic'];
$wwwImg = substr($pic, 13);
//echo "<span id=aCard><img src=" . $wwwImg . " height ='250px'>";
//echo "<span class=text>" . $row['fullCardInfo'] . "</span></span>";
echo "<div class=fullContainer><div class='imgContainer'><img class=image src=" . $wwwImg ."></div><p class=text>" . $row['fullInfo'] . "</p></div>";
} // End while of results
echo "</div>";// End results div//</div>";
//echo '<div class="modal"><p id="anID"></p></div>';
} // End else of "if results"
mysqli_close($conn);
?>
<script>
</script>
<div>
<p id="anID">This in a div</p>
</div>
<!--<span>
<p id="anID">This in a span</p>
</span>-->
</body>
</html>中编写JavaScript来更改get.php标签中的文本。我已经在<p id="anID">的几个不同区域中尝试过JS，从get.php标记中开始，到在PHP的<head>标记中回显，再到纯<script>标记之后， PHP语句已完成（我想我将它们留在了几个不同的地方）。
问题是我所做的任何工作都无法更改我引用的<script>标记中的文本。此外，<p>标头中的console.log语句似乎也没有被调用。
从长远来看，我将添加JS查询选择器以在单击图像时显示更大的图像，但是现在我只是想在get.php中使任何JavaScript正常工作，而我正在努力这样做。我看不到这是建议重复的重复方式，因为我不是在这里尝试通过innerHTML运行JavaScript