我在左边有搜索结果。单击结果时，它需要在右侧的 div 中加载 - 无需刷新或加载新页面。搜索给出三个搜索结果并有分页。无论我单击哪个搜索结果，都会加载相同的 ID。谁能看到我哪里出错了？索引.php //get rows $query = $db->query("SELECT * FROM posts ORDER BY id DESC LIMIT $limit"); if($query->num_rows > 0){ ?> <div class="posts_list"> <?php while($row = $query->fetch_assoc()){ $postID = $row['id']; ?> <div class="list_item"><h2><?php echo $row["title"] ?></h2><button type="button" onclick="loadDoc()" >View</button></div> <div class="item_viewer"><p id="demo"></p></div> <script type="text/javascript"> function loadDoc() { var xhttp = new XMLHttpRequest(); xhttp.onreadystatechange = function() { if (this.readyState == 4 && this.status == 200) { document.getElementById("demo").innerHTML = this.responseText; } }; xhttp.open("GET", "file/file.php?id=<?php echo $row['id'] ?>", true); xhttp.send(); } </script> <?php } ?> </div> <?php echo $pagination->createLinks(); ?> <?php } ?>文件.php<body> <div class="container"> <div class="box"><?PHP//create connection$connect = new mysqli($host, $dbUsername, $dbPassword, $dbname);if (mysqli_connect_error()) {die('Connect Error('. mysqli_connect_errno().')'. mysqli_connect_error());} else {$id = $_GET['id'];$id = mysqli_real_escape_string($connect,$id);$query = "SELECT * FROM `posts` WHERE `id`='" . $id . "'";$result = mysqli_query($connect,$query);while($row = mysqli_fetch_array($result)) {echo "Company: <b>" .$row['title']. "</b>";echo "<br/>";echo "ID: <b>" .$row['id']. "</b>";?><a href="<?PHP echo $row[''];}}?>"><button class="btn success">View</button></a> <a href="" ><button class="btn default">Back</button></a> </div> </div></body>
1 回答
慕慕森
TA贡献1856条经验 获得超17个赞
您的代码loadDoc在循环内生成多个函数，因此每个按钮都使用相同的函数（相同的 rowId 静态放置在其中）。
如果您只有一个函数并将 id 作为参数传递，那就更好了，如下所示：
<?
$query = $db->query("SELECT * FROM posts ORDER BY id DESC LIMIT $limit");
if($query->num_rows > 0){ ?>
<div class="posts_list">
<?php
while($row = $query->fetch_assoc()){
$postID = $row['id'];
?>
<div class="list_item">
<h2><?php echo $row["title"] ?></h2>
<!-- Pass Id into the function -->
<button type="button" onclick="loadDoc(<?php echo $row['id'] ?>)" >View</button>
</div>
<div class="item_viewer"><p id="demo"></p></div>
<?php } ?>
</div>
<script type="text/javascript"> // Move JavaScript function out of loop, so that it only exists once in the document
function loadDoc(rowId) { // Pass rowId as argument
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("demo").innerHTML = this.responseText;
}
};
xhttp.open("GET", "file/file.php?id=" + rowId, true);
xhttp.send();
}
</script>
<?php echo $pagination->createLinks(); ?>
<?php } ?>
添加回答
举报
0/150
提交
取消