我的网站上有一个表，它从数据库中获取数据，在表中我有一行的delete按钮，我刚刚用jquery使表可以搜索，但是现在我不知道如何保留delete按钮。
索引.php：
<!DOCTYPE html>
<html lang="en">
<head>
<meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1">
<meta name="viewport" content="width=device-width, initial-scale=1, shrink-to-fit=no">
<meta charset="utf-8">
<meta name="description" content="Inventar">
<meta name="author" content="Martin Eide Bjørndal">
<!-- Latest compiled and minified CSS -->
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/4.5.0/css/bootstrap.min.css">
<link rel="stylesheet" href="/src/css/style.css">
<!-- jQuery library -->
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.5.1/jquery.min.js"></script>
<!-- Popper JS -->
<script src="https://cdnjs.cloudflare.com/ajax/libs/popper.js/1.16.0/umd/popper.min.js"></script>
<!-- Latest compiled JavaScript -->
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/4.5.0/js/bootstrap.min.js"></script>
</head>
<body>
<button class="back" onclick="window.location='../'">
<img src="/src/icons/back.png">
</button> <!-- Tilbake knapp -->
<button class="ny_inventar" onclick="window.location='/admin/brukere/ny'">
<img src="/src/icons/add.png">
</button> <!-- Ny Bruker knapp -->
<div class="container-fluid" id="utskjekkcontainer">
<div class="row justify-content-center">
<div class="col-mv-10 bg-light mt-5 rounded p-3">
<h1 class="text-primary p-2">Brukere</h1>
<hr>
<div class="form-inline">
<label for="search" class="font-weight-bold lead text-dark">Søk</label>
<input type="text" name="search" id="search_text" class="form-control form-control-lg rounded-0 border-primary" placeholder="Søk...">
</div>
<hr>
<?php
include "../../src/fn/init.php";
$stmt=$conn->prepare("SELECT * FROM brukere");
$stmt->execute();
$result=$stmt->get_result();
?>
<table class="table table-hover table-light table-striped" id="table_data">
<thead>
<tr>
<th>#</th>
<th>Fornavn</th>
<th>Etternavn</th>
<th>Utskjekket</th>
<th>Admin</th>
<th>Slett</th>
</tr>
</thead>
<tbody>
<?php while($row=$result->fetch_assoc()){ ?>
<tr>
<td><?php echo $row["id"]; ?></td>
<td><?php echo $row["fornavn"]; ?></td>
<td><?php echo $row["etternavn"]; ?></td>
<td><?php echo $row["utskjekket"]; ?></td>
<td><?php echo $row["is_admin"]; ?></td>
<td><a href="delete.php?id=<?php echo urlencode($row['id']); ?>" onclick="return confirm('Er du sikker?');">Slett</a></td>
</tr>
<?php }
$conn->close();
?>
</tbody>
</table>
</div>
</div>
</div>
<script type="text/javascript">
$(document).ready(function(){
$('#search_text').keyup(function(){
var search = $(this).val();
$.ajax({
url:'action.php',
method:'post',
data:{query:search},
success:function(response){
$('#table_data').html(response);
}
});
});
});
</script>
</body>
</html>动作.php：
<?php
include "../../src/fn/init.php";
$output = "";
if(isset($_POST["query"])){
$search = $_POST["query"];
$stmt = $conn->prepare("SELECT * FROM brukere WHERE fornavn LIKE CONCAT('%',?,'%') OR etternavn LIKE CONCAT('%',?,'%') OR id LIKE CONCAT('%',?,'%')");
$stmt->bind_param("sss",$search, $search, $search);
} else {
$stmt=$conn->prepare("SELECT * FROM brukere");
};
$stmt->execute();
$result=$stmt->get_result();
if($result->num_rows>0){
$output = "<thead>
<tr>
<th>#</th>
<th>Fornavn</th>
<th>Etternavn</th>
<th>Utskjekket</th>
<th>Admin</th>
<th>Slett</th>
</tr>
</thead>
<tbody>";
while($row=$result->fetch_assoc()){
$output .= "
<tr>
<td>".$row["id"]."</td>
<td>".$row["fornavn"]."</td>
<td>".$row["etternavn"]."</td>
<td>".$row["utskjekket"]."</td>
<td>".$row["is_admin"]."</td>
<td>".<a href="delete.php?id=<?php echo urlencode($row['id']); ?>" onclick="return confirm('Er du sikker?');">Slett</a>."</td>
</tr>";
};
$output .= "</tbody>";
echo $output;
} else {
echo "<h3>No match found!</h3>";
};
?>这是搜索脚本，问题是在while循环中添加了链接，我不知道如何使它工作。
3条答案
按热度按时间piv4azn71#
这个
a标记不会作为纯文本添加到output字符串，您需要重写该行，将字符串部分正确地连接到变量部分。尝试替换：
使用：
wnrlj8wa2#
要允许php从表结果回显id，只需执行php回显即可
您不必在这个阶段进行url编码，因为您只是简单地构建a:链接的href
fjaof16o3#
你可以用它