我一直在关注有关在网站上显示数据库信息的在线教程，但是当我尝试使用 localhost/app_location 加载它时，它只显示标题栏。我不知道为什么它不起作用。有什么建议 ？
这是我关注的教程' https://www.youtube.com/watch?v=oxZj82kh4FA&t=195s '
这是我对代码的看法:
进程.php
<?php
$mysqli = new mysqli('localhost', 'root', '', 'testing');
if (mysqli_connect_errno()) {
echo json_encode(array('mysqli' => 'Failed to connect to MySQL: ' . mysqli_connect_error()));
exit;
}
$page = isset($_GET['p'])? $_GET['p'] : '';
if($page=='view'){
$result = $mysql->query("SELECT * FROM tabledit WHERE deleted != '1'");
while($row = $result->fetch_assoc()){
?>
<tr>
<td><?php echo $row['bookingId']; ?></td>
<td><?php echo $row['description']; ?></td>
<td><?php echo $row['bookingStart']; ?></td>
<td><?php echo $row['bookingEnd']; ?></td>
</tr>
<?php
})
}else {
header('Content-Type: application/json');
$input = filter_input_array(INPUT_POST);
if ($input['action'] == 'edit') {
$mysqli->query("UPDATE tabledit SET description='" . $input['description'] . "', bookingStart='" . $input['bookingStart'] . "'
, bookingEnd='" . $input['bookingEnd'] . "' WHERE bookingId='" . $input['bookingId'] . "'");
} else if ($input['action'] == 'delete') {
$mysqli->query("UPDATE tabledit SET deleted=1 WHERE bookingId='" . $input['bookingId'] . "'");
} else if ($input['action'] == 'restore') {
$mysqli->query("UPDATE tabledit SET deleted=0 WHERE bookingId='" . $input['bookingId'] . "'");
}
mysqli_close($mysqli);
echo json_encode($input);
}
?>
index.html:
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8"/>
<meta name="viewpoint" content="width=device-width, initial-scale=1">
<title>Table</title>
<link rel="stylesheet" href="https://stackpath.bootstrapcdn.com/bootstrap/4.3.1/css/bootstrap.min.css" >
<link rel="stylesheet" href="https://stackpath.bootstrapcdn.com/font-awesome/4.7.0/css/font-awesome.min.csst"/>
</head>
<body onload="viewData()">
<div class="container" style="margin-top:35px">
<h4>Table Quick Edit</h4>
<table id="tabledit" class="table table-bordered table-striped">
<thead>
<tr>
<th>BookingId</th>
<th>Description</th>
<th>BookingStart</th>
<th>BookingEnd</th>
</tr>
</thead>
<tbody></tbody>
</table>
</div>
<script src="jquery-3.4.1.min.js"></script>
<script src="js/bootstrap.js"></script>
<script src="js/jquery.tabledit.js"></script>
<script>
function viewData(){
$.ajax({
url: 'process.php?p=view',
method:'GET'
}).done(function(data){
$('tbody').html(data)
tableData()
})
}
function tableData(){
$('#tabledit').Tabledit({
url: 'process.php',
eventType: 'dblclick',
editButton: true,
deleteButton: true,
hideIndentifier: false,
buttons: {
edit: {
class: 'btn btn-sm btn-warning',
html: '<span class="glyphicon glyphicon-pencil"></span> Edit',
action: 'edit'
},
delete: {
class: 'btn btn-sm btn-danger',
html: '<span class="glyphicon glyphicon-trash"></span> Trash',
action: 'delete'
},
save: {
class: 'btn btn-sm btn-success',
html: 'Save'
},
restore: {
class: 'btn btn-sm btn-warning',
html: 'Restore',
action: 'restore'
},
confirm: {
class: 'btn btn-sm btn-default',
html: 'Confirm'
}
},
columns: {
identifier: [0, 'bookingId'],
editable: [[1, 'description'], [2, 'bookingStart'], [3, 'bookingEnd']]
},
onSuccess: function(data, textStatus, jqXHR) {
viewData
},
onFail: function(jqXHR, textStatus, errorThrown) {
console.log('onFail(jqXHR, textStatus, errorThrown)');
console.log(jqXHR);
console.log(textStatus);
console.log(errorThrown);
},
onAjax: function(action, serialize) {
console.log('onAjax(action, serialize)');
console.log(action);
console.log(serialize);
}
});
}
</script>
</body>
</html>
我仔细检查了每一行。我每次都运行 xampp。它连接到数据库，但没有显示任何内容。
最佳答案
你正在混合使用 mysql 和 mysqli
$result = $mysql->query("SELECT * FROM tabledit WHERE deleted != '1'");
应该是
$result = $mysqli->query("SELECT * FROM tabledit WHERE deleted != '1'");
关于php - 来自数据库的信息应该显示在网站上，但没有，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/56927229/