这个问题在这里已经有答案了：
$stmt->execute（）：如何知道db insert是否成功(6个答案）
两年前关门了。
<?php
require_once("./include/membersite_config.php");
if(!$fgmembersite->CheckLogin())
{
$fgmembersite->RedirectToURL("index.php");
exit;
}
include 'header.php';
include './include/sql/connect.php';
if(isset($_POST['submit']))
{
$insert = "INSERT INTO `customers`(`bb_id`, `name`, `email`, `phone`, `circle`, `ssa`, `sdca`) VALUES (?,?,?,?,?,?,?)";
$stmt = mysqli_prepare($connect, $insert);
$stmt->bind_param('sssisss', $_POST['bb_id'],$_POST['name'],$_POST['email'],$_POST['phone'],$_POST['circle'],$_POST['ssa'],$_POST['sdca']);
$stmt->execute();
if (!$stmt)
{
printf("Error: %s\n", mysqli_error($connect));
exit();
}
}
$cust = "SELECT * FROM `customers`";
$qry = mysqli_query($connect,$cust);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en-US" lang="en-US">
<head>
<meta http-equiv='Content-Type' content='text/html; charset=utf-8'/>
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>Customers</title>
<!--link rel="stylesheet" type="text/css" href="style/fg_membersite.css"-->
<link rel="stylesheet" type="text/css" href="style/tableview.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
</head>
<body>
<br>
<div id='tableview'>
<div class='ext-box'>
<form action="" method="post">
<input type="text" name="bb_id" placeholder="Broadband ID">
<input type="text" name="name" placeholder="Name">
<input type="text" name="email" placeholder="Email">
<input type="text" name="phone" placeholder="Phone">
<input type="text" name="circle" placeholder="Circle">
<input type="text" name="ssa" placeholder="SSA">
<input type="text" name="sdca" placeholder="SDCA"><br><br>
<input type="submit" name="searchsubmit" value="Search Records">
<input type="submit" name="submit" value="Update Records">
</form>
</div>
<br><br><br>
</div>
<table class="blueTable">
<thead>
<tr>
<th>Broadband ID</th>
<th>Name</th>
<th>Email</th>
<th>Phone</th>
<th>Circle</th>
<th>SSA</th>
<th>SDCA</th>
</tr>
</thead>
<tfoot>
<tr>
<td colspan="7">
<div class="links"><a href="#">«</a> <a class="active" href="#">1</a> <a href="#">2</a> <a href="#">3</a> <a href="#">4</a> <a href="#">»</a></div>
</td>
</tr>
</tfoot>
<tbody>
<?php while($row = mysqli_fetch_array($qry)):?>
<tr>
<td><?php echo $row['bb_id'];?></td>
<td><?php echo $row['name'];?></td>
<td><?php echo $row['email'];?></td>
<td><?php echo $row['phone'];?></td>
<td><?php echo $row['circle'];?></td>
<td><?php echo $row['ssa'];?></td>
<td><?php echo $row['sdca'];?></td>
</tr>
<?php endwhile;?>
</tbody>
</table>
</body>
</html>上面的代码应该将表单中填写的数据插入数据库。但奇怪的是有时候。
就像我尝试插入数据，因为2小时使用铬，但它不会这样做。突然，我对ie做了同样的事情，它拿走了插页。我想也许浏览器兼容性，但当我再次尝试与ie它停止插入后，做了一次。
1条答案
按热度按时间jhdbpxl91#
你没有检查执行是否成功。
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