我知道这已经被问过好几次了。其中之一就是这个:https://stackoverflow.com/questions/18869513/php-insert-into-not-working但是尝试为这个解决方案提供的解决方案对我不起作用。所以我有这个主页:
<html>
<head>
<title>My First PHP Website</title>
</head>
<?php
session_start();
if($_SESSION['user']){}
else{header("location:index.php");}
$user = $_SESSION['user'];
?>
<body>
<h2>Homepage</h2>
<p>Hello
<?php Print "$user"?>
</p>
<a href="logout.php">Click Here to Logout</a>
<br>
<br>
<form action="add.php" method="POST">
Add more to list:
<input type="text" name="details">
<br> Public Post?
<input type="checkbox" name="public[]" value="yes">
<br>
<input type="submit" value="Add to List">
</form>
<h2 align="center">My List</h2>
<table border="1px" width="100%">
<tr>
<th>ID</th>
<th>Details</th>
<th>Edit</th>
<th>Delete</th>
<th>Public Posts</th>
</tr>
<?php
$connect = mysqli_connect("localhost","root","") or die(mysqli_error());
mysqli_select_db($connect, "first_db") or die ("Cannot Connect to Database");
$query = mysqli_query($connect,"SELECT * from list");
while($row = mysqli_fetch_array($query))
{
Print "<tr>";
Print '<td align="center">'. $row['id'] . "</td>";
Print '<td align="center">'. $row['details'] . "</td>";
Print '<td align="center">'. $row['date_posted']." - ".$row['time_posted']."</td>";
Print '<td align="center">'. $row['date_edited']." - ".$row['time_edited']."</td>";
Print '<td align="center"><a href="edit.php">edit</a> </td>';
Print '<td align="center"><a href="delete.php">delete</a> </td>';
Print '<td align="center">'. $row['public']. "</td>";
Print "</tr>";
}
?>
</table>
</body>
</html>
和这个操作页面:
<?php
session_start();
if($_SESSION['user']){
}
else{
header("location:index.php");
}
if($_SERVER["REQUEST_METHOD"] == "POST"){
$connect = mysqli_connect("localhost","root","");
$link = mysqli_connect("localhost","root","","first_db");
$details = mysqli_real_escape_string($connect, $_POST['details']);
$time = strftime("%X");
$date = strftime("%B %d, %Y");
$decision ="no";
mysqli_connect("localhost", "root","") or die(mysqli_error());
mysqli_select_db($connect,"first_db") or die("Cannot connect to database");
foreach($_POST['public'] as $each_check)
{
if($each_check !=null ){
$decision = "yes";
}
}
mysqli_query ($link,"INSERT INTO list (details) VALUES ('$details')");
header("location: home.php");
}
else
{
header("location:home.php");
}
?>
这应该是将输入的数据插入我的数据库，但不幸的是它不会做任何事情。我想不出这个的解决方案。我希望你们能帮我解决这个问题。提前致谢!
最佳答案
您不需要两个连接变量 $connect 和 $link。使用任何一个。
此外，您可以尝试调试。
mysqli_query ($link,"INSERT INTO list (details) VALUES ('".$details."')") or die(mysqli_error($link));
关于php - "INSERT INTO"没有向数据库中插入数据，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/36277771/