Question

我正在尝试通过mysqli 在同一页面上输入和显示数据，但它显示错误＆＃34; 无法在线获取mysqli ＆＃34;并且无法显示记录。

<?php

$db=mysqli_connect("localhost","root","","abc")  or die("Not connected".mysqli_error());
$database=mysqli_select_db($db,'abc')  or die("Database not found".mysqli_error());

if(isset($_POST['submit'])){
    $roll=$_POST['roll'];
    $name=$_POST['name'];
$ins=mysqli_query($db,"insert into abc1 (roll,name)values('$roll','$name')");
}
mysqli_close($db);

?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
<form method="post">
Roll No <input type="text" name="roll" />
Name <input type="text" name="name" />
<input type="submit" name="submit" value="Submit" />
</form>
<table><tr>
<td>Roll No.</td>
<td>Name</td>
</tr>
<?php $query=mysqli_query($db,"select * from abc1");
$result=mysqli_query($db,$query);
$id=0;
while($row=mysqli_fetch_array($result,MYSQLI_ASSOC))
{
?>

<tr><td><?php echo $row['roll']; ?></td>
<td><?php echo $row['name']; ?></td>
<?php } ?>
</tr></table>
</body>

</html>

任何帮助将不胜感激。在此先感谢

Answer 1

您正在调用mysqli_query函数两次：

$query = mysqli_query($db,"select * from abc1");
$result = mysqli_query($db,$query);

错误：无法在线获取mysqli

意思是，您在mysqli_fetch_array函数中传递资源ID而不是Query Statement。所以你只需要将其用作：

$query = "select * from abc1";
$result = mysqli_query($db,$query);

旁注：

无需再次连接数据库：

$database=mysqli_select_db($db,'abc')

mysqli_connect()已经连接了您的数据库。

Answer 2

你正在做两次mysqli_query。试试这个。

$query = "select * from abc1"; $result = mysqli_query($db, $query);

mysqli获取数组错误无法获取mysqli

2 个答案: