我有这段代码可以在表格中显示网站上的两条记录。目前无法使用。
<?php
//make connection
mysqli_connect('localhost', 'root', 'password', 'databasename')or die('Connection could not be established With msql_connect');
$sql = "SELECT * FROM players ORDER BY bankacc DESC LIMIT 10";
$records = mysqli_query($sql) or die('Connection could not be established With mysqli_query');
?>
要创建表,我需要以下PHP代码:
<?php
while($players=mysqli_fetch_assoc ($records)){
echo"<tr>";
echo"<td>","<center>".$players['name'];"</center>";"<td>";
echo "<td>","<center>","$",number_format($players['bankacc']);"</center>";"<td>";
echo"</tr>";
}//end while
?>
当我尝试运行它时,它说:
无法使用mysqli_query建立连接
希望我的网页上有人帮助我。
最佳答案
您会感到困惑/不了解以程序方式或面向对象方式使用mysqli的两种方式。
如果看到代码,您会发现mysqli_connect
没有存储在任何地方,因此mysqli_query
也会引发错误,必须将两者链接在一起。
编辑:从您问题中的代码,您正在使用一种程序方法,因此,这里是用于修复您的错误的解决方案:
<?php
//make connection
$conn = mysqli_connect('localhost', 'root', 'password', 'databasename')or die('Connection could not be established With msql_connect');
$sql = "SELECT * FROM players ORDER BY bankacc DESC LIMIT 10";
$records = mysqli_query($conn, $sql) or die('A query error has occurred: ' . mysqli_error($conn));
?>
关于php - PHP mysqli代码不起作用,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/45524566/