以下是包含在文件'Column_details.php'中的一段代码。目标是在用户单击表名时显示表的所有属性。表名在$ id中检索。问题是这个查询在phpmyadmin上成功运行。但在localhost上执行Column_details.php时不会运行。sql查询不在本地主机上执行
$id = $_GET['id'];
echo $id;
$result = mysqli_query($conn, "show fields from '$id'");
$row = mysqli_num_rows($result);
if($row<=0){
echo " No such columns";
}
else{
echo "<table border='1'>";
while($row=mysqli_fetch_array($result)){
$col_name = $row['Field'];
$click = "<a href='Column_details.php?mv= ".$col_name."'>" . "</a>";
echo "<tr>";
echo "<td>" . $col_name . "</td>";
echo "<td>" . $click . "</td>";
}
echo "</table>";
}
$conn->close();
?>
</body>
</html>
您可以发布您的ConnectionString – maSTAShuFu
连接@maSTArHiAn连接字符串 $康恩=新的mysqli($服务器名,用户名$ ,$ password,$ dbname); ($ conn-> connect_error){ die(“Connection failed:”。$ conn-> connect_error); } –
和呼应$ conn你会得到什么? – maSTAShuFu