Question

我想从我的数据库中获取数据。上传文件时页面不会更改。我哪里错了？

verifycheck.php

<?php
$con=mysql_connect ("###", "###", "###");
mysql_select_db ("db_name", $con);


$result = mysql_query($con,"SELECT * FROM db_tablename");


echo "<table border='1'>
<tr>
<th>username</th>
<th>email</th>
<th>password</th>
<th>confirm_password</th>
</tr>";

while($row = mysql_fetch_assoc($result)) {
  echo "<tr>";
  echo "<td>" . $row['username'] . "</td>";
  echo "<td>" . $row['email'] . "</td>";
  echo "<td>" . $row['password'] . "</td>";
  echo "<td>" . $row['confirm_password'] . "</td>";
  echo "</tr>";
}

echo "</table>";

mysql_close($con);
?>

Answer 1

连接对象应该是第二个参数，查询字符串应该是第一个参数。

试试这个

$result = mysql_query("SELECT * FROM db_tablename",$con);

而不是

$result = mysql_query($con,"SELECT * FROM db_tablename");

不推荐使用Mysql函数，将来会删除 Mysqli 或 PDO 以防止sql注入

Answer 2

从$con,

中删除mysql_query

由于mysql_query()期望第一个参数是SQL查询，而不是SQL连接。

$result = mysql_query("SELECT * FROM db_tablename");

Answer 3

请使用您的连接参数检查为localhost，用户名密码，数据库以及更改

  $result = mysql_query($con,"SELECT * FROM db_tablename"); 
   as 
   $result = mysql_query("SELECT * FROM db_tablename",$con);

用php从数据库中获取数据

3 个答案: