1

所以，这是我的代码：

$query = "SELECT `id` FROM `users` WHERE `username`='$username' AND `password`='$password_hash'";
echo $query;
echo mysql_result(mysql_query($query), 0);

当我在 PHP 中运行此代码时，我执行了查询，并在 phpMyAdmin 中运行它，它显示 id 1。但它不输出任何内容，当运行完整代码时，我收到此错误：

Warning: mysql_result() expects parameter 1 to be resource
4

4 回答 4

2

从 php 手册：

mysql_query() will also fail and return FALSE if the user does not have permission to access the table(s) referenced by the query.

您确定您已连接到数据库并且拥有良好的权限吗？

于 2012-12-04T18:21:42.263 回答
1

确保您正确设置连接和查询

 $link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
 if (!$link) {
     die('Could not connect: ' . mysql_error());
 }
 if (!mysql_select_db('database_name')) {
    die('Could not select database: ' . mysql_error());
 }
 $result = mysql_query('SELECT * FROM MyTable');
 if (!$result) {
    die('Could not query:' . mysql_error());
}
echo mysql_result($result, 0); // outputs first row

mysql_close($link);
于 2012-12-04T19:12:50.850 回答
0 
        $query = "SELECT `id` FROM `users`
                  WHERE `username`='$username'
                  AND `password`='$password_hash'";
        echo $query;
        $sql_result=mysql_query($query) or die(mysql_error());
        echo mysql_result($sql_result, 0);
于 2012-12-04T18:23:54.087 回答
0

使用此代码并根据要求对其进行编辑，并检查数据库与 utf8_general_ci 的联合，并将表保存为一种 innodB ...

  <?php
   $mysqli_hostname = "localhost";
   $mysqli_user = "root";
   $mysqli_password = "krunal";
   $mysqli_database = "krunal";
   $bd = mysqli_connect($mysqli_hostname, $mysqli_user,$mysqli_password,$mysqli_database);  
   if(mysqli_connect_errno()){die("database connection  failed");}
   ?>


   <?php 
   $sql= "SELECT * FROM `done`;";
   $result=mysqli_query($bd,$sql);
   if(!$result){
   die("database query failed". mysqli_connect_error()."(".mysqli_connect_errno().")");}?>



   <?php while($row=mysqli_fetch_row($result)){
   var_dump($row); }?>

   <?php mysqli_close($bd);?>
于 2017-03-05T14:06:34.017 回答