Question

我正在尝试从数据库（产品列表）中检索数据，但在执行代码时我有一个空白页面。

<?php


$DB_HOST = "@@@";
$DB_NAME = "@@@";
$DB_PASS = "@@@";
$DB_USER = "@@@";


$db_obj = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME);
if($db_obj->connect_errno > 0) {
  die('Connection failed [' . $db_obj->connect_error . ']');
}

if($db_obj->connect_errno > 0){
    die('Unable to connect [' . $db_obj->connect_errno . ']');
}


?>



<html>
<head>
<title>List of products</title>
</head>
<body>
<?php


$query="SELECT * FROM product"; 
$result_obj = $db_obj->query($query);
$row=mysql_fetch_row($result);

echo $row;




?>

</body>
</html>

我做得对吗？问题究竟在哪里？是问题可以从服务器端？

新尝试：

<?php
// session_start();
// require("mysqli.php"); 

$DB_HOST = "webdev.cs.kent.edu";
$DB_NAME = "ralsuhai";
$DB_PASS = "810646396";
$DB_USER = "ralsuhai";


$db_obj = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME);
if($db_obj->connect_errno > 0) {
  die('Connection failed [' . $db_obj->connect_error . ']');
}

if($db_obj->connect_errno > 0){
    die('Unable to connect [' . $db_obj->connect_errno . ']');
}


?>



<html>
<head>
<title>List of products</title>
</head>
<body>
<?php


$query="SELECT * FROM product"; 
$result_obj = $db_obj->query($query);
//first try
//while($row=$result_obj->fetch_row() )
//{
  //   var_dump($row);
 //}

 //second try
//$res = array();
//while($row = $result->fetch_array(MYSQLI_ASSOC))
//{
  //$res[] = $row;
//}
//print_r($res);




?>
<table><tr><th>Product Id</th><th>Name</th><th>Price</th></tr>
<?php
$result_obj = $db_obj->query($query);
while($row=$result_obj->fetch_row() )
{ ?>
    <tr>
     <td><?=implode('</td><td>',$row)?></td>
    </tr>
<?php
 }
?>
</table>
<?php










?>

</body>
</html>

Answer 1

你在这里遇到了一些问题。首先，变量$db_obj是一个mysqli对象，并且您尝试在其上使用mysql_fetch行函数。这不会起作用，接下来即使它确实有效，$ row也是一个阵列。这个......

?>
<table><tr><th>Product Id</th><th>Name</th><th>Price</th></tr>
<?php
$result_obj = $db_obj->query($query);
while($row=$result_obj->fetch_row() )
{ ?>
    <tr>
     <td><?php echo implode('</td><td>',$row); ?></td>
    </tr>
<?php
 }
?>
</table>
<?php

这应该转储查询返回的每一行，你可以看到发生了什么。请注意，$ db_obj-＆gt; query（$ query）的结果是一个对象。这应该打印一张包含数据的表格。

Answer 2

<?php
   $DB_HOST = "@@@";
   $DB_NAME = "@@@";
   $DB_PASS = "@@@";
   $DB_USER = "@@@";

   $db_obj = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME);
   if($db_obj->connect_errno) {
   die('Connection failed [' . $db_obj->connect_error . ']');
   }
?>
<html>
  <head>
    <title>List of products</title>
      </head>
      <body>
      <?php
         $query="SELECT * FROM product";
         $result_obj = $db_obj->query($query);
         $row=$result_obj->fetch_assoc();
         print_r($row);
      ?>
  </body>
 </html>

使用PHP代码从数据库中检索数据

2 个答案: