Question

我正在为我的注册码制作收据，并且我一直收到此错误：

mysql_fetch_array（）期望参数1是资源，在
中给出布尔值

<?php
// (2)gather details of CustomerID sent
$customerId = $_GET['CustomerID'] ;
// (3)create query
$query = "SELECT * FROM Customer WHERE CustomerID = $customerId";
// (4) Run the query on the customer table through the connection
$result = mysql_query ($query);
// (5) print message with ID of inserted record
if ($row = mysql_fetch_array($result))
{
print "The following Customer was added";
print "<br>Customer ID: " . $row["CustomerID"];
print "<br>First Name: " . $row["Firstnames"];
print "<br>Surname: " . $row["Surname"];
print "<br>User Name: " . $row["Username"];
print "<br>Email: " . $row["Email"];
print "<br>Password: " . $row["Password"];
}
?>

Answer 1

替换它：

$result = mysql_query ($query);

有了这个，看看发生了什么错误：

$result = mysql_query ($query) or die(mysql_error());

Answer 2

该错误消息表示$result各自mysql_query()未返回有效资源。

请尝试mysql_query($query) or die(mysql_error());。顺便说一句，我看不到mysql_connect()和mysql_select_db()！

警告：

您的代码非常不安全（ - ＆gt; SQL注入）！

请通过$_GET或$_POST（如果是整数！）来逃避来自mysql_real_escape_string()或intval的所有数据。

Answer 3

首先必须连接到数据库服务器并选择要使用的数据库。

<?php
if ( isset($_GET['CustomerID']) )
{
  $customerId = $_GET['CustomerID'] ;

  $con = mysql_connect("your_db_address", "db_username", "db_password");  
  if(!$con)
  {
     echo "Unable to connect to DB: " . mysql_error();
     exit;
  }

  $db  = mysql_select_db("your_db_name");
  if (!$db)
  {
     echo "Unable to select DB: " . mysql_error();
     exit;
  }

  /* Rest of your code starting from $query */
}
?>

用户收据问题php

3 个答案: