Question

我知道我在这里遇到语法错误，但我无法解决这个问题。

我正在尝试将数据库中的单个特定记录的结果显示在PHP变量中。看起来很简单，但显然不适合我。这是我正在尝试运行的代码：

    // Create connection
    $con=mysqli_connect($mysqlserver,$mysqluser,$mysqlpass,$mysqldb);

    // Check connection
    if (mysqli_connect_errno()) {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    $SQL = "SELECT * FROM players_brixx_gastonia WHERE email='catheygreaney@example.com'"; 
    $result = mysqli_query($SQL,$con);
    $result1 = mysqli_fetch_assoc($result);
    $prevpoints1 = $result1['points'];
    echo $prevpoints1;

    mysqli_close($con);

我的页面正在输出：“警告：mysql_fetch_assoc（）期望参数1是资源，布尔值在”

中给出

这通常意味着我的查询失败，但是当在phpmyadmin中单独运行查询时，它就完成了显示我需要的行。

这是怎么回事？

更新：

KaNch解决了我的问题。谢谢！

Answer 1

 // Create connection
$con=mysqli_connect($mysqlserver,$mysqluser,$mysqlpass,$mysqldb);

// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$SQL = "SELECT * FROM players_brixx_gastonia WHERE email='catheygreaney@example.com'"; 
$result = mysqli_query($con,$SQL);
$result1 = mysqli_fetch_assoc($result);
$prevpoints1 = $result1['points'];
echo $prevpoints1;

mysqli_close($con);

Answer 2

使用此功能，您尚未创建与数据库的连接。首先按如下方式创建与数据库的连接

$con = mysqli_connect("localhost","usrname","password","db_name");
// Check connection
if (mysqli_connect_errno())
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

执行以下查询

$result = mysqli_query($con, "SELECT * FROM players_brixx_gastonia WHERE email='catheygreaney@example.com'");
while($row = mysqli_fetch_array($result))
{
    echo $row['points'];
}

Answer 3

mysql_query如果查询成功执行则返回资源，如果出现错误则返回FALSE。由于mysql_fetch_assoc抱怨参数是布尔值，因此您的查询可能有错误。

MySQL数据输出在可打印的PHP变量中

3 个答案: