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Mysqli_query返回false,而Mysqli_error返回NULL?

[英]Mysqli_query returns false and Mysqli_error returns NULL?

我一直在创建预订系统并创建约会,但是我的SQL语句不起作用。 我一直在寻找解决方案,但无济于事。

下面列出的是我的PHP代码。 我的第一个SQL语句可以正常工作并返回正确的ClientID,但是,第二个SQL语句并未将其全部插入数据库中。 我对结果做了var_dumps,返回bool(false),对结果做了mysqli_error,返回null。 我的错误消息最后仅显示echo'd消息,也没有mysqli_error或错误号。

(注意:为了保护数据,某些值已更改/删除)

<?php
    session_start();
    if(! $_SESSION['Username']) {
        header("location:Index.php");
    }    
    $servername = "localhost";
    $username = "root";
    $password = "";
    $dbname = "";
    $tablename = "appointmentinformation";
    $tablenamed = "clientinformation";

    $connection = mysqli_connect("$servername", "$username", "$password", "$dbname") or die("Could not connect to the database");

    $clientusername = $_SESSION['Username'];
    $sql = "SELECT ClientID FROM $tablenamed WHERE Username = '$clientusername' LIMIT 1";
    $results = mysqli_query($connection, $sql);
    if (! $results) {
        echo ("Could not select the data : " . mysql_error());
    } else {
        $datarows = mysqli_fetch_row($results);
        $clientid = $datarows[0];
    }

    $date = $_POST["Date"];
    $month = $_POST["Month"];
    $year = $_POST["Year"];
    $time = $_POST["Time"];
    $length = $_POST["Length"];

    $date = stripslashes($date);
    $month = stripslashes($month);
    $year = stripslashes($year);
    $time = stripslashes($time);
    $length = stripslashes($length);

    $date = mysqli_real_escape_string($date);
    $month = mysqli_real_escape_string($month);
    $year = mysqli_real_escape_string($year);
    $time = mysqli_real_escape_string($time);
    $length = mysqli_real_escape_string($length);

    $query = "INSERT INTO appointmentinformation (ClientID, Length, Date, Month, Year, Time, Price) VALUES ('$clientid', '$length', '$date', '$month', '$year', '$time', '$price')";
    $result = mysqli_query($connection, $query);
    if ($result) {
        header("Location:UserCP.php");
    } else {
        echo ("Could not insert data : " . mysqli_error($result) . " " . mysqli_errno());
    }
?>
$query = "INSERT INTO appointmentinformation (ClientID, Length, Date, Month, Year, Time, Price) VALUES ('$clientid', '$length', '$date', '$month', '$year', '$time', '$price')";
    $result = mysqli_query($connection, $query);
    if ($result) {
        header("Location:UserCP.php");
    } else {
        echo ("Could not insert data : " . mysqli_error($result) . " " . mysqli_errno());
    }

mysqli_query返回的结果为空。 这会将您发送到代码的else分支。 然后,您编码了mysqli_error($ result)等于mysqli_error(null)。

我读过的文档说要使用与查询的“链接”初始化变量。 您这样做的原因是:

 $connection = mysqli_connect("$servername", "$username", "$password", "$dbname") or die("Could not connect to the database");

现在,您要将其编码为mysqli_error($ connection)和mysqli_errno($ connection)。

另一个建议。 在mysqli_connect语句之后立即添加此代码。

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

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