Mysqli_query returns false 和 Mysqli_error returns NULL？

Question

我一直在创建预订系统和约会，但我的 SQL 语句不起作用。我一直在努力寻找解决方案，但无济于事。

下面列出的是我的 php 代码。我的第一个 SQL 语句工作得很好并且 returns 正确的 ClientID，但是，第二个 SQL 语句没有将它全部插入到数据库中。我对结果做了 var_dumps，返回 bool(false)，对结果做了 mysqli_error，返回 null。我最后的错误消息只显示回显消息，而不是 mysqli_error 或错误号。

（注：为了保护数据，部分值为changed/removed）

<?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());
    }
?>

Answer 1

$query = "INSERT INTO appointmentinformation (ClientID, Length, Date, Month, Year, Time, Price) VALUES ('$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。这会将您带到代码的 else 分支。然后你编码 mysqli_error($result) 等于 mysqli_error(null).

我读过的文档说要用查询的 "link" 初始化一个变量。你这样做了：

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

您现在要将其编码为 mysqli_error($connection) 和 mysqli_errno($connection)。

补充一个建议。在 mysqli_connect 语句之后添加此代码。

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

Mysqli_query returns false 和 Mysqli_error returns NULL？

Mysqli_query returns false and Mysqli_error returns NULL?

php

mysql

mysqli

insert