Question

我目前正在处理我的简单信息系统项目。我的代码有错误，我已经在网上搜索了一些答案，但没有任何效果。错误说“注意：未定义的变量：第15行的C：\ wamp \ www \ Members Information System \ dependents_table.php中的mysqli”

这是我的代码：

<body>
    Log In
    <p><%=  messages %></p>
    <form action="/login" method = "post">
       username : <input type="text" name = "username"> <br>
       password : <input type="text" name = "password">
       <input type="submit" value = "Submit">
    </form>
</body>

第15行是：

<?php
    $member_query = $mysqli->query("select *from members order by member_id DESC")or die(mysql_error());
    $member_row = $member_query->fetch_array();
    $member_id = $member_row['member_id'];
    $query = $mysqli->query("select * from dependents where member_id = '$member_id'") or die(mysql_error());
    while ($row = $query->fetch_array()) {
    $id = $row['dependent_id'];
    ?>

我的数据库是：

$member_query = $mysqli->query("select *from members order by member_id DESC")or die(mysql_error());

我希望你能帮助我。谢谢大家。

Answer 1

给出*和from之间的空格。

$ member_query = $ mysqli-＆gt;查询（＆＃34;从成员订单中选择*由member_id DESC＆＃34;）或死（mysql_error（））;

Answer 2

您必须通过wamp manager启用mysqli扩展

https://john-dugan.com/wp-content/uploads/2014/07/apc-for-wamp-enable-apc-extension.png

Answer 3

您正在使用pdo进行数据库连接，但未使用它来查询字符串

这是一个区分pdo和mysqli

的简单示例

// mysqli
$mysqli = new mysqli("example.com", "user", "password", "database");
$result = $mysqli->query("SELECT 'Hello, dear MySQL user!' AS _message FROM DUAL");
$row = $result->fetch_assoc();
echo htmlentities($row['_message']);


// PDO
$pdo = new PDO('mysql:host=example.com;dbname=database', 'user', 'password');
$statement = $pdo->query("SELECT 'Hello, dear MySQL user!' AS _message FROM DUAL");
$row = $statement->fetch(PDO::FETCH_ASSOC);
echo htmlentities($row['_message']);

Answer 4

尝试这个我只是用双引号编辑第15行

 $member_query = $mysqli->query("select *from members order by member_id DESC")or die(mysql_error());
$member_row = $member_query->fetch_array();
$member_id = $member_row['member_id'];
$query = $mysqli->query("select * from dependents where member_id = '".$member_id."'") or die(mysql_error());
while ($row = $query->fetch_array()) {
$id = $row['dependent_id'];

注意：未定义的变量：mysqli

4 个答案: