如何修复mysqli_query（）和mysqli_fetch_array（）错误

Question

我正在尝试创建一个PHP脚本，在按钮单击时将文本从文件打印到网站。

这是我的代码：

static System.Byte[] test()
{
    str _filePath = "C:\\Users\\Harry\\Desktop\\ABC.PDF";

    System.Byte[]           pdfBuffer;
    System.IO.FileInfo      fileInfo;
    System.IO.FileStream    fs;
    int                     size;
    Set                     permissionSet = new Set(Types::Class);

    permissionSet.add(new FileIOPermission(_filePath,'r'));
    permissionSet.add(new InteropPermission(InteropKind::ClrInterop));

    CodeAccessPermission::assertMultiple(permissionSet);

    //Load the file
    fileInfo = new System.IO.FileInfo(_filePath);
    //Initiallize the byte array by setting the length of the file
    size = int642int(fileInfo.get_Length());
    pdfBuffer = new System.Byte[size]();
    // Stream the file
    fs = new System.IO.FileStream(fileInfo.get_FullName(), System.IO.FileMode::Open, System.IO.FileAccess::Read);
    fs.Read(pdfBuffer, 0, pdfBuffer.get_Length());
    fs.Close();
    fs.Dispose();

    //Revert the access
    CodeAccessPermission::revertAssert();

    return pdfBuffer;    
}

这是每次单击按钮时出现的错误：

任何人都知道如何解决这个问题？

Answer 1

您需要在修改过的脚本中引入大多数mysqli_ *函数所需的数据库连接变量{1}}：

$con

此外，在这种情况下，您需要传递mysqli_query（），<?php $con = mysqli_connect("host", "username", "password", "dbname") or die(mysqli_error($con)); if (isset($_POST['button'])) { $result = mysqli_query($con, "SELECT name,lastname FROM users ORDER BY RAND()"); $name = null; $lastname = null; while ($row = mysqli_fetch_array($result)) { $name = $row['name']; $lastname = $row['lastname']; } echo '<p><small>Output:</small></p><pre style="min-width:auto;display:table;">' . $name . '_' . $lastname . '</pre>Format: <b>name_lastname</b> <span style="width:310px;">'; } ?> <form method="POST"> <button type="submit" class="btn btn-default" name="button">button</button> <br> </form> 返回的结果集标识符。一切都可以在W3Schools找到。

Answer 2

第一次错误

1. 您需要确保自己拥有MySql
的访问权限

2. 确保您已在

   中提供了所有参数（主机，用户名，密码，dbname）的所有正确值

   mysqli_connect("host", "username", "password", "dbname");

第二次错误 在

mysqli_query("SELECT name,lastname FROM users ORDER BY RAND()");
mysqli_connect有两个参数dbconnection＆amp; query。请参阅mysqli_query

while ($row = mysqli_fetch_array()) {
中的

第三次错误

mysqli_fetch_array()需要至少1个参数（例如，mysqli_query()返回结果。请检查此mysqli_fetch_array

修改后的代码

<?php
    $conn = mysqli_connect("host", "username", "password", "dbname") or die(mysqli_error($conn));
    if (isset($_POST['button'])) {
        $result =  mysqli_query($conn, "SELECT name,lastname FROM users ORDER BY RAND()");
        $name = null;
        $lastname = null;
        while ($row = mysqli_fetch_array($result)) {
            $name = $row['name'];
            $lastname = $row['lastname'];
        }
        echo '<p><small>Output:</small></p><pre style="min-width:auto;display:table;">' . $name . '_' . $lastname . '</pre>Format: <b>name_lastname</b> <span style="width:310px;">';
    }
?>
<form method="POST">
    <button type="submit" class="btn btn-default" name="button">button</button>
    <br>
</form>