2

我花了几个小时阅读并试图弄清楚为什么这不起作用但下拉菜单没有填充。我认为这很简单，但我就是看不到它。任何人？

dbconn.php

<?php

define('DB_NAME' , 'artprints');
define('DB_USER' , 'root');
define('DB_PASS' , '');
define('DB_HOST' , 'localhost');

函数.php

<?php

include_once 'dbconn.php';

function connect(){
    $connection = mysqli_connect(DB_HOST, DB_USER, DB_PASS) or die ('Could not connect to the database' . mysl_error());
    mysqli_select_db($connection, DB_NAME);
}

function close(){
    mysql_close();
}

function query(){
    $myData = mysql_query("SELECT * FROM artists");
    while ($record = mysql_fetch_array($myData)){
    echo '<option value="' . $record['artistID'] . '">' . $record['artistID'] . '</option>';
    }
}

测试.php

<?php
 include_once 'func.php';
 connect();
 ?>

 <html>
 <head>
 <title>Drop down testing</title>
 </head>
 <body>
 <select name='artist'>
 <?php query() ?>
 </select>
 <?php close() ?>
 </body>
 </html>
4

2 回答 2

8

你正在混合mysqli_*和mysql_*

mysqli_connect和mysqli_select_db

反对

mysql_query和mysql_fetch_array

于 2013-11-14T13:24:55.813 回答
1

将其复制并粘贴到您的 function.php 中

include_once ('dbconn.php');

function connect()
{
    $connection = mysqli_connect(DB_HOST, DB_USER, DB_PASS) or die ('Could not connect to the database' . mysl_error());
    mysqli_select_db($connection, DB_NAME);
}

function close(){
    mysqli_close();
}

function query(){
    $myData = mysqli_query("SELECT * FROM artists");
    while ($record = mysql_fetch_array($myData)){
    echo '<option value="' . $record['artistID'] . '">' . $record['artistID'] . '</option>';
    }
}
于 2013-11-14T13:34:22.980 回答