下载菜单加载了数据库项 [英] Dropdown menu loaded with database items
本文介绍了下载菜单加载了数据库项的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我正在处理一个页面,它将我的数据库中的项目加载到下拉列表(选项)菜单。
我的代码:
I'm currently working on a page that loads items from my database into a dropdown (option) menu.
My code:
<html>
<?php
$con=mysqli_connect("localhost","root","","mysql");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_select_db($con,"mysql");
$query = "SELECT * FROM artikel";
$rs1 = mysql_fetch_array($con,$query);
?>
<select>
<option>Selecteer een artikel:</option>
<?php
foreach ($rs1 as $row) {
echo '<option value=\"{$row['Omschrijving']}\">{$row['Omschrijving']}</option>';
}
?>
</select>
</html>
错误:
Error:
Parse error: syntax error, unexpected 'Omschrijving' (T_STRING), expecting ',' or ';' in D:\Program Files\xampp\htdocs\Week 5\opgave 1.9.1.php on line 20
任何人都知道我的回声出了什么问题?
Anyone any idea what's wrong with my echo?
推荐答案
con = mysqli_connect( localhost,root,,mysql);
// 检查 connection
if (mysqli_connect_errno()) {
echo 失败 to connect < span class =code-summarycomment> to MySQL: 。 mysqli_connect_error();
}
mysqli_select_db(
con=mysqli_connect("localhost","root","","mysql"); // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } mysqli_select_db(
con,mysql) ;
con,"mysql");
query < /跨度> = SELECT * FROM artikel < span class =code-summarycomment>;
query = "SELECT * FROM artikel";
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