0

PHP

$con=mysqli_connect("localhost","root","","mmogezgini");
$menuler = mysqli_query($con,"SELECT * FROM menuler");
while($menu = mysqli_fetch_array($menuler))
{
    echo "<button class=\"ust_link\"
        onClick=\"window.location.href='Menu_Url.php'\">Menu_Name</button>";
}
mysqli_close($con);

MySQL代码已经准备好并且MySQL连接成功,我只是不知道如何放置menu['name']andmenu['url']而不是Menu_NameandMenu_Url

4

4 回答 4

0

你有两种可能性:

echo "<button class=\"ust_link\" onClick=\"window.location.href='". $menu['url'].".php'\">". $menu['name'] ."</button>";

或者 :

echo "<button class=\"ust_link\" onClick=\"window.location.href='{$menu['url']}'\">{$menu['name']}</button>";

PHP:回声 - 手册

于 2013-09-30T08:58:53.730 回答
0

像这样:

$con=mysqli_connect("localhost","root","","mmogezgini");
    $menuler = mysqli_query($con,"SELECT * FROM menuler");
    while($menu = mysqli_fetch_array($menuler))
      {
      echo "<button class=\"ust_link\" onClick=\"window.location.href='".$menu['url'].".php'\">".menu['php']."</button>";
      }
    mysqli_close($con);
于 2013-09-30T08:59:00.023 回答
0

试试这个代码:

<?php
while($menu = mysqli_fetch_array($menuler)) {
   $name = $menu["Name'"];
   $url = $menu["url"];
?>
   <button class="ust_link" onclick="window.location.href='<?php echo $url; ?>'"><?php echo $name; ?></button>
<?php
 }
?>

由于您正在使用,因此很难一次回显所有 html

onclick="location.href=''"

它结合了单引号和双引号

于 2013-09-30T09:01:16.347 回答
0

试试这个:

    $con=mysqli_connect("localhost","root","","mmogezgini");
    $menuler = mysqli_query($con,"SELECT * FROM menuler");
    while($menu = mysqli_fetch_array($menuler))
    {
        echo "<button class=\"ust_link\"onClick=\"window.location.href='".$menu['url'].".php'\">". $menu['name'] ."</button>";
    }
    mysqli_close($con);
于 2013-09-30T09:01:51.227 回答