Question

我有两个SQL查询显示与我的数据库不同的结果。我在导航栏中将这些结果用作导航链接。目前，所有结果都显示为单行文本，我希望将每个SQL查询显示为下拉菜单，并将查询中的所有结果作为下拉列表中的选项。

这是我正在使用的代码：

<?php

$q = "SELECT cat_id, cat_name FROM Category";
$result = mysqli_query($_SESSION['conn'], $q);
while ($row = mysqli_fetch_row($result)) {
    echo "<a href='category.php?id=$row[0]'>$row[1]</a> ";
    //display product categories
}
mysqli_free_result($result); //free result

$q = "SELECT brand_id, brand_name FROM Brand";
$result = mysqli_query($_SESSION['conn'], $q);
while ($row = mysqli_fetch_row($result)) {
    echo "<a href='brand.php?id=$row[0]'>$row[1]</a> ";
    //display product Brands
}
?>

Answer 1

使用select option

echo "<select name='category'>";
while ($row = mysqli_fetch_row($result)){
   echo "<option value='$row[0]'>$row[1]</option> ";
}
echo "</select>";

Answer 2

$q="SELECT cat_id, cat_name FROM Category";
$result = mysqli_query($_SESSION['conn'],$q);
$option1.="<select name='category'>";
while ($row = mysqli_fetch_row($result)){
   $option1.="<option value='$row[0]'>$row[1]</option> ";
}
$option1.="</select>";

第二个相同在视图文件中打印此$ option1值

Answer 3

<select onchange="document.location.href='category.php?id='+this.value;">  
<?php   
    $result = mysqli_query($_SESSION['conn'],$q);
    while ($row = mysqli_fetch_row($result)):?>
    <option value="<?=$row[0];?>"><?=$row[1];?></option>
    <?php endwhile;?>
</select>

使SQL查询成为下拉菜单

3 个答案: