做一个SQL查询下拉菜单 [英] Make an SQL query a drop down menu
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问题描述
目前所有的结果显示为单行文本,我想让每个SQL查询显示为一个下拉菜单,所有查询结果作为下拉菜单中的一个选项。
这是我使用的代码:
<?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>;
//显示产品类别
}
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>;
//显示产品品牌
}
?>
解决方案
$ 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>;
对于第二个
打印此视图文件中的$ option1值
I have two SQL queries that display different results from my database. I'm using these results as navigation links in my nav bar. At the moment all the results display as a single line of text, I want to make each SQL query display as a drop down menu, with all the results from the query as an option in the drop down.
Here's the code I'm using:
<?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
}
?>
解决方案
$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>";
same for second print this $option1 value in your view file
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