PHP的显示输出基于从MySQL下拉 [英] php display output based on drop down from mysql

问题描述

我知道这已经存在，但是我们在那篇文章中没有得到答案. 我一直遇到以下问题.我能够从数据库中检索数据并将其显示在下拉菜单中.但是我不知道如何使用与输入相同的下拉列表并将其嵌入到form.My代码中如下所示.这里从数据库中检索项目名称，并以下拉列表的形式显示.现在，我需要使用相同的下拉列表(HTML表单)，该表单将用于将数据输入到另一个表中.下面是我的代码

I know this is already have but we are not getting answer in that post. i have been having the following problem .I am able to retrieve the data from database and display it in a drop down menu.But i dont know how to use the same drop down list as the input and embed it in a form.My code is as follows.Here project names are retrieved from the database and are displayed in the form of drop down list.Now i need to use the same drop down list in the form(HTML form) which will be used to input data into another table. below is my code

 <div id="footer"><?php
    //Include database configuration file
    include('dbConfig.php');

    //Get all state data
    $query = $db->query("SELECT * FROM state WHERE status = 1 ORDER BY state_name ASC");

    //Count total number of rows
    $rowCount = $query->num_rows;
    ?>
    <select name="state" id="state">
        <option value="">Select state</option>
        <?php
        if($rowCount > 0){
            while($row = $query->fetch_assoc()){ 
                echo '<option value="'.$row['state_id'].'">'.$row['state_name'].'</option>';
            }
        }else{
            echo '<option value="">state not available</option>';
        }
        ?>
    </select></div>

if($rowCount > 0){
        echo '<option value="">Select district</option>';
        while($row = $query->fetch_assoc()){ 
            echo '<option value="'.$row['district_id'].'">'.$row['district_name'].'</option>';
        }
    }else{
        echo '<option value="">district not available</option>';
    }

推荐答案

您需要将Select标记带到if语句的外面，在分支结束时，您需要关闭select标记.而if-else取决于您检索的值.

You need to bring your Select tag out side of the if statement like and at the ending of branching you need to close your select tag. And if-else depends on your retrieved value.

echo '<select name="test">';
    if($rowCount > 0){
        echo '<option value="">Select district</option>';
        while($row = $query->fetch_assoc()){ 
            echo '<option value="'.$row['district_id'].'">'.$row['district_name'].'</option>';
        }
    }else{
        echo '<option value="">district not available</option>';
    }
    echo '</select>';

这篇关于PHP的显示输出基于从MySQL下拉的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！