php 从下拉菜单中保存用户选择并运行查询答案

【问题标题】：php saving user selection from a dropdown menu and running a queryphp 从下拉菜单中保存用户选择并运行查询
【发布时间】：2011-04-16 07:23:39
【问题描述】：

我有一个连接到我的数据库的 php 脚本，并且有两个下拉菜单。现在我想做的是，根据用户选择，对用户选择运行查询。下拉菜单中的选项是不同的国家/地区。

我想保存此输入并使用该值来运行查询。 IE。如果用户选择 USA 和 Brazil，我将运行一些查询，例如 select * from my database where country == selection 1 and selection 2（巴西和美国）。

我如何从下拉菜单中保存用户选择并使用它来运行这样的查询？我更关心实际设置查询而不是编写查询。

任何帮助将不胜感激！

到目前为止我的代码：

<html>
<head>
<title> Welcome! </title>
<link rel="stylesheet" type="text/css" href="style.css"/>
</head>
<Form Name ="form1" Method ="POST" ACTION = "page1.php">
<?php

$link = mysql_connect('localhost', 'root', '');
if (!$link)
{
 $output = 'Unable to connect to the database server.';
 include 'output.html.php'; 
 exit();
}

mysql_select_db('top recipes');
if (!mysql_select_db('top recipes'))
{
 $output = 'Unable to locate the joke database.';
 include 'output.html.php';
 exit();
}

function dropdown( $name, array $options, $selected=null )
{
    /*** begin the select ***/
    $dropdown = '<select name="'.$name.'" id="'.$name.'">'."\n";

    $selected = $selected;
    /*** loop over the options ***/
    foreach( $options as $key=>$option )
    {
        /*** assign a selected value ***/
        $select = $selected==$key ? ' selected' : null;

        /*** add each option to the dropdown ***/
        $dropdown .= '<option value="'.$key.'"'.$select.'>'.$option.'</option>'."\n";
    }

    /*** close the select ***/
    $dropdown .= '</select>'."\n";

    /*** and return the completed dropdown ***/
    return $dropdown;
}

function dropdowntwo( $nametwo, array $optionstwo, $selectedtwo=null )
{
    /*** begin the select ***/
    $dropdowntwo = '<select name="'.$nametwo.'" id="'.$nametwo.'">'."\n";

    $selectedtwo = $selectedtwo;
    /*** loop over the options ***/
    foreach( $optionstwo as $key=>$option )
    {
        /*** assign a selected value ***/
        $select = $selectedtwo==$key ? ' selectedtwo' : null;

        /*** add each option to the dropdown ***/
        $dropdowntwo .= '<option value="'.$key.'"'.$select.'>'.$option.'</option>'."\n";
    }

    /*** close the select ***/
    $dropdowntwo .= '</select>'."\n";

    /*** and return the completed dropdown ***/
    return $dropdowntwo;
}
?>

<form>

<?php
$name = 'my_dropdown';
$options = array( 'USA', 'Brazil', 'Random' );
$selected = 1;

echo dropdown( $name, $options, $selected );

$nametwo = 'my_dropdowntwo';
$optionstwo = array( 'USA', 'Brazil', 'Random' );
$selectedtwo = 1;

echo dropdowntwo( $nametwo, $optionstwo, $selectedtwo );
?>
<INPUT TYPE = "Submit" Name = "Submit1" VALUE = "Select">
</form>

【问题讨论】：

标签： php database drop-down-menu

【解决方案1】：

这取决于您想要运行查询的时间。如果查询将在 normal 发送刚刚处理发送的请求之后运行。

<?php
if (!empty($_POST['my_dropdown'])) {
  $country1 = $_POST['my_dropdown'];
  // validate if $country1 is in allowed values or use at least
  $country1 = mysql_real_escape_string($country1);
}
// similar for my_dropdowntwo => $country2

// process only with both values?
if (!empty($country1) && !empty($country2)) {
    // you can strore it into $_SESSION if you want - you need to run session_start() before headers!
    $_SESSION['country1'] = $country1;
    // or store only to cookies if it should be perzistent
    setcookie("country1", $country1, time()+3600,, '/');
    // or you can use use variables directly if you want to run query now -> use $country1 or $country2 variabes
}

// if you want to use later stored variables
// first test if you have both variables ...
// then
$query = sprintf("SELECT something FROM someTable where country = '%s' AND country2 = '%s', $_SESSION['country1'], $_SESSION['country2']);
// or use $_COOKIE['country1'] if you use cookies instead
// ...
?>

如果你的应用程序会比这个单页大，你可能应该阅读一些关于当今流行的 MVC 或使用一些 php 框架而不是像这样混合代码:)

【讨论】：