php從下拉菜單中保存用戶選擇並運行查詢

我有一個連接到我的數據庫的php腳本，並有兩個下拉菜單。現在我想要做的是，根據用戶選擇的內容，對用戶選擇運行查詢。下拉菜單中的選項是不同的國家/地區。php從下拉菜單中保存用戶選擇並運行查詢

我想保存此輸入並使用該值運行查詢。即如果用戶選擇美國和巴西，我會運行一些查詢，如select *從我的數據庫中選擇國家==選擇1和選擇2（巴西和美國）。

我該如何從下拉菜單保存用戶選擇並使用它來運行這樣的查詢？我更關心實際設置查詢而不是編寫查詢。

任何幫助，將不勝感激！

我迄今爲止代碼：

<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>

來源

2011-04-16 Eric

這取決於當你想運行查詢。如果查詢將在正常之後運行，發送剛剛發送的請求。

<?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 
// ... 
?>

它你的應用程序將是較大的那麼這個單頁，你應該閱讀一些有關當今流行MVC或使用一些PHP框架，而不是像這樣:)

混合代碼

來源

2011-04-16 07:48:07 Zap

php從下拉菜單中保存用戶選擇並運行查詢

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