【发布时间】：2015-11-04 18:20:16
【问题描述】：
我仍在学习，如果有任何帮助，我们将不胜感激。 HTML 非常简单。还没有样式或类似的东西，都在 localhost 上进行测试。所以我没有现场设置示例。
我只使用 PHP 和 phpmyadmin 中的一个表。我拥有的文件是index.php（表格在这里），search.php（搜索代码存储在这里），provider1.php，provider2.php，provider3.php，provider4.php。
到目前为止，我有一个存储有以下列的表：提供者、州全名、城市、州、邮政编码。我有一个搜索表单，用户输入邮政编码，如果邮政编码与表中的一行匹配，它会查看提供者，检查它是哪一个（提供者1，提供者2，提供者3，提供者4）。在search.php上回显他们有多少选择，城市所在的邮政编码，然后还回显他们输入的邮政编码。
我的问题是我如何告诉它，如果该邮政编码有提供程序，则使用该 provider.php。示例：
92804 = provider1，然后使用provider1.php，并且仍然回显出与该邮政编码对应的城市和州。
index.php:
<!doctype html>
<html>
<head>
<meta charset="UTF-8">
<title>search</title>
</head>
<body>
<form action="search.php" method="get">
<input type="text" name="search" placeholder="zipcode..."/>
<input type="submit" value=">>"/>
</form>
</body>
</html>
搜索.php：
<?php
mysql_connect("localhost","root","root") or die("could not connect");
mysql_select_db("zipcodes") or die("could not find db");
//collect
if (isset($_GET['search'])) {
$searchq = $_GET['search'];
$searchq = preg_replace("#[^0-9a-z]#i","",$searchq);
$query = mysql_query("SELECT * FROM `TABLE 1` WHERE `Zip Code` = '$searchq'") or die("could not search");
$count = mysql_num_rows($query);
if($count == 0) {
$output = 'There was no search results!!';
}else {
while($row = mysql_fetch_array($query)) {
$city = $row['City'];
$state = $row['State'];
$provider = $row['Provider Name'];
$output = '<div> '.$provider.' '.$city.' '.$state.' </div>';
}
}
}
switch ($provider) {
case "provider1":
//put code here that tells it to use provider1.php
break;
case "provider2":
//put code here that tells it to use provider2.php
break;
case "provider3":
//put code here that tells it to use provider3.php
break;
case "provider4":
//put code here that tells it to use provider4.php
break;
default:
echo "No provider, try again";
}
?>
<!doctype html>
<html>
<head>
<meta charset="UTF-8">
<title><?php echo $provider ?> in <?php echo $city ?>, <?php echo $state ?></title>
</head>
<body>
This is just a results page for testing. This page needs to know what zip to match to a provider, then go to that page.
<p>You have <?php echo $count ?> choices</p>
<p>Your provider is <?php echo $provider ?></p>
<p>You are located in <?php echo $city ?>, <?php echo $state ?> </p>
<p>The zip code you searched was <?php echo $searchq ?></p>
</body>
</html>
provider1.php：
<!doctype html>
<html>
<head>
<meta charset="UTF-8">
<title><?php echo $provider ?> in <?php echo $city ?>, <?php echo $state ?></title>
</head>
<body>
This would be a template for provider1
<p>You have <?php echo $count ?> choices</p>
<p>You are located in <?php echo $city ?>, <?php echo $state ?> </p>
<p>The zip code you searched was <?php echo $searchq ?></p>
</body>
</html>
这可能吗？还是我需要使用 laravel 之类的东西来启动它或某种类型的框架。
【问题讨论】：