PHP函数 - 将HTML表单输入转换为多个输入的可恢复函数

时间:2016-06-24 21:06:34

标签: php html xml function zillow

我正在使用Zillow API来从用户输入中获取一些数据。到目前为止,我已经能够正确地使用API​​从单个输入中获取所需的数据。

我遇到的问题是当我尝试将我的简单代码块转换为可以多次重复使用多次输入的函数时。最后,我希望用户能够上传CSV文件或其他内容并通过多个输入运行此功能。

以下是我的代码正常运行:

HTML

<form action="logic.php" method="post">
    <p>Address: <input type="text" name="address"></p>
    <p>City/State: <input type="text" name="csz"></p>
    <input type="submit">
</form>

PHP

//API Key
$api_key = 'XXXxxXXX';


//User Inputs
$search = $_POST['address'];
$citystate = $_POST['csz'];
//User Inputs Fromatted
$address = urlencode($search);
$citystatezip = urlencode($citystate);



//Get Address ID From API
$url = "http://www.zillow.com/webservice/GetSearchResults.htm?zws-id=".$api_key."&address=".$address."&citystatezip=".$citystatezip;
$result = file_get_contents($url);
$data = simplexml_load_string($result);
$addressID = $data->response->results->result[0]->zpid;



//Get Estimate from API (using adressID)
$estimate_url = "http://www.zillow.com/webservice/GetZestimate.htm?zws-id=".$api_key."&zpid=".$addressID;
$estimate_result = file_get_contents($zurl);
$estimate_data = simplexml_load_string($zresult);
$estimate = $zdata->response->zestimate->amount;

echo $estimate;

现在的问题是当我尝试将这两个函数包装成两个单独的函数时,将它们用于多个输入。

$api_key = 'XXXxxXXX';

//User Inputs
$search = $_POST['address'];
$citystate = $_POST['csz'];
//User Inputs Fromatted
$address = urlencode($search);
$citystatezip = urlencode($citystate);


function getAddressID($ad,$cs){

    //Get Address ID From API
    $url = "http://www.zillow.com/webservice/GetSearchResults.htm?zws-id=".$api_key."&address=".$ad."&citystatezip=".$cs;
    $result = file_get_contents($url);
    $data = simplexml_load_string($result);
    $addressID = $data->response->results->result[0]->zpid;
    return $addressID;

}

$addressID = getAddressID($address, $citystatezip);




function getEstimate($aID){
    //Get Estimate from API (using adressID)
    $estimate_url = "http://www.zillow.com/webservice/GetZestimate.htm?zws-id=".$api_key."&zpid=".$aID;
    $estimate_result = file_get_contents($estimate_url);
    $estimate_data = simplexml_load_string($estimate_result);
    $estimate = $estimate_data->response->zestimate->amount;
    return $estimate;

}

echo getEstimate($addressID); //Calling function doesn't return anything

如果基本上我正在做与第一个PHP示例相同的事情。为什么这不能在函数内运行?我忽视了什么吗?

对此非常感谢Ant的帮助。

1 个答案:

答案 0 :(得分:1)

问题是你在两个函数中使用$api_key变量,那个变量不可用。 PHP的工作方式与其他语言略有不同。您可以在此处阅读:http://php.net/manual/en/language.variables.scope.php

我建议你提取一个调用api的函数。这样你就可以在该函数中声明api键。它还允许您更轻松地维护代码(您可以通过添加一些错误处理或切换到curl或其他东西来改善您的api调用)。程序员的黄金法则,不要重复自己。

代码看起来像这样(未经测试):

//User Inputs
$search    = $_POST['address'];
$citystate = $_POST['csz'];
//User Inputs Fromatted
$address      = urlencode($search);
$citystatezip = urlencode($citystate);

function callZillow($endpoint, array $params)
{
    $params['zws-id'] = 'XXX'; // this would be your api_key

    $url    = 'http://www.zillow.com/webservice/' . $endpoint . '.htm?' . http_build_query($params);
    $result = file_get_contents($url);

    return simplexml_load_string($result);
}


function getAddressID($ad, $cs)
{
    //Get Address ID From API
    $data      = callZillow('GetSearchResults', ['address' => $ad, 'citystatezip' => $cs]);
    $addressID = $data->response->results->result[0]->zpid;

    return $addressID;
}

$addressID = getAddressID($address, $citystatezip);

function getEstimate($aID)
{
    //Get Estimate from API (using adressID)
    $estimate_data = callZillow('GetZestimate', ['zpid' => $aID]);
    $estimate      = $estimate_data->response->zestimate->amount;

    return $estimate;
}

echo getEstimate($addressID);