php 如何使用 curl GET 而不是 POST
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How to use curl GET instead of POST
提问by Plummer
I'm tyring to use curl to print a return from a url. The code I have so far looks like this:
我很想使用 curl 来打印来自 url 的返回值。到目前为止，我的代码如下所示：
<?php
$street = $_GET['street'];
$city = $_GET['city'];
$state = $_GET['state'];
$zip = $_GET['zip'];
$url = 'http://eligibility.cert.sc.egov.usda.gov/eligibility/eligibilityservice';
$query = 'eligibilityType=Property&requestString=<?xml version="1.0"?><Eligibility xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:noNamespaceSchemaLocation="/var/lib/tomcat5/webapps/eligibility/Eligibilitywsdl.xsd"><PropertyRequest StreetAddress1="'.$street.'" StreetAddress2="" StreetAddress3="" City="'.$city.'" State="'.$state.'" County="" Zip="'.$zip.'" Program="RBS"></PropertyRequest></Eligibility>';
$url_final = $url.''.$url_query;
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_POST, 1);
curl_setopt($ch, CURLOPT_POSTFIELDS,$query);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
$return = curl_exec ($ch);
curl_close ($ch);
echo $return;
?>
the only obvious problem I know of it that the server being queried uses GETinstead of POST. Are there GETalternatives to this method?
唯一明显的问题，我知道它的服务器所查询的用途GET，而不是POST。GET这种方法有替代方法吗？
回答by AlienWebguy
curl_setopt($ch, CURLOPT_POST, 0);
Curl uses GET by default. You were setting it to POST. You can override it if you ever need to with curl_setopt($ch, CURLOPT_HTTPGET, 1);
Curl 默认使用 GET。您将其设置为 POST。如果您需要，您可以覆盖它curl_setopt($ch, CURLOPT_HTTPGET, 1);
回答by Oyeme
Use file_get_contents() function
file_get_contents
使用 file_get_contents() 函数
file_get_contents
Or curl_setopt($ch, CURLOPT_HTTPGET, 1);
或者 curl_setopt($ch, CURLOPT_HTTPGET, 1);
回答by Antara Das
use
用
curl_setopt_array($ch, array(
CURLOPT_RETURNTRANSFER => 1,
CURLOPT_URL => "http://yourlink.com",
CURLOPT_USERAGENT => 'Codular Sample cURL Request'));
回答by miken32
All these years and nobody's given the right answer; the way to build a query string is to use http_build_query()with an array. This automatically escapes everything and returns a simple string.
这些年来，没有人给出正确的答案；构建查询字符串的方法是使用http_build_query()数组。这会自动转义所有内容并返回一个简单的字符串。
$xml = '<?xml version="1.0"?><Eligibility xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:noNamespaceSchemaLocation="/var/lib/tomcat5/webapps/eligibility/Eligibilitywsdl.xsd"><PropertyRequest StreetAddress1="'.$street.'" StreetAddress2="" StreetAddress3="" City="'.$city.'" State="'.$state.'" County="" Zip="'.$zip.'" Program="RBS"></PropertyRequest></Eligibility>';
$data = [
"eligibilityType" => "Property",
"requestString" => $xml
];
$query = http_build_query($data);
$url .= "?$query";
回答by edwardmp
You are missing a question mark in the URL. Should be like:
您在 URL 中缺少一个问号。应该是这样的：
$query = '?eligibilityType=Property&...';
Also, that XML in your URL needs encoding, e.g. use the urlencode()function in PHP.
此外，您的 URL 中的 XML 需要编码，例如在 PHP 中使用urlencode()函数。
