我必须在 PHP 中模拟 AJAX 请求,就像在 jQuery 中一样。我当前的代码在这里:
原始 AJAX 调用(不得修改)
$.ajax({
type: "POST",
url: "/someFile.php",
data: data,
success: function(response) {
some_code_here;
},
error: function() {
some_code_here;
}
});
当前 PHP 代码 - 尝试模拟上面的 JS 代码行为
function _misc_test() {
$data = json_decode("xxx"); // The "xxx" is placeholder for the same string, as is in data var in JS above
$ajaxResponse = _make_post_request('/someFile.php', $data);
print_r($ajaxResponse);
}
function _make_post_request($url, $data) {
$ch = curl_init($url);
curl_setopt($ch, CURLOPT_POST, TRUE);
curl_setopt($ch, CURLOPT_POSTFIELDS, $data);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
$response = curl_exec($ch);
curl_close($ch);
return $response;
}
不幸的是,PHP 代码似乎不会生成与 JS 代码完全相同的数据包 - 这正是我所需要的。有人可以帮我吗?
<小时/>编辑:也许很重要,JS 中的 data
变量保存像这样的复杂 JS 对象:
{"options":{"userIP":"89.102.122.16","playerType":"flash","playlistItems":[{"Type":"Archive","Format":"MP4_Web","Identifier":"209 452 80139\/0042","Title":"Nezn\u00e1m\u00ed hrdinov\u00e9","Region":"","SubtitlesUrl":"http:\/\/img2.ceskatelevize.cz\/ivysilani\/subtitles\/209\/209452801390042\/subtitles-1.txt","Indexes":null,"Gemius":{"Param":[{"Name":"materialIdentifier","Value":"209 452 80139\/0042"},{"Name":"testParam","Value":"testValue"}]}}],"previewImageURL":null}}
最佳答案
在js中:data: $('form').serialize();
在 PHP 中: How to post data in PHP using file_get_contents?
$jsonstr = '{"options":{"userIP":"89.102.122.16","playerType":"flash","playlistItems":[{"Type":"Archive","Format":"MP4_Web","Identifier":"209 452 80139\/0042","Title":"Nezn\u00e1m\u00ed hrdinov\u00e9","Region":"","SubtitlesUrl":"http:\/\/img2.ceskatelevize.cz\/ivysilani\/subtitles\/209\/209452801390042\/subtitles-1.txt","Indexes":null,"Gemius":{"Param":[{"Name":"materialIdentifier","Value":"209 452 80139\/0042"},{"Name":"testParam","Value":"testValue"}]}}],"previewImageURL":null}}';
print_r(
$data = json_decode($jsonstr ,true)
);
$data_url = http_build_query ($data);
$data_url = str_replace("amp;","",$data_url); //fix for & to &
$data_len = strlen ($data_url);
$url = 'http://domain.com/returnPost.php';
$result = file_get_contents ($url, false,
stream_context_create (
array ('http'=>
array ('method'=>'POST'
, 'header'=>"Connection: close\r\nContent-Length: $data_len\r\n"
, 'content'=>$data_url
))
)
);
print_r(
$result
);
在 returnPost.php
print_r($_POST);
关于php - 在 PHP 中模拟 jQuery.ajax 请求,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/5918401/