-d不是數組而是字符串時，將命令行cURL轉換為php cURL

Question

這是我的命令行cURL代碼

curl -v --insecure -XPOST -H 'X-USER: nxxx' -H 'X-SIGNATURE: dxxx' -H "Content-type: application/json" -d '444' 'https://cxxx.com/api/balance'

它運作完美...

我的問題是我試圖將其轉換為php cURL，但我不斷收到{“ error”：“ AUTHENTICATION ERROR”}

這是我對php的最新嘗試...

<?php

$json_url = "https://cxxx.com/api/balance";

// -d variable from cURL
$json_string = "444";

$headers = array();
$headers[] = 'X-USER: nxxx';
$headers[] = 'X-SIGNATURE: dxxx';
$headers[] = 'Content-Type: application/json';



$ch = curl_init($json_url);


// Configuring curl options
$options = array(
CURLOPT_RETURNTRANSFER => true,
CURLOPT_VERBOSE => true,
CURLOPT_SSL_VERIFYPEER => true,
CURLOPT_POST => true,
CURLOPT_HTTPAUTH => true,
CURLOPT_HTTPHEADER => $headers,
CURLOPT_POSTFIELDS => $json_string,//encoding for -d variable cURL
);

// Setting curl options
curl_setopt_array( $ch, $options );

// Getting results
$result = curl_exec($ch); 
echo ($result);
?>

Answer 1

其原因是CURLOPT_SSL_VERIFYPEER設置為true，您可以看到curl命令包含--insecure然后必須將此CURLOPT_SSL_VERIFYPEER設置為false，我還編寫了一個代碼，發送與您發布的curl命令相同的請求，請嘗試

<?php

$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, "http://cxxx.com/api/balance");
curl_setopt($ch, CURLOPT_POST, TRUE);
curl_setopt($ch, CURLOPT_POSTFIELDS, "444");
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, FALSE); //since you put --insecure
curl_setopt($ch, CURLOPT_HTTPHEADER,["X-USER: nxxx","X-SIGNATURE: dxxx","Content-type: application/json"]);
$response = curl_exec($ch);

echo $response;