Question

我将参数数组传递给api并尝试以json格式返回相同的数组元素

我创建了非常简单的2页

1。用于发送带有api的数组

<?php
$array['Phone'] =  123456;
$array['name'] = 'name1';
$array['email'] = 'testemail';
$array['password'] = 'password';
$array['api_key'] = $api_key;
$array['referral_code'] = 'kk';

$url = "http://getsjobs.esy.es/registerapi.php?Array[]=$array";

$ch = curl_init();
curl_setopt($ch,CURLOPT_URL,$url);
curl_setopt($ch,CURLOPT_RETURNTRANSFER,1);
curl_setopt($ch,CURLOPT_CONNECTTIMEOUT, 4);
$json = curl_exec($ch);
if(!$json) {
    echo curl_error($ch);
}

curl_close($ch);

$data = json_decode($json,true);


foreach ($data as $item) {

    echo $data;

}
?>php

和

尝试返回与JSON格式相同的数组（用于测试）

<?PHP

if(isset($_GET['Array[]'])) {

$_Phone= $array['Phone'];
$_name = $array['name'];
$_email = $array['email'];
$_password = $array['password'];
$api_key = md5(uniqid(rand(), true));
$_referral_code = $array['referral_code'];


$posts = $_GET['Array[]'];

header('Content-type: application/json');


}


?>

但SomeHow Json数据无法从api获取数组。我试过通过以下示例仔细创建代码，但我觉得这里缺少一些东西。

只需要一个提示可能是阵列没有通过或接收的原因？任何帮助都会被欣赏

Answer 1

传递数据的方式不正确，PHP可能会在此行上抛出一个Array to String转换错误

$url = "http://getsjobs.esy.es/registerapi.php?Array[]=$array";

你需要这样做

<?php
$api_key = '';

$params = [
    'Array[Phone]' => 123456,
    'Array[name]' => 'name1',
    'Array[email]' => 'testemail',
    'Array[password]' => 'password',
    'Array[api_key]' => $api_key,
    'Array[referral_code]' => 'kk',
];
$url = "http://localhost/test.php?".http_build_query($params);

// echo $url;

$ch = curl_init();
curl_setopt($ch,CURLOPT_URL,$url);
curl_setopt($ch,CURLOPT_RETURNTRANSFER,1);
curl_setopt($ch,CURLOPT_CONNECTTIMEOUT, 4);
$json = curl_exec($ch);
if(!$json) {
    echo curl_error($ch);
}

curl_close($ch);

$data = json_decode($json,true);

foreach ($data as $item) {

    echo $item."<br/>\n";

}

要检索数据：

<?php
if(isset($_GET['Array'])) {
    $array = $_GET['Array'];

    $_Phone= $array['Phone'];
    $_name = $array['name'];
    $_email = $array['email'];
    $_password = $array['password'];
    $api_key = md5(uniqid(rand(), true));
    $_referral_code = $array['referral_code'];

    echo json_encode($array);
}

从api检索数据

1 个答案: