这是我的php文件代码,我在其中从thecatapi中获取数据 我需要数据表的特定格式
$xml=file_get_contents('https://thecatapi.com/api/images/get?format=xml&results_per_page=3');
$xml = new SimpleXMLElement($xml);
$datas=array();
$datas['rows'] = $xml->data->images;
echo json_encode($datas, true);
输出为:
{"total":50,
"rows":{"image":[
{"url":"https:\/\/thecatapi.com\/api\/images\/get.php?id=e0i",
"id":"e0i",
"source_url":"http:\/\/thecatapi.com\/?id=e0i"
},
{"url":"https:\/\/thecatapi.com\/api\/images\/get.php?id=MTYwNDE0MQ",
"id":"MTYwNDE0MQ",
"source_url":"http:\/\/thecatapi.com\/?id=MTYwNDE0MQ"
},{
"url":"https:\/\/thecatapi.com\/api\/images\/get.php?id=bon",
"id":"bon",
"source_url":"http:\/\/thecatapi.com\/?id=bon"
}
]
}
}
但是我想要这种json形式
{
"total": 800,
"rows": [
{
"url": 0,
"id": "Item 0",
"source_url": "$0"
},
{
"url": 0,
"id": "Item 0",
"source_url": "$0"
},
{
"url": 0,
"id": "Item 0",
"source_url": "$0"
},
答案 0 :(得分:-1)
只是想出了一个替代解决方案。更改此行:
$datas['rows'] = $xml->data->images;
对此:
$datas['rows'] = $xml->data->images->image;
否则您仍然可以这样做
将json字符串解码为一个对象,然后根据需要对其进行修改。
$stdClassObject = json_decode($jsonData);
$stdClassObject->rows = $stdClassObject->rows->image;
$newJsonData = json_encode($stdClassObject, JSON_PRETTY_PRINT);
echo '<pre>';
print_r($newJsonData);
echo '</pre>';
新输出:
{
"total": 50,
"rows": [
{
"url": "https:\/\/thecatapi.com\/api\/images\/get.php?id=e0i",
"id": "e0i",
"source_url": "http:\/\/thecatapi.com\/?id=e0i"
},
{
"url": "https:\/\/thecatapi.com\/api\/images\/get.php?id=MTYwNDE0MQ",
"id": "MTYwNDE0MQ",
"source_url": "http:\/\/thecatapi.com\/?id=MTYwNDE0MQ"
},
{
"url": "https:\/\/thecatapi.com\/api\/images\/get.php?id=bon",
"id": "bon",
"source_url": "http:\/\/thecatapi.com\/?id=bon"
}
]
}