我在RN中有以下代码:
postToServer(){
const requestBody = 'pin=1&status=false';
return fetch('https://192.168.10.200/writeStatus.php', {
method: 'POST',
headers: {'Content-Type': 'application/x-www-form-urlencoded'},
body: requestBody
}).then((response) => response.json())
.then((responseJson) => {
console.log(responseJson);
});
}
此外,我还有以下php文件:
<?php
header('Access-Control-Allow-Origin: *');
$data = json_decode(file_get_contents('php://input'), true);
print_r($data);
?>
当我运行postToServer时,我会收到以下消息:
可能的未处理的承诺拒绝(id: 0):TypeError:网络请求失败的TypeError:网络请求在XMLHttpRequest.dispatchEvent (modules/expo/AppEntry.bundle?platform=android&dev=true&minify=false&hot=false&assetPlugin=P:\sandbox\MojDom1\node_modules\expo\tools\hashAssetFiles:12942:35)的XMLHttpRequest.xhr.onerror上失败.(这里留下的一些行--我对链接有限制)在MessageQueue.callFunctionReturnFlushedQueue (modules/expo/AppEntry.bundle?platform=android&dev=true&minify=false&hot=false&assetPlugin=P:\sandbox\MojDom1\node )_modules\expo\tools\hashAssetFiles:2122:12)
有什么建议吗?
发布于 2018-01-02 06:05:52
添加错误的catch
方法:
postToServer(){
const requestBody = 'pin=1&status=false';
return fetch('https://192.168.10.200/writeStatus.php', {
method: 'POST',
headers: {'Content-Type': 'application/x-www-form-urlencoded'},
body: requestBody
}).then((response) => console.log(response))
}).catch(e => console.log(e));
https://stackoverflow.com/questions/48055748
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