javascript - 微信接入JS时遇到这个问题
Warning: fwrite() expects parameter 1 to be resource, boolean given in jssdk.php on line 80
Warning: fclose() expects parameter 1 to be resource, boolean given in jssdk.php on line 81
Warning: fopen(jsapi_ticket.json) [function.fopen]: failed to open stream: Permission denied in jssdk.php on line 56
Warning: fwrite() expects parameter 1 to be resource, boolean given in jssdk.php on line 57
Warning: fclose() expects parameter 1 to be resource, boolean given in jssdk.php on line 58
回复内容:
Warning: fopen(access_token.json) [function.fopen]: failed to open stream: Permission denied in jssdk.php on line 79
Warning: fwrite() expects parameter 1 to be resource, boolean given in jssdk.php on line 80
Warning: fclose() expects parameter 1 to be resource, boolean given in jssdk.php on line 81
Warning: fopen(jsapi_ticket.json) [function.fopen]: failed to open stream: Permission denied in jssdk.php on line 56
Warning: fwrite() expects parameter 1 to be resource, boolean given in jssdk.php on line 57
Warning: fclose() expects parameter 1 to be resource, boolean given in jssdk.php on line 58
是什么环境下。如果是Linux环境就把,就把access_token.json和jsapi_ticket.json的读写权限打开。
别告诉我你用的是SAE。 SAE不支持本地写权限,得修改代码。
你可以把access token存储到SAE的memcache里面
如果是其他的系统,请把写权限打开。
这个是在SAE作为测试环境的原因, 今天刚好我也遇到了. SAE是没有文件的读写权限的.
节哀
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