我想使用以下文件夹结构在谷歌云上传和运行我的php应用程序:
yaml文件如下所示:
string strDSN = "Provider=Microsoft.ACE.OLEDB.12.0;Data Source=" + Application.StartupPath + "\\Db1.accdb";
string strSQL = "SELECT * FROM Tbl1" ;
// create Objects of ADOConnection and ADOCommand
OleDbConnection myConn = new OleDbConnection(strDSN);
OleDbDataAdapter myCmd = new OleDbDataAdapter( strSQL, myConn );
myConn.Open();
DataSet dtSet = new DataSet();
myCmd.Fill( dtSet, "Tbl1" );
DataTable dTable = dtSet.Tables[0];
foreach( DataRow dtRow in dTable.Rows )
{
listBox1.Items.Add( dtRow["id"].ToString());
listBox2.Items.Add( dtRow["nm"].ToString());
listBox3.Items.Add(dtRow["ag"].ToString());
}
当我部署它时,运行index.php文件,但不附加任何css或js文件。什么应该是yaml文件的结构,以便它接受来自css和js文件夹的css和js文件?
由于
答案 0 :(得分:0)
试试这个(处理程序的顺序很重要)
application: <your-app-id-goes-here>
runtime: php55
api_version: 1
handlers:
- url: /css
static_dir: css
- url: /js
static_dir: js
- url: /images
static_dir: images
- url: .*
script: index.php