求助:dbase_open打开dbf文件,只有2行代码~
注:extension=php_dbase.dll 已经默认开启
代码如下:
PHP code
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<?php $file_dbf = 'test.dbf';
$db = dbase_open($file_dbf,0);
?>
出现错误提示:
Warning: dbase_open() [function.dbase-open]: unable to open database test.dbf in F:\web\AppServ\www\test\dbf.php on line 3
我试了很多路径都出现同样的警告:(路径是正确的)
$file_dbf = 'test.dbf';
$file_dbf = '/test.dbf';
$file_dbf = ./test.dbf';
$file_dbf = 'http://localhost/test/test.dbf';
$file_dbf = 'F:\web\AppServ\www\test\test.dbf';
$file_dbf = 'F:/web/AppServ/www/test/test.dbf';
请问这是什么原因?应该如何解决?谢谢~
------解决方案--------------------
用 ODBC
VFP(.dbc, .dbf)
标准连接:
"Provider=vfpoledb;Data Source=C:\DatabasePath\MyDatabase.dbc; "
或
"Provider=VFPOLEDB.1;Data Source=d:\temp;Password= " ";Collating Sequence=MACHINE "
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