小弟初学PHP,请教各位高手一个问题:
如何在网页中显示数据库里类型为Bolb的二进制富文本
小弟使用的一个系统,原来是用delphi做的,字段保存的是一个二进制的richtext
现在想到网页上显示出来,要如何实现,请教了!
如何在网页中显示数据库里类型为Bolb的二进制富文本
小弟使用的一个系统,原来是用delphi做的,字段保存的是一个二进制的richtext
现在想到网页上显示出来,要如何实现,请教了!
所以应输出到div或iframe
$styno = '021659';
$sql = "select a.styleno,b.name,c.name,a.grpno,a.goodsname,d.name,
e.name,f.name,g.name,a.orderdate,a.makemethod from workorder a
left join sampletype b on b.sampletypeid = a.sampletypeid
left join season c on c.seasonid = a.seasonid
left join depart d on d.departid = a.departid
left join customer e on e.custid = a.providerid
left join users f on f.userid = a.userid
left join users g on g.userid = a.helper
where a.styleno like '%".$styno."%' order by a.styleno"; $result = ibase_query($res,$sql) or die(ibase_errmsg());
while($row = ibase_fetch_row($result))
{
$styleno = $row[0];
$sampletype = $row[1];
$season = $row[2];
$contractno = $row[3];
$goodsname = $row[4];
$departname = $row[5];
$providername = $row[6];
$kaidan = $row[7];
$gendan = $row[8];
$orderdate = $row[9];
$makemethod = $row[10];
Header( "Content-type: text/rtf");
echo $makemethod;
......网页上显示的是:0x %I
请教,为什么,应该如何才是正确的做法?
require_once('./data/config.php');
$res = ibase_pconnect($fdb_host.':'.$fdb_dbname,$fdb_user,$fdb_pwd) or die('<br>'.ibase_errmsg()); $sql = "select photo from workorder where styleno='021659'";
$set = ibase_query($sql);
while($row = ibase_fetch_object($set))
{
header("Content-type:image/jpeg");
ibase_blob_echo($row->PHOTO);
}
//free $set
ibase_free_result($set);
?>代码如上,错误如下:
Warning: ibase_blob_echo() [function.ibase-blob-echo]: Unrecognized BLOB ID in D:\AppServ\www\ags\blob.php on line 13
换成php5.2.6搞定了。。