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ถ้าเราจะเรียก รูปจาก MSSQL <Binary data> โดยไม่ต้องมีหน้ารับจะทำได้ไหมครับ |
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น่าจะต้องน่ะครับ
Code (PHP)
<?
$Connect = mssql_connect("yourhost","user","pass") or die("Error Connect to DB");
$DB = mssql_select_db("Database");
$SQL = "SELECT * FROM Images WHERE Img_Id = '$Img_Id' ";
$Query = mssql_query($SQL) or die ("Error Query [".$SQL."]");
$Result = mssql_fetch_array($Query);
$Type=$Result["Img_Type"];
header("Content-type: $type ");
echo $Result["Img_Data"];
?>
Go to : PHP SQL Server BLOB การ Upload และ Insert รูปภาพลงใน SQL Server mssql (BLOB,Image DataType)
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Date :
2011-12-26 08:35:08 |
By :
webmaster |
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แต่ทำไม่มันถึง ใช้คำสั่ง SQL ไม่ได้ละครับ พี่ วินCode (PHP)
<?
$Connect = mssql_connect("TOON","sa","0000") or die("Error Connect to DB");
$DB = mssql_select_db("inv");
$SQL = "SELECT * FROM pd_pic a WHERE a.pd_cod = '%PM%' ";
$Query = mssql_query($SQL) or die ("Error Query [".$SQL."]");
$Result = mssql_fetch_array($Query);
header("Content-type: $type ");
echo $Result["pic_front"];
?>
มันไม่ออก
แต่ถ้า......................
Code (PHP)
<?
$Connect = mssql_connect("TOON","sa","0000") or die("Error Connect to DB");
$DB = mssql_select_db("inv");
$SQL = "SELECT * FROM pd_pic a WHERE a.pd_cod = 'PMHPCNQL3100C' ";
$Query = mssql_query($SQL) or die ("Error Query [".$SQL."]");
$Result = mssql_fetch_array($Query);
header("Content-type: $type ");
echo $Result["pic_front"];
?>
มันออก ครับตัวที่เรียกตัวเดียว นะครับ
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Date :
2011-12-26 09:58:58 |
By :
time.toon |
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จะต้องเรียกทีล่ะตัวครับ 
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Date :
2011-12-26 13:49:20 |
By :
webmaster |
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Load balance : Server 00
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