เพิ่มรูปไม่ลงฐานข้อมูลอ่ะ ช่วยดูให้หน่อยคะ รูปไม่ไปฐานข้อมูลอ่ะ
หน้ารับข้อมูลCode (PHP) <?
include("connection.php");
?>
<?
$build_id=$_POST[build_id];
$build_number=$_POST[build_number];
$build_running=$_POST[build_running];
$build_name=$_POST[build_name];
$build_address=$_POST[build_address];
$build_owner=$_POST[build_owner];
$build_engineer=$_POST[build_engineer];
$build_date=$_POST[build_date];
$build_finished=$_POST[build_finished];
//echo $user,$pass,$pass1,$name,$sername,$sex,$posion,$person,$day,$mont,$year,$tims,$tel,$job,$email ;
$sql="Insert into buildings (build_id , build_number , build_running , build_name ,
build_address , build_owner , build_engineer , build_date , build_finished)
values('$build_id' ,'$build_number' ,'$build_running' ,
'$build_name' , '$build_address', '$build_owner', '$build_engineer' ,'$build_date' , '$build_finished' )";
$result = mysql_query($sql) or die(mysql_error());
$sql="SELECT * from buildings order by build_id desc "; // ส่ง id มาแล้ว ทำไมต้อง select อีก
$result = mysql_query($sql);
if($row=mysql_fetch_assoc($result))
{
$build_id=$row["build_id"];
?>
<script language="javascript">
function fncCreateElement(){
var mySpan = document.getElementById('mySpan');
var myElement1 = document.createElement('input');
myElement1.setAttribute('type',"pic_file");
myElement1.setAttribute('name',"filUpload[]");
//myElement1.setAttribute('id',"filUpload[]");
mySpan.appendChild(myElement1);
//*** Remove Element ***//
/*
var deleteEle = document.getElementById('txt1');
mySpan.removeChild(deleteEle);
*/
var myElement2 = document.createElement('<br>');
mySpan.appendChild(myElement2);
}
</script>
<meta http-equiv="Content-Type" content="text/html; charset=windows-874"></head>
<body>
<form name="frmMain" method="post" action="addwork3.php" enctype="multipart/form-data">
<input type="file" name="filUpload[]">
<input name="btnButton" id="btnButton" type="button" value="+" onClick="JavaScript:fncCreateElement();">
<br>
<span id="mySpan"></span>
<input name="build_id" type="hidden" value="<? echo $build_id; ?>">
<input name="build_id" type="hidden" value="<? echo $build_id; ?>">
<input name="pic_file" type="hidden" value="<? echo $pic_file; ?>">
<input name="btnSubmit" type="submit" value="Submit">
</form><? } ?>
</table>
Code (PHP) <html>
<head>
<title>อัพโหลดรูป</title>
<meta http-equiv="Content-Type" content="text/html; charset=windows-874"></head>
<body>
<?
$objConnect = mysql_connect("localhost","root","1234") or die("Error Connect to Database");
$objDB = mysql_select_db("bloo");
$build_id = $_POST[build_id];
/*$pic_id = $_POST[pic_id];
$pic_file= $_POST[pic_file];
*/
echo $build_id;
for($i=0;$i<count($_FILES["filUpload"]["name"]);$i++)
{
if($_FILES["filUpload"]["name"][$i] != "")
{
if(copy($_FILES["filUpload"]["tmp_name"][$i],"myfile/".$_FILES["filUpload"]["name"][$i]))
{
//*** Insert Record ***//
$name = $_FILES["filUpload"]["name"][$i];
echo $name;
$strSQL = "INSERT INTO pictures ";
$strSQL .="(pic_file,build_id,pic_id) VALUES ('$name','$build_id','$pic_id')";
$objQuery = mysql_query($strSQL);
}
}
}
echo "Copy/Upload Complete<br>";
?>
<a href="showdetail.php">View files</a>
</body>
</html>
Tag PHP, MySQL
Date :
2011-04-03 21:12:13By :
blookaView :
687Reply :
2
// ส่ง id มาแล้ว ทำไมต้อง select อีก
Date :
2011-04-03 21:21:54By :
hnudka
อ่าาาาาาาาาาาาาา
Date :
2011-04-03 21:33:35By :
blooka
Server 05
Load balance : Server 05
