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ทำหน้าโชว์รูปที่เราอัพโหลดไว้ในฐานข้อมูลยังไงคะ ช่วยแนะนำหน่อยค่ะ หรือตัวอย่างให้ดูก็ได้ค่ะ |
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Code (PHP)
<html>
<head>
<title>ThaiCreate.Com Tutorial</title>
</head>
<body>
<?
mysql_connect("localhost","root","root") or die(mysql_error());
mysql_select_db("mydatabase");
$strSQL = "SELECT * FROM gallery";
$objQuery = mysql_query($strSQL);
$Num_Rows = mysql_num_rows($objQuery);
$Per_Page = 4; // Per Page
$Page = $_GET["Page"];
if(!$_GET["Page"])
{
$Page=1;
}
$Prev_Page = $Page-1;
$Next_Page = $Page+1;
$Page_Start = (($Per_Page*$Page)-$Per_Page);
if($Num_Rows<=$Per_Page)
{
$Num_Pages =1;
}
else if(($Num_Rows % $Per_Page)==0)
{
$Num_Pages =($Num_Rows/$Per_Page) ;
}
else
{
$Num_Pages =($Num_Rows/$Per_Page)+1;
$Num_Pages = (int)$Num_Pages;
}
$strSQL .=" order by GalleryID ASC LIMIT $Page_Start , $Per_Page";
$objQuery = mysql_query($strSQL);
echo"<table border=\"0\" cellspacing=\"1\" cellpadding=\"1\"><tr>";
$intRows = 0;
while($objResult = mysql_fetch_array($objQuery))
{
echo "<td>";
$intRows++;
?>
<center>
<img src="thaicreate/<?=$objResult["Picture"];?>"><br>
<?=$objResult["GalleryName"];?>
<br>
</center>
<?
echo"</td>";
if(($intRows)%2==0)
{
echo"</tr>";
}
}
echo"</tr></table>";
?>
<br>
Total <?= $Num_Rows;?> Record : <?=$Num_Pages;?> Page :
<?
if($Prev_Page)
{
echo " <a href='$_SERVER[SCRIPT_NAME]?Page=$Prev_Page'><< Back</a> ";
}
for($i=1; $i<=$Num_Pages; $i++){
if($i != $Page)
{
echo "[ <a href='$_SERVER[SCRIPT_NAME]?Page=$i'>$i</a> ]";
}
else
{
echo "<b> $i </b>";
}
}
if($Page!=$Num_Pages)
{
echo " <a href ='$_SERVER[SCRIPT_NAME]?Page=$Next_Page'>Next>></a> ";
}
?>
</body>
</html>
<?
mysql_close();
?>
พอจะเป็นตัวอย่างได้ครับ
Go to : PHP MySQL Upload File to MySQL - Multiple Dynamic CreateElement Input File Upload
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| Date :
2010-09-05 22:16:36 |
By :
webmaster |
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ทำโชว์รูปแต่ละรูปได้แล้วค่ะ
แต่ว่าจะโชว์อัลบั้มรูปอ่ะค่ะ แบบใน facebook อ่ะค่ะ
ตรงที่โชว์อัลบั้มรูป พอกดเข้าไปก็แสดงแต่รูปในอัลบั้ม
เหมือนในเว็บนี้ www.tapluz.multiply.com
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| Date :
2010-09-05 22:29:35 |
By :
nongjar |
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Code (PHP)
<html>
<head>
<title>ThaiCreate.Com Tutorial</title>
</head>
<body>
<?
$objConnect = mysql_connect("localhost","root","root") or die("Error Connect to Database");
$objDB = mysql_select_db("mydatabase");
$strSQL = "SELECT * FROM album";
$objQuery = mysql_query($strSQL) or die ("Error Query [".$strSQL."]");
echo"<table border=\"1\" cellspacing=\"1\" cellpadding=\"1\"><tr>";
$intRows = 0;
while($objResult = mysql_fetch_array($objQuery))
{
echo "<td>";
$intRows++;
?>
<center>
<a href="show_gallery.php?AlbumID=<?=$objResult["AlbumID"];?>"><img src="myfile/<?=$objResult["AlbumShot"];?>" width="100" height="100"></a><br>
<?=$objResult["AlbumName"];?>
<br>
</center>
<?
echo"</td>";
if(($intRows)%3==0)
{
echo"</tr>";
}
}
echo"</tr></table>";
mysql_close($objConnect);
?>
</body>
</html>
Go to : PHP Create Album Images Gallery / jQuery Lightbox
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| Date :
2011-09-16 23:27:42 |
By :
thaicreate |
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 Load balance : Server 03
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