【发布时间】:2011-04-12 03:21:45
【问题描述】:
所以直到插入语句的所有内容都可以完美运行。我知道数据库正在连接,因为我可以使用前两个语句从数据库中选择信息。我也知道 execute_statment3 可以工作,因为没有打印错误,并且当它被放入 sql 时,语句会以应有的方式插入。因此问题出在脚本和 phpmyadmin 之间的通信上。请帮助我已经盯着这个问题两天了,我快疯了。
<?php
session_start();
$hostname = 'localhost';
$username = '####';
$password = '####';
$connection = mysql_connect($hostname, $username, $password)
or die ('Connection error!!!');
$database = '####';
mysql_select_db($database);
$uid = $_SESSION['ID'];
$album = $_POST['albumname'];
$description = $_POST['description'];
$filename = $_FILES["upload_file"]["name"];
$filetype = $_FILES["upload_file"]["type"];
$filesize = $_FILES["upload_file"]["size"];
$file_on_server = $_FILES["upload_file"]["tmp_name"];
if ($filetype == "image/jpeg") {
$file_copy_name = date(m.d.y_H.i.s) . ".jpg";
copy($file_on_server, "uploads/" . $file_copy_name);
print "<br>";
print "<img src = \"uploads/$file_copy_name\">";
print "<br>";
$ret = system("pwd");
$picture = "uploads/$file_copy_name";
}
$execute_statement = "SELECT * FROM ImageAlbums WHERE Album = '$album'";
$results = mysql_query($execute_statement) or die ('Error executing SQL statement!!!');
while($item = mysql_fetch_array($results))
{
$album2 = $item['Album'];
}
if ($album2 == $album)
{
$execute_statement2 = "SELECT * FROM ImageAlbums WHERE Album = '$album'";
$results2 = mysql_query($execute_statement2) or die ('Error executing SQL statement2!!!');
while ($row2 = mysql_fetch_array($results2)) {
$AID = $row2["AlbumID"];
}
$execute_statement3 = "INSERT INTO Images (`ImageID`, `AlbumID`, `Description`, `Extensions`) VALUES ('NULL', '$AID', '$description', '$file_copy_name')";
($execute_statement3) or die ('Error executing SQL statement3!!!');
}
print "<br>";
print "<br>";
print $execute_statement3;
print "<br>";
print "<br>";
print $AID;
print "<br>";
print "<br>";
print $picture;
?>
我为此脚本使用了两个数据库,其中一个数据库称为 ImageAlbums,并且有两列称为 AlbumID 和 Album(AlbumID 是主键)。第二个表称为 Images,有四列 ImageID(主键)、AlbumID(外键)、Description 和 Extensions。
【问题讨论】: