【发布时间】:2012-10-12 09:36:14
【问题描述】:
我有一个简单的表格,有人填写,然后另一个表格运行查询以检索数据。
一直在考虑在表单中包含上传的图像,然后让表格显示包含此图像的表单数据。
这是我的一些代码
if ($_SERVER['REQUEST_METHOD'] == 'POST'){
$uname = $_POST['username'];
$password = $_POST['password'];
$uname = htmlspecialchars($uname);
$password = htmlspecialchars($password);
//==========================================
// CONNECT TO THE LOCAL DATABASE
//==========================================
$user_name = "xxxxxx";
$pass_word = "xxxxxx";
$database = "xxxxx";
$server = "xxxxxx";
$db_handle = mysql_connect($server, $user_name, $pass_word);
$db_found = mysql_select_db($database, $db_handle);
if ($db_found) {
$SQL = "SELECT * FROM students WHERE L1 = '$uname' AND L2 = '" .md5 ($_POST['password'])."'";
$result = mysql_query($SQL);
$num_rows = mysql_num_rows($result);
//====================================================
// CHECK TO SEE IF THE $result VARIABLE IS TRUE
//====================================================
if ($result) {
if ($num_rows > 0) {
$color="1";
$result = mysql_query("SELECT * FROM entry, students WHERE entry.studentName = students.studentName AND students.L1='$uname' ")
or die(mysql_error());
echo "<p>You records as of ";
echo date('l jS \of F Y h:i:s A');
echo "<table border='1' cellpadding='2' cellspacing='0'>";
echo "<tr> <th>Date</th><th>Student Name</th> <th>Tutor name</th> <th>Procedure name</th> <th>Grade</th><th>Student Reflection</th><th>Tutor Comments</th><th>Professionalism</th> <th>Communication</th> <th>Alert</th> <th>Dispute</th><th>Username</th> <th>Image</th></tr>";
// keeps getting the next row until there are no more to get
while($row = mysql_fetch_array( $result )) {
if($color==1){
echo "<tr bgcolor= >
<td>".$row['date']."</td><td>".$row['studentName']."</td><td>".$row['tutorName']."</td><td>".$row['procedureName']."</td><td>".$row['grade']."</td><td>".$row['studentReflection']."</td><td>".$row['tutorComments']."</td><td>".$row['professionalism']."</td><td>".$row['communication']."</td><td>".$row['alert']."</td><td>".$row['dispute']."<td>".$row['L1']."</td></td>
<td><img src='images/".$row['studentImage']."'></td>;
</tr>";
// Set $color==2, for switching to other color
$color="2";
}
// When $color not equal 1, use this table row color
else {
echo "<tr bgcolor='#4eb557'>
<td>".$row['date']."</td><td>".$row['studentName']."</td><td>".$row['tutorName']."</td><td>".$row['procedureName']."</td><td>".$row['grade']."</td><td>".$row['studentReflection']."</td><td>".$row['tutorComments']."</td><td>".$row['professionalism']."</td><td>".$row['communication']."</td><td>".$row['alert']."</td><td>".$row['dispute']."<td>".$row['L1']."</td></td>
<td><img src='images/".$row['studentImage']."'></td>;
</tr>";
// Set $color back to 1
$color="1";
}
}
echo '</table>';
当结果回来时,我已经获取了除了图像之外的所有数据,图像只是显示为大量字符。
我哪里错了?
【问题讨论】:
-
您应该将图像作为文件上传并将其路径存储在数据库中。
-
您是将图像数据存储在数据库中还是实际图像的路径?
-
我已经尝试了这两种方法,但似乎都不起作用。因此,我将尝试通过将图像存储在数据库中来使其工作
-
永远不要将图像直接存储在数据库中。始终将它们存储在一个文件夹中,并将图像的路径存储在数据库中。完成后,检查“studentImage”字段中存储的内容。然后将该链接放入浏览器,看看您是否看到图像。让我们知道会发生什么。