我对php还不熟悉,所以我一直在尝试制作一个php页面,将信息添加到Mysql数据库中。但是当我在表单上按下submit时,它会指向php页面,而什么也不会发生。空白页。
这是我目前所拥有的。
形式
<table class="table" >
<h2>Resturants</h2>
<tbody>
<form data-toggle="validator" enctype="multipart/form-data" action="editRestaurants.php" method="post">
<tr>
<td><label for="inputName" class="control-label" style="font-size:17px;">Title :</label></td>
<td><input type="text" class="form-control" id="title" name="title" required></td>
</tr>
<tr>
<td><label for="inputName" class="control-label" style="font-size:17px;">Text :</label></td>
<td><textarea style="width:300px;height:100px" class="form-control" id="text" name="text" required></textarea><br/></td>
</tr>
<tr>
<td><label for="inputName" class="control-label" style="font-size:17px;">Link :</label></td>
<td><input type="text" class="form-control" id="link" name="link" required></td>
</tr>
<tr>
<td><label for="inputName" class="control-label" style="font-size:17px;">Photo:</label></td>
<td><input type="file" id="image" class="form-control" name="image" accept="image/jpeg" required></td>
</tr>
<tr>
<td><input type="submit" value="Submit" style="font-size:17px;" id="submit"></td>
</tr>
</form>
</tbody>
</table>
这是php文件
include('connection.php');
session_start();
$title=$_POST['title'];
$text=$_POST['text'];
$link=$_POST['link'];
$image=$_FILES['image'];
$imagename=$image['name'];
if(empty($imagename))
{
$imagename="defualt.jpg";
}
if(!empty($title) && !empty($text))
{
$query=mysqli_query($GLOBALS["___mysqli_ston"], 'INSERT INTO `resturant`(`ID`, `Title`, `ImagePath`, `Text`, `Link`, `Date`) VALUES (NULL,'$title','$imagename','$text','$link',CURRENT_DATE())');
if(!empty($image))
{
$target_dir= "NewsImages/";
$target_file = $target_dir . basename($imagename);
move_uploaded_file($image["tmp_name"], $target_file);
}
$_SESSION['status']="Successful";
}
else
{
$_SESSION['status']="Please check all fields";
}
header('Location:cms.php');
最佳答案
您的$query
有语法错误。以下代码将起作用:
$query=mysqli_query($GLOBALS["___mysqli_ston"], "INSERT INTO `resturant`(`ID`, `Title`, `ImagePath`, `Text`, `Link`, `Date`) VALUES (NULL,'$title','$imagename','$text','$link',CURRENT_DATE())");
我简化了你的代码。现在,它将在重定向到
cms.php
之前检查文件是否已成功上载。if(empty($imagename)) {
$imagename="defualt.jpg";
}
if(!empty($title) && !empty($text)) {
$query=mysqli_query($GLOBALS["___mysqli_ston"], "INSERT INTO `resturant`(`ID`, `Title`, `ImagePath`, `Text`, `Link`, `Date`) VALUES (NULL,'$title','$imagename','$text','$link',CURRENT_DATE())");
$target_dir= "NewsImages/";
$target_file = $target_dir . basename($imagename);
if(move_uploaded_file($image["tmp_name"], $target_file)) {
$_SESSION['status']="Successful";
header('Location:cms.php');
exit;
} else {
$_SESSION['status']="Something went wrong while uploading the image.";
}
} else {
$_SESSION['status']="Please check all fields";
}