我有这样的表格：
<form method="post" enctype="multipart/form-data">
<input type="text" id="title" placeholder="Project Title"/><br />
<input type="text" id="vurl" placeholder="If You have any video about project write your video url path here" style="width:435px;"/><br />
<textarea id="prjdesc" name="prjdesc" rows="20" cols="80" style="border-style:groove;box-shadow: 10px 10px 10px 10px #888888;"placeholder="Please describe Your Project"></textarea>
<label for="file">Filename:</label>
<input type="file" name="file" id="file" /><br>
<input type="button" name="submit" value="Submit" id="update"/>
</form>
单击提交时，数据存储在数据库中并使用 Ajax 调用显示这是我的 js 代码：
$("#update").click(function(e) {
alert("update");
e.preventDefault();
var ttle = $("#title").val();
alert(ttle);
var text = $("#prjdesc").val();
var vurl = $("#vurl").val();
var img = $("#file").val();
alert(vurl);
var dataString = 'param='+text+'¶m1='+vurl+'¶m2='+ttle+'¶m3='+img;
$.ajax({
type:'POST',
data:dataString,
url:'insert.php',
success:function(id) {
alert(id);
window.location ="another.php?id="+id;;
}
});
});
在这里，我使用 insert.php 存储数据并使用 another.php 显示，但是当进入图像部分时，我不明白如何将图像存储在 db 中的文件夹和路径中，我的意思是我对在 insert.php 中集成代码感到有点困惑
插入.php
$host="localhost";
$username="root";
$password="";
$db_name="geny";
$tbl_name="project_details";
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$name = $_POST['param'];
$video = $_POST['param1'];
$title = $_POST['param2'];
$sql="INSERT INTO $tbl_name (title, content, video_url) VALUES ('$title','$name','$video')";
if(mysql_query($sql)) {
echo mysql_insert_id();
} else {
echo "Cannot Insert";
}
如果我分开然后图像存储在文件夹中..
如果我分开，那么表单代码是：
<form action="upload_file.php" method="post"
enctype="multipart/form-data">
<label for="file">Filename:</label>
<input type="file" name="file" id="file"><br>
<input type="submit" name="submit" value="Submit">
</form>
上传文件.php：
<?php
$allowedExts = array("gif", "jpeg", "jpg", "png");
$extension = end(explode(".", $_FILES["file"]["name"]));
if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/jpg")
|| ($_FILES["file"]["type"] == "image/pjpeg")
|| ($_FILES["file"]["type"] == "image/x-png")
|| ($_FILES["file"]["type"] == "image/png"))
&& ($_FILES["file"]["size"] < 50000)
&& in_array($extension, $allowedExts))
{
if ($_FILES["file"]["error"] > 0)
{
echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
}
else
{
echo "Upload: " . $_FILES["file"]["name"] . "<br>";
echo "Type: " . $_FILES["file"]["type"] . "<br>";
echo "Size: " . ($_FILES["file"]["size"] / 1024) . " kB<br>";
echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br>";
if (file_exists("C:/wamp/www/WebsiteTemplate4/upload/" . $_FILES["file"]["name"]))
{
echo $_FILES["file"]["name"] . " already exists. ";
}
else
{
move_uploaded_file($_FILES["file"]["tmp_name"],
"C:/wamp/www/WebsiteTemplate4/upload/" . $_FILES["file"]["name"]);
// echo "Stored in: " . "upload/" . $_FILES["file"]["name"];
$tmp = "C:/wamp/www/WebsiteTemplate4/upload/" . $_FILES["file"]["name"];
echo $tmp;
}
}
}
else
{
echo "Invalid file";
}
?>
这是完美的工作......
我的问题是如何将此代码集成到 insert.php 中...请帮助我...