我有一个这样的表格：
<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>
点击提交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;;
}
});
});
这里我使用insert.php存储数据并使用another.php显示 但是当我来到图像部分我不明白如何将文件存储在db中的文件夹和路径中，我的意思是我有点困惑，在insert.php中集成代码
insert.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，那么图像存储在文件夹中..
如果我分开，那么表格代码是：
<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>
upload_file.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 $tmp;
}
}
}
else
{
echo "Invalid file";
}
?>
这完美地运作......
我的问题是如何在insert.php中集成此代码... 请帮帮我......
答案 0 :(得分：1)
此代码可以在不使用javascript的情况下正常工作。但请确保更改“2”和“66”分别的目录和表名称和字段。
我们将创建一个隐藏的textarea，它将创建数据字符串，我们将使用$ _GET获取所有参数
<script>
$("#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 textareastring = $('#string').val();
var dataString = 'textareastring' = textareastring;
$.ajax({
type:'POST',
data:dataString,
url:'insert.php?param='+text+'¶m1='+vurl+'¶m2='+ttle+'¶m3='+img',
success:function(id) {
alert(id);
window.location ="another.php?id="+id;;
}
});
});
</script>
<textarea id="string" style="display:none;">aa</textarea>
<?php
$name = $_GET['param3'];
// The name.n ow replacing all the $file_name with $name
$url = $_GET['param1'];
$text = $_GET['param'];
$title = $_GET['param2'];
$upload_dir = $url;
$num_files = 1;
//the file size in bytes.
$size_bytes =104857600; //51200 bytes = 50KB.
//Extensions you want files uploaded limited to.
$limitedext = array(".tif",".gif",".png",".jpeg",".jpg");
//check if the directory exists or not.
if (!is_dir("$upload_dir")) {
die ("Error: The directory <b>($upload_dir)</b> doesn't exist. ");
}
//check if the directory is writable.
if (!is_writeable("$upload_dir")){
die ("Error: The directory <b>($upload_dir)</b> . ");
}
if (isset($_POST['upload_form'])){
echo "<h3>Upload results:</h3><br>";
//do a loop for uploading files based on ($num_files) number of files.
for ($i = 1; $i <= $num_files; $i++) {
//define variables to hold the values.
$new_file = $_FILES['file'.$i];
$name = $new_file['name'];
//to remove spaces from file name we have to replace it with "_".
$name = str_replace(' ', '_', $name);
$file_tmp = $new_file['tmp_name'];
$file_size = $new_file['size'];
#-----------------------------------------------------------#
# this code will check if the files was selected or not. #
#-----------------------------------------------------------#
if (!is_uploaded_file($file_tmp)) {
//print error message and file number.
echo "File: Not selected.<br><br>";
}else{
#-----------------------------------------------------------#
# this code will check file extension #
#-----------------------------------------------------------#
$ext = strrchr($name,'.');
if (!in_array(strtolower($ext),$limitedext)) {
echo "File $i: ($name) Wrong file extension. <br><br>";
}else{
#-----------------------------------------------------------#
# this code will check file size is correct #
#-----------------------------------------------------------#
if ($file_size > $size_bytes){
echo "File : ($name) Faild to upload. File must be no larger than <b>100 MB</b> in size.";
}else{
#-----------------------------------------------------------#
# this code check if file is Already EXISTS. #
#-----------------------------------------------------------#
if(file_exists($upload_dir.$name)){
echo "File: ($name) already exists. <br><br>";
}else{
#-------------------------------#
# this function will upload the files. #
#-------------------------------#
if (move_uploaded_file($file_tmp,$upload_dir.$name)) {
$sql = "INSERT INTO table_name(field1, field2) VALUES('$field1', '$field2');";
echo "File: ($name) has been uploaded successfully." . "<img src='uploads/$name'/>";
}else{
echo "File: Faild to upload. <br><br>";
}#end of (move_uploaded_file).
}#end of (file_exists).
}#end of (file_size).
}#end of (limitedext).
}#end of (!is_uploaded_file).
}#end of (for loop).
# print back button.
////////////////////////////////////////////////////////////////////////////////
//else if the form didn't submitted then show it.
}else{
echo "<form method=\"post\" action=\"$_SERVER[PHP_SELF]\" enctype=\"multipart/form- data\">";
// show the file input field based on($num_files).
for ($i = 1; $i <= $num_files; $i++) {
echo "<b>Image: </b><input type=\"file\" size=\"70\" name=\"file". $i ."\" style=\"width:45%\">";
}
echo " <input type=\"hidden\" name=\"MAX_FILE_SIZE\" value=\"$size_bytes\">
<input type=\"submit\" name=\"upload_form\" value=\"Upload\">
</form>";
}
?>
答案 1 :(得分：1)
从此链接中参考如何使用ajax上传图像。它会对你有所帮助。
http://www.9lessons.info/2011/08/ajax-image-upload-without-refreshing.html
答案 2 :(得分：0)
...您可以查看本文中提到的此代码，（http://jagdeepmalhi.blogspot.ru/2011/02/phpmysql-store-file-path-in-database.html） 它有示例代码，如果出现问题，请不要忘记阅读有关它的注释！
答案 3 :(得分：0)
如果您的意思是在单击提交按钮后如何调用insert.php，则在此行中
<form method="post" enctype="multipart/form-data">
你必须添加这个
<form method="post" enctype="multipart/form-data" action="insert.php">
答案 4 :(得分：-1)
<?php
/*dont use path like this C:/wamp/www/WebsiteTemplate4/upload/ becuase you are working at localhost server*/
if (file_exists("upload/" . $_FILES["file"]["name"])){
echo $_FILES["file"]["name"] . " already exists. ";
}else{
$file = $_FILES["file"]["name"]
$filePath = "upload/" . $file;
if(move_uploaded_file($_FILES["file"]["tmp_name"], $filePath)){
/*prepare sql query here and insert*/
$sql = "INSERT INTO table_name(field1, field2) VALUES('$field1', '$field2');";
if(mysql_query($sql)){
echo "File saved in database successfully <strong>{$filePath}</strong>";
}else{
echo "File not uploaded there are an error <strong>{$filePath}</strong>";
}
}else{
echo "File not uploaded there are an error <strong>{$file}</strong>";
}
} ?>
如果您有任何疑问或代码无法正常工作，请尝试此代码，然后再次询问我。 感谢