使用PHP动态创建的表单将不同的值存储在数据库中吗? [英] Store different value in the database from the forms created dynamically using PHP?
问题描述
请看下面的屏幕截图.
Have a look at the screenshot below.
根据屏幕截图，我将数据提取到基于数据库的不同块中.例如，基于用户名和密码，这些值将从数据库中获取并显示在不同的块中.我有PHP可以将值存储到数据库中，但是我面临的问题是，当我尝试从其他块上载它时，它仍然保存了第一个块中的值. 代码如下:
As per the screenshot, I have the data fetched into the different blocks based on the database. For example, based on the username and password, these values are fetched from the database and displayed in different blocks. I have the PHP to store the value into the database, but the problem that I am facing is that when i try to upload it from other block, it still saves the value from the first block. Codes as as below:
<?php
include('includes/config.php');
$upload = 'uploads/';
session_start();
$_SESSION['$userid'];
$sql = "SELECT * FROM tbl_store INNER JOIN tbl_job ON tbl_store.store_code = tbl_job.store_code WHERE username = '$userid'";
$result = mysqli_query($conn,$sql);
$rowcount=mysqli_num_rows($result);
// echo "$rowcount";
$stores = array();
$stores_add = array();
$stores_chain = array();
$job = array();
$client = array();
$brand = array();
$week= array();
$x = 1;
$imgCnt =1;
while($row = mysqli_fetch_array($result)){
echo "工作".'<br/>';
echo $row['jobs'].'<br/>'.'<br/>';
$job[] = $row['jobs'];
echo "客戶".'<br/>';
echo $row['Client'].'<br/>'.'<br/>';
$client[] = $row['Client'];
echo "牌子".'<br/>';
echo $row['Brand'].'<br/>'.'<br/>';
$brand[] = $row['jobs'];
echo "週數".'<br/>';
echo $row['week'].'<br/>'.'<br/>';
$week[] = $row['week'];
$target = $upload.'/'.$row['Client'].'/'.$row['Brand'].'/'.$row['jobs'].'/'.$row['store_code'].'/';
$testpath = $row['Client'].'/'.$row['Brand'].'/'.$row['jobs'].'/'.$row['store_code'].'/';
$_SESSION['target1'.$x] = $target;
if(!file_exists($target))
{
mkdir($target,0777,true);
}
?>
<form id='uploadForm-<?php echo $imgCnt; ?>' action = '' enctype='multipart/form-data' method = 'POST' class="form<?php echo $imgCnt; ?>">
<input type="file" class="image<?php echo $imgCnt; ?>" name="img" onChange="readURL(this);" />
<img id="blah" src="#" alt="your image" /><br/><br/>
<input type='button' id = '<?php echo $imgCnt; ?>' class='uploadPicture<?php echo $imgCnt; ?> btn btn-primary' value = '上載'>
<!-- <input type="button" value="上載" class="uploadPicture" id="upload_btn<?php echo $imgCnt; ?>"/> -->
</form>
<form enctype="application/x-www-form-urlencoded">
<table width="200" border="1">
<tr>
<td>Product</td>
<td>Promotional Price</td>
<td>Regular Price</td>
<td>Stacking</td>
</tr>
<tr>
<td><input type="text" id="product"></td>
<td><input type="text" id="pp1"></td>
<td><input type="text" id="rp1"></td>
<td><input type="text" id="stacking"></td>
</tr>
</table>
<div id ="div1">
<input type="button" value="Submit" onClick="PostData();"/><br/>
</div>
</form>
<script> src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<script type="text/javascript">
function PostData() {
// 1. Create XHR instance - Start
var xhr;
if (window.XMLHttpRequest) {
xhr = new XMLHttpRequest();
}
else if (window.ActiveXObject) {
xhr = new ActiveXObject("Msxml2.XMLHTTP");
}
else {
throw new Error("Ajax is not supported by this browser");
}
// 1. Create XHR instance - End
// 2. Define what to do when XHR feed you the response from the server - Start
xhr.onreadystatechange = function () {
if (xhr.readyState === 4) {
if (xhr.status == 200 && xhr.status < 300) {
document.getElementById('div1').innerHTML = xhr.responseText;
}
}
}
// 2. Define what to do when XHR feed you the response from the server - Start
var product = document.getElementById("product").value;
var pp1 = document.getElementById("pp1").value;
var rp1 = document.getElementById("rp1").value;
var stacking = document.getElementById("stacking").value;
// var image = document.getElementById("image").value;
// 3. Specify your action, location and Send to the server - Start
xhr.open('POST', 'report.php');
//xhr.open('POST', 'config.php');
xhr.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
xhr.send("product=" + product + "&pp1=" + pp1 + "&rp1=" + rp1 + "&stacking=" + stacking);
//xhr.send("&pid=" + pid);
// 3. Specify your action, location and Send to the server - End
}
</script>
<?php
echo "-------------------------------------------------------".'<br/>';
$x = $x+1;
$imgCnt++;
}
?>
我已经从其中删除了用于图像上传的代码，因为它可以正常工作.问题是来自另一个块的数据没有存储到数据库中.即使第一个块的值也被第二次存储.如何解决这个问题.
I have removed the code for image upload from it as it works completely fine. The problem is the data from the other block is not stored to the database. only the value for the first block is stored second time even. How to solve this problem.
用于存储数据的PHP:
PHP to store data:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "testing";
$conn = new mysqli($servername, $username, $password,
$dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO tbl_report (product,pp1, rp1,stacking)
VALUES ('$product', '$pp1', '$rp1','$stacking')";
if ($conn->query($sql) === TRUE) {
echo "Successful";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
?>
推荐答案
在@Logan Wayne指出的内容上进行扩展...
To expand on what @Logan Wayne pointed out...
一个ID在一个页面中应该是唯一的.但是，如果存在多个具有指定ID的元素，则getElementById()方法将返回源代码中的第一个元素.
An ID should be unique within a page. However, if more than one element with the specified ID exists, the getElementById() method returns the first element in the source code.
因此，在JavaScript中，当您获取对表数据元素的引用时，将始终获得具有您提供的ID的Document对象的FIRST实例.
So, in your JavaScript, when you grab references to your table data elements, you'll always get the FIRST instance of a Document object with whatever id you provide.
// 2. Define what to do when XHR feed you the response from the server - Start
var product = document.getElementById("product").value; <-- will always return the same element
var pp1 = document.getElementById("pp1").value; <-- will always return the same element
var rp1 = document.getElementById("rp1").value; <-- will always return the same element
var stacking = document.getElementById("stacking").value; <-- will always return the same element
您将不得不为您的td对象分配唯一的ID，或者再次如@Logan Wayne所述，利用HTML DOM对象的class属性.
You'll either have to assign unique ids to your td objects, or, again as @Logan Wayne mentioned, utilize the class property of HTML DOM objects.
类可用于对类似元素进行分组.在为表中的不同列分配类名称后(产品，促销价格，常规价格，堆叠)您可以使用 getElementsByClassName()获取td元素的数组.
Classes can be used to group like elements. After assigning class names to the different columns in your table (Product, Promotional Price, Regular Price, Stacking) you can use getElementsByClassName() to get an array of the td elements.
...
var products = document.getElementsByClassName("product"); <-- array of product td elements
...
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