我的 PHP 页面中有一个表,位于 PHP 的 HTML block 内。下面是该表的代码。
<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>
将其发送到另一个 PHP 的 JavaScript 代码如下。
<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);
}
</script>
将值存储到数据库的 PHP 如下。
<?php
//Updated after the answer from AnikIslam
$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;
}
?>
如屏幕截图所示,我有 2 个 block 来输入数据。现在数据仅从第一个 block 输入数据库。我希望数据从第二个 block 存储到数据库中的第二行。如何实现。
最佳答案
尝试这样
xhr.send("product=" + product + "&pp1=" + pp1 + "&rp1=" + rp1 + "&stacking=" + stacking);
对于 PHP 部分
$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;
}
编辑
每个表单都应该有唯一的 ID
在提交按钮中传递 block 名称
像这样
第一个区 block
onClick="PostData('Block1');"
第二 block
onClick="PostData('Block2');"
JS
function PostData(block){
if(block==="Block1"){
// fetech data from block1
}
else if(block==="Block2"){
// fetech data from block2
}
}
关于javascript - 使用 Javascript 输入数据未存储在数据库中?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31869515/