我现在的问题是，当我点击页面上的图片时，它将第一次显示出来。但它将第二次显示失败。这个过程将从将数据发送到ajax开始，然后ajax（prosec.js）将其发送到php页面（process1.php）
当我删除blockquote（$query="SELECT…"）中的代码时，它将运行，但如果不运行，它将显示失败。
process1.php
<?php
include 'session.php';
include 'connection.php';
if(isset($_POST['dataS'])) {
$table = $_POST['table'];
$concat = "";
$serial = $_POST['dataS'];
$query = "SELECT * FROM product WHERE serialNum = '$serial'";
$result = mysqli_query($conn, $query);
$row = mysqli_fetch_assoc($result);
if($row) {
$prodName = $row['prodName'];
$quanProd = 1;
$priceProd = $_POST['total'] + $row['salePrice'];
if($table == "") {
$query = "SELECT * FROM product WHERE serialNum = '$serial'";
$result = mysqli_query($conn, $query);
$row = mysqli_fetch_assoc($result);
}
else{
$DOM = new DOMDocument;
$DOM->loadHTML($table);
$items = $DOM->getElementsByTagName('tr');
$check = 0;
$check_one = 0;
$y=0;
function tdrows($elements,$check,$serial,$prodName,$y) {
$quantity="";
$item = "";
$price = "";
$delete = "";
$x = 0;
foreach($elements as $element) {
if($x == 0)
$delete = $element->nodeValue;
else if($x == 1)
$item = $element->nodeValue;
else if($x == 2)
$quantity = $element->nodeValue;
else if($x == 3)
$price = $element->nodeValue;
$x++;
}
**$query = 'SELECT prodName FROM product WHERE prodName = "$item"';
$search = mysqli_query($conn, $query) or die(mysqli_error());
$row = mysqli_fetch_assoc($search);
$s = $row['prodName'];**
if($prodName == $s) {
$quantity++;
$check = 1;
}
else {
$check = 0;
}
return $check;
}
foreach ($items as $node) {
$check = tdrows($node->childNodes,$check,$serial,$prodName,$y);
$y++;
}
}
$priceProd = number_format((float)$priceProd, 2, '.', '');
echo json_encode (
array ( //this array is used to send the data back to ajax.
"success" => "1",
"concat" => $concat,
"quantity" => $quanProd,
"price" => $priceProd,
)
);
}
else {
echo json_encode (
array ( //this array is used to send the data back to ajax.
"success" => "0",
)
);
}
}
?>
process.js
$(document).ready(
function() {
$("body").on("click","#product .add",
function(e) {
var total = document.getElementById("total").value;
var table = document.getElementById('table-list').innerHTML;
table = (table.trim) ? table.trim() : table.replace(/^'s+/,'');
var serial = $(this).attr('id');
var totalQ = document.getElementById("totalQ").value;
if(total == "")
total = 0;
else
total = parseFloat(total);
if(totalQ == "")
totalQ = 0;
else
totalQ = parseInt(totalQ);
var dataS = serial;
e.preventDefault();
$.ajax({
type : "POST",
url : "process1.php",
crossDomain: true,
data : {dataS : dataS, table : table, total : total},
dataType : 'json',
})
.done(function(html) {
if(html.success == 1) {
console.log('done: %o', html);
$("#table-list").html(html.concat).show();
document.getElementById('totalQuantity').innerHTML = html.quantity;
document.getElementById("total").value = html.price;
document.getElementById("payment").value = html.price;
document.getElementById('totalQ').value = html.quantity;
document.getElementById('title').innerHTML = html.price;
document.getElementById('input').value='';
$("#input").focus();
}
else {
alert("Wrong serial number!");
document.getElementById('input').value='';
$("#input").focus();
}
})
.fail(function(html) {
console.info('fail: %o', html);
alert("fail");
});
return false;
});
});
connection.php
<?php
$conn = mysqli_connect('localhost','root','','rds');
?>
您的查询错误：请尝试此
$query = "SELECT prodName FROM product WHERE prodName = '".$item."'";
根据您的图片，您的问题是数据库连接不正确。当您执行第一个请求时，它不会进行任何数据库交互（因为在blocknotes之外）。您将发送table数据的第二个请求将执行查询。因此，第一个请求将成功，而第二个请求将在mysqli（$conn）对象上给您一个错误。
if($table == "") {
//Database interaction
$query = "SELECT * FROM product WHERE serialNum = '$serial'";
$result = mysqli_query($conn, $query);
$row = mysqli_fetch_assoc($result);
}
else{
//No database interaction because of the blocknotes
$DOM = new DOMDocument;
$DOM->loadHTML($table);
$items = $DOM->getElementsByTagName('tr');
$check = 0;
$check_one = 0;
$y=0;
function tdrows($elements,$check,$serial,$prodName,$y) {
$quantity="";
$item = "";
$price = "";
$delete = "";
$x = 0;
foreach($elements as $element) {
if($x == 0)
$delete = $element->nodeValue;
else if($x == 1)
$item = $element->nodeValue;
else if($x == 2)
$quantity = $element->nodeValue;
else if($x == 3)
$price = $element->nodeValue;
$x++;
}
**$query = 'SELECT prodName FROM product WHERE prodName = "$item"';
$search = mysqli_query($conn, $query) or die(mysqli_error());
$row = mysqli_fetch_assoc($search);
$s = $row['prodName'];**
if($prodName == $s) {
$quantity++;
$check = 1;
}
else {
$check = 0;
}
return $check;
}
foreach ($items as $node) {
$check = tdrows($node->childNodes,$check,$serial,$prodName,$y);
$y++;
}
}
请检查您的用户名、密码和数据库名称。我敢肯定你在这里用错了什么。如connection.php文件中所述，您不使用密码。您确定用户root没有密码吗？你能用phpMyAdmin这样的MySQL管理工具访问数据库吗？