[英]mysqli query not working in foreach php
我的問題是,當我單擊頁面上的圖片時,它將第一次顯示出來。 但是第二次它將顯示失敗。 該過程將從將數據發送到ajax開始,然后ajax(prosess.js)將其發送到php頁面(process1.php)。
當我刪除blockquote($ query =“ SELECT ...”)中的代碼時,它將運行,但是如果沒有,它將顯示失敗。
<?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",
)
);
}
}
?>
$(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;
});
});
<?php
$conn = mysqli_connect('localhost','root','','rds');
?>
您的查詢有誤:請嘗試
$query = "SELECT prodName FROM product WHERE prodName = '".$item."'";
根據您的圖片,您的問題是數據庫連接不正確。 當您執行第一個請求時,它不會進行任何數據庫交互(因為關閉了注釋)。 您將發送第二個請求table
數據,它將執行查詢。 因此,第一個請求將成功,而第二個請求將使您的mysqli
( $conn
)對象出現錯誤。
if($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++;
}
}
檢查您的用戶名,密碼和數據庫名稱。 我敢肯定你在這里用錯了。 如在connection.php
文件中所述,您不使用密碼。 您確定用戶root
用戶沒有密碼嗎? 您可以使用MySQL管理工具(例如phpMyAdmin)訪問數據庫嗎?
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