我遇到一个问题,用户 ID 的数据正在抓取该用户 ID 下输入的所有数据。我只需要获取使用相同 tr_id 输入数据库的当前数据,而不是用户在表中拥有的所有数据。
example
***How I want it to show.***
tr_id 1234 product:pens qty:2
tr_id 1234 product:pencils qty:2
***not Like this***
tr_id 1234 product:pens qty:2
tr_id 1234 product:pencils qty:2
tr_id 3456 product:papers qty:2
tr_id 3456 product:markers qty:2
测试.php
<?php
include('dbconnect.php');
session_start();
if(!isset($_SESSION['uid'])){
header('Location:index.php');
}
$uid=$_SESSION['uid'];
$sql="SELECT * FROM customer_order WHERE uid='$uid' order by id desc";
$run_query=mysqli_query($conn,$sql);
while ($row = mysqli_fetch_array($run_query))
{
echo $row['pid'] . " " . $row['p_name'] . " " . $row['tr_id'] . "<br/>";
}
?>
action.php
if(isset($_POST['payment_checkout'])){
$uid=$_SESSION['uid'];
$sql="SELECT * FROM cart WHERE user_id='$uid'";
$run_query=mysqli_query($conn,$sql);
$i=rand();
while($cart_row=mysqli_fetch_array($run_query))
{
$cart_prod_id=$cart_row['p_id'];
$cart_prod_title=$cart_row['product_title'];
$cart_qty=$cart_row['qty'];
$cart_price_total=$cart_row['total_amount'];
$sql2="INSERT INTO customer_order (uid,pid,p_name, p_price,p_qty,p_status,tr_id) VALUES ('$uid','$cart_prod_id','$cart_prod_title','$cart_price_total','$cart_qty','CONFIRMED','$i')";
$run_query2=mysqli_query($conn,$sql2);
}
最佳答案
麦克雷德的评论让我得到了这个答案
<?php
include('dbconnect.php');
session_start();
if(!isset($_SESSION['uid'])){
header('Location:index.php');
}
$uid=$_SESSION['uid'];
$sql="SELECT * FROM customer_order WHERE uid='$uid' order by id desc";
$run_query=mysqli_query($conn,$sql);
$row = mysqli_fetch_array($run_query);
$tr_id = $row['tr_id'];
$sql="SELECT * FROM customer_order WHERE uid='$uid' and tr_id='$tr_id' order by id desc";
$run_query=mysqli_query($conn,$sql);
while ($row = mysqli_fetch_array($run_query))
{
echo $row['pid'] . " " . $row['p_name'] . " " . $row['tr_id'] . "<br/>";
}
?>
关于php - 仅在 while 循环中获取最后一个事务 ID 信息时遇到问题,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/46309616/