Question

所以我的代码有两个版本

哪个不起作用
哪个有效...

此代码不起作用：

<?php
$status=0;
$user_id=$friend_id="";
$user_name=$_COOKIE['usernam'];
$friend_name=$_POST['text'];
$connect=mysqli_connect("localhost","root","*censored*","mydbrun");
$sql1="SELECT id FROM users WHERE username='".$user_name."'";
$result=mysqli_query($connect,$sql1);
if(mysqli_num_rows($result)>0)
{
    $row=mysqli_fetch_array($result);
    $user_id=$row['id'];
}
$connect1=mysqli_connect("localhost","root","*censored*","mydbrun");
$sql2="SELECT id FROM users WHERE username='".$friend_name."'";
$result1=mysqli_query($connect1,$sql2);
if(mysqli_num_rows($result1)>0)
{
    $row1=mysqli_fetch_array($result1);
    $friend_id=$row1['id'];
}
$connect2=mysqli_connect("localhost","root","*censored*","mydbrun");
$sql="INSERT INTO friends(person_one,person_two,status)           VALUES('".$user_id."','".$friend_id."','".$status."')";
if(mysqli_query($connect2,$sql)){
    echo "Request sent";
}
else{
    echo "Request could not be sent";
}
?>

显示“请求无法发送” 但是当我尝试这段代码时......

<?php
$status=0;
$user_id=$friend_id="";
$user_name="rishav37";
$friend_name="vivek";
$connect=mysqli_connect("localhost","root","*censored*","mydbrun");
$sql1="SELECT id FROM users WHERE username='".$user_name."'";
$result=mysqli_query($connect,$sql1);
if(mysqli_num_rows($result)>0)
{
    $row=mysqli_fetch_array($result);
    $user_id=$row['id'];
}
$connect1=mysqli_connect("localhost","root","*censored*","mydbrun");
$sql2="SELECT id FROM users WHERE username='".$friend_name."'";
$result1=mysqli_query($connect1,$sql2);
if(mysqli_num_rows($result1)>0)
{
    $row1=mysqli_fetch_array($result1);
    $friend_id=$row1['id'];
}
$connect2=mysqli_connect("localhost","root","*censored*","mydbrun");
$sql="INSERT INTO friends(person_one,person_two,status)           VALUES('".$user_id."','".$friend_id."','".$status."')";
if(mysqli_query($connect2,$sql)){
    echo "Request sent";
}
else{
    echo "Request could not be sent";
}
?>

但是此代码有效！我甚至检查了$_COOKIE['usernam']和$_POST['text']会发生什么，他们已经存储了“rishav37”和“vivek”......我不知道为什么以前的代码不起作用！有人帮忙，先谢谢。

PHP代码无法正常工作！无法执行mysqli查询

0 个答案: