php - mysqli查询(选择)不起作用(选择)php

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两年前关闭。
我只是想做一个简单的选择，但我不知道为什么它不起作用。
只是在这个页面上不起作用。我确信我的mysql数据库中有这些数据，我认为这应该起作用。。

<?php 
require_once 'includes/db_connection.php';
require_once 'includes/functions.php'; 
require_once 'includes/profile/get_the_id.php';
require_once 'app/profile/header/head-profile.php'; 

$query = "SELECT * FROM `description` WHERE 'user_id'=$id";
$result = mysqli_query($conn,$query) or die(mysqli_error($conn));

while ($row = mysqli_fetch_assoc($result)){ 
    $intro = $row['introducere'];
}

$query = "SELECT * FROM `images` WHERE 'user_id' = '$id'";
$result = mysqli_query($conn,$query) or die(mysqli_error($conn));
while ($row = mysqli_fetch_assoc($result)){
        $img = $row['img'];
    }

?>
错误消息是：

 Notice: Undefined variable: intro in C:\xampp\htdocs\New\profile.php on line 33

Notice: Undefined variable: img in C:\xampp\htdocs\New\profile.php on line 33

在这个页面中，我也有两个必需的页面，这是html部分所必需的，我正在尝试建立一个像facebook这样的网站，这些页面是描述和横幅和个人资料图像部分。

<?php
$conn = mysqli_connect('localhost','root','','forum');
$username = $_SESSION['username'];
require_once 'get_the_id.php';

$target =" ";
if (isset($_POST['upload'])) {
    $target = "./img/" . basename($_FILES['image']['name']);

    $image = $_FILES['image']['name'];

    $query = "INSERT INTO `images` VALUES('','$id','$image')";
    mysqli_query($conn,$query) or die(mysqli_error($conn));

    $query = "UPDATE `users` SET `img_profil` = '$image' WHERE `users`.`id` = '$id'";
    mysqli_query($conn,$query) or die(mysqli_error($conn));
    $msg="";
    if (move_uploaded_file($_FILES['image']['tmp_name'], $target)){
        $msg="Success";
        }else{
            $msg="Fail";
        }
    $_SESSION['image'] = $image;
}

$query = "SELECT * FROM `images` WHERE 'user_id'='$id'";
$result = mysqli_query($conn,$query);
while ($row = mysqli_fetch_assoc($result)){
    $img = $row['img'];
}?>

蚂蚁这个：

<?php
$conn = mysqli_connect('localhost','root','','forum');
require 'includes/profile/get_the_id.php';

$desc = $_SESSION['desc'];  
$username = $_SESSION['username'];
$_SESSION['intro'] = " ";
if(isset($_POST['save'])){
    $intro =mysqli_real_escape_string($conn,trim($_POST['intro']));
    if ($desc == 1) {
        //daca are deja o descriere se actualizeaza 
        $query = "UPDATE `description` SET `introducere` = '$intro' WHERE `description`.`user_id` ='$id'";
        mysqli_query($conn,$query) or die(mysqli_error($conn));

    }elseif ($desc == 0) {
        //daca nu are descriere se insereaza descrierea facuta acum in db
        $query = "INSERT INTO `descripbion` VALUES('','$id','$username','$intro','','','')";
        mysqli_query($conn,$query);
        //desc devine 1 deoarece a facut o descriere
        $query = "UPDATE `users` SET `desc` = '1' WHERE `users`.`id` = '$id'";
        mysqli_query($conn,$query) or die(mysqli_error($conn));

    }
    $_SESSION['intro'] = $intro;
}

最佳答案

您的代码中一定存在引用错误的问题。
尝试使用，

$query="SELECT * FROM images WHERE user_id= $id"; 

而不是，

$query = "SELECT * FROM `images` WHERE 'user_id'='$id'";

关于php - mysqli查询(选择)不起作用(选择)php ，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/43548359/

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Answer 1

您的代码中一定存在引用错误的问题。
尝试使用，

$query="SELECT * FROM images WHERE user_id= $id"; 

而不是，

$query = "SELECT * FROM `images` WHERE 'user_id'='$id'";