mysqli查詢（選擇）不工作（選擇）php

我想做一個簡單的SELECT，但我不知道爲什麼它不工作。就在這個頁面是不是working.I相信，我有我的MySQL數據庫的數據，我想這應該工作..mysqli查詢（選擇）不工作（選擇）php

<?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']; 
    }

錯誤消息是：？

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

在這個頁面中我也有2次需要的頁面，這都需要在html part.I想使Facebook這樣的網站，而這些網頁是描述和橫幅&輪廓圖像部分。

<?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']; 
}?>

螞蟻這樣的：

<?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; 
}

來源

2017-04-21 Belciug Ovidiu-Mihai

你在這裏有一些不正確的引用''user_id'= $ id'。列名稱不應加引號。 http://stackoverflow.com/questions/11321491/when-to-use-single-quotes-double-quotes-and-backticks-in-mysql#11321508 –

使用像這樣'「。$ _ POST ['id']。「」。 – lalithkumar

Ommmmg是啊這麼多邁克爾.. –

不正確的引用必須在你的代碼的問題。

嘗試使用， $ query =「SELECT * FROM images WHERE user_id = $ id」; 而不是， $ query =「SELECT * FROM images WHERE'user_id'='$ id'」;

來源

2017-04-21 17:25:11

是啊....太多了！ –

mysqli查詢（選擇）不工作（選擇）php

回答

相關問題