Question

我正在开发一个评论系统，其中我将要从两个表中提取信息，问题是来自第一个查询（找到的ID）的结果影响第二个查询的输出。我的桌子看起来像这样。

成员

Id，First_name，Last_name，电子邮件，密码，Img_url，Activation_No，已激活和日期

帖子

PostId，Email，Text，ForumId，DatePosted，Likes，Dislikes

我使用的代码是;

<?php
    // retrive post
    include('php/config.php');
    include ('php/function.php');

    // retrive comments with post id

$stmt = $mysqli->prepare("SELECT (posts.Email,posts.Text) FROM posts WHERE posts.ForumId = '$forumId' LEFT JOIN  SELECT (members.Img_url) FROM members WHERE members.Email = (posts.Email)");
$stmt->execute();
$stmt->bind_result($PostId,$Email,$Text,$ForumId,$DatePosted,$Likes,$Dislikes,$usr_img);
$stmt->fetch();
    print "<div class=comment-item>";
    print   "<div class=comment-avatar>";
    print   "<img src=".$usr_img."alt=avatar>";
    print   "</div>";
    print   "<div class=comment-post>";
    print       "<h4>" .$Email. "<span> said....</span></h4>";
    print       "<p>" .$Text. "</p>";
    print   "</div>";
    print"</div>";
$stmt->close();
?>

我收到的错误讯息是

致命错误：在第90行的C：\ Users \ PC \ Documents \ XAMPP \ htdocs \ post.php中的非对象上调用成员函数execute（）

Answer 1

您的SQL查询有错误。试试这个

SELECT posts.Email, posts.Text, members.Img_url
FROM posts 
LEFT JOIN members 
    ON members.Email = posts.Email
WHERE posts.ForumId = '$forumId'

Answer 2

您的prepare()查询失败，因此$stmt是非对象。看起来您的查询语法有误，因为LEFT JOIN之后您无法WHERE，而JOIN使用ON而不是WHERE。试试这样的事情 -

$stmt = $mysqli->prepare("SELECT posts.Email,posts.Text, members.Img_url FROM posts LEFT JOIN   members ON members.Email = posts.Email  WHERE posts.ForumId = '$forumId'") ;

Answer 3

SELECT posts.Email, posts.Text, members.Img_url
FROM posts
LEFT JOIN members ON members.Email = posts.Email
WHERE posts.ForumId = '$forumId'

你应该看看mysql的语法..你在这里犯了一些错误;）我希望它能像这样工作。

PHP加入了MySQL查询

3 个答案: