php - 查询失败 您的 SQL 语法有错误;检查与您的 MariaDB 服务器版本相对应的手册以获取正确的语法

标签 php mysql database crud

我在将内容更新到数据库时遇到问题。错误是:

Query FailedYou have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'post_content = 'kjgljkkjhklj ', post_image = '45739895_2381062595269282_8123898' at line 1

这是我的整个代码:

if (isset($_GET['p_id'])) {
    $the_post_id = $_GET['p_id'];
}

$query = "SELECT * FROM posts WHERE post_id = $the_post_id";
$select_posts_by_id = mysqli_query($connection, $query);

while ($row = mysqli_fetch_assoc($select_posts_by_id)) {
    $post_id = $row['post_id'];
    $post_author = $row['post_author'];
    $post_title = $row['post_title'];
    $post_category_id = $row['post_category_id'];
    $post_status = $row['post_status'];
    $post_image = $row['post_image'];
    $post_content = $row['post_content'];
    $post_tags = $row['post_tags'];
    $post_comment = $row['post_comment_count'];
    $post_date = $row['post_date'];
}


// if update post button is clicked
if (isset($_POST['update_post'])) {

    $post_author = $_POST['post_author'];
    $post_title = $_POST['post_title'];
    $post_category_id = $_POST['post_category'];
    $post_status = $_POST['post_status'];
    $post_image = $_FILES['image']['name'];
    $post_image_temp = $_FILES['image']['tmp_name'];
    $post_content = $_POST['post_content'];
    $post_tags = $_POST['post_tags'];


    move_uploaded_file($post_image_temp, "../images/{$post_image}");

    if (empty($post_image)) {
        $query = "SELECT * FROM posts WHERE post_id = $the_post_id ";

    }

    $query = "UPDATE posts SET ";
    $query .= "post_title = '{$post_title}', ";
    $query .= "post_category_id = '{$post_category_id}', ";
    $query .= "post_date = now(), ";
    $query .= "post_author = '{$post_author}', ";
    $query .= "post_status = '{$post_status}', ";
    $query .= "post_tags = '{$post_tags}' ";
    $query .= "post_content = '{$post_content}', ";
    $query .= "post_image = '{$post_image}' ";
    $query .= "WHERE post_id = {$the_post_id}"; // this is from the get request


    $update_post = mysqli_query($connection, $query);

    confirmQuery($update_post);

}

这是下面的表格:

<form action="" method="POST" enctype="multipart/form-data">

    <div class="form-group">
        <label for="title">Post Title</label>
        <input value="<?php echo $post_title; ?>" type="text" name="post_title" class="form-control" required="true">
    </div>

    <div class="form-group">
        <label for="post_category">Post Categories</label>
        <select name="post_category" id="" class="form-control form-control-md">
            <?php 
                $query = "SELECT * FROM categories";
                $select_categories = mysqli_query($connection, $query);

                confirmQuery($select_categories); // this is from functions.php

                while ($row = mysqli_fetch_assoc($select_categories)) {
                        $cat_id = $row['cat_id'];
                        $cat_title = $row['cat_title'];

                        echo "<option value='{$cat_id}'>{$cat_title}</option>";
                    }   
             ?>
        </select>
    </div>

    <div class="form-group">
        <label for="author">Post Author</label>
        <input value="<?php echo $post_author; ?>" type="text" name="post_author" class="form-control" required="true">
    </div>

    <div class="form-group">
        <label for="post_status">Post Status</label>
        <input value="<?php echo $post_status; ?>" type="text" name="post_status" class="form-control" required="true">
    </div>

    <div class="form-group">
        <label for="image">Post Image</label>
        <img width="100px" src="../images/<?php echo $post_image; ?>">
        <input type="file" name="image" required="true">
    </div>

    <div class="form-group">
        <label for="post_tags">Post Tags</label>
        <input value="<?php echo $post_tags; ?>" type="text" name="post_tags" class="form-control" required="true">
    </div>

    <div class="form-group">
        <label for="post_content">Post Content</label> 
        <textarea name="post_content" class="form-control" id="" cols="30" rows="10" required="true">
            <?php echo $post_content; ?>
        </textarea>
    </div>

    <div class="form-group">
        <input type="submit" name="update_post" class="btn btn-primary" value="Update Post">
    </div>
</form>

最佳答案

FWIW,我发现这更容易阅读:

$query = "
UPDATE posts 
   SET post_title = '$post_title'
     , post_category_id = '$post_category_id'
     , post_date = now()
     , post_author = '$post_author'
     , post_status = '$post_status'
     , post_tags = '$post_tags'
     , post_content = '$post_content'
     , post_image = '$post_image'
 WHERE post_id = $the_post_id;
";

...但用正确参数化的查询替换这些字符串也很重要

关于php - 查询失败 您的 SQL 语法有错误;检查与您的 MariaDB 服务器版本相对应的手册以获取正确的语法,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54452765/

相关文章:

php - 电子邮件营销统计实现

database - 规范化两列表

javascript - 如何使用 Javascript 在服务器上保存来自 jsPDF 的 PDF 文件?

php - 使用 csv 文件更新 mysql 表。我需要调试器

php - 如何在 PHP 中输​​出 HackerRank 的简单数组求和挑战?

mysql/mariadb - 创建表列索引不工作

php - 如何在跨三个表的 mysql 查询中使用联接

mysql - 无法在 RDS 实例上启用加密

带有连接或子查询的 MySQL 查询

mysql - 查询最多 'regular'消费客户的方法(AdventureWorks)