php - 如果我们将查询存储到变量中，MySQL REPLACE 将不起作用

Question

我的查询是这样的:

<?php
$sql = "UPDATE macs_temp SET home_page_heading='$home_page_heading', home_page_sub_heading='$home_page_sub_heading', home_page_description='$home_page_description', about_us_heading='$about_us_heading', about_us_description='$about_us_description', business_description='$business_description', we_do_description='$we_do_description', video_description='$video_description', photography_description='$photography_description', music_description='$music_description', web_and_app_description=REPLACE($web_and_app_description, '\'', '') WHERE id=1";

if ($conn->query($sql) === TRUE) {
//    echo "Record updated successfully";
header("Location: index.php");
   exit;
} else {
    echo "Error updating record(Contact System Admin): " . $conn->error;
    echo "Inside else error..!";
}
?>

错误如下: 更新记录时出错(联系系统管理员):您的 SQL 语法有错误；检查与您的 MySQL 服务器版本相对应的手册，了解在第 1 行第 1 行的 ''j, '\'', '') WHERE id=1' 附近使用的正确语法，其他错误..!

注意:我的表单包含 11 个字段，其中任何/多个字段可能包含撇号，因此我想使用所有撇号来更新 mysql 数据库中的所有数据。

Answer 1

尝试一下，您需要将 $web_and_app_description 括在 '' 中，以便将其视为字符串

$sql = "UPDATE macs_temp SET home_page_heading='$home_page_heading', 
home_page_sub_heading='$home_page_sub_heading', 
home_page_description='$home_page_description', 
about_us_heading='$about_us_heading', about_us_description='$about_us_description', 
business_description='$business_description', 
we_do_description='$we_do_description', video_description='$video_description', 
photography_description='$photography_description', 
music_description='$music_description', 
web_and_app_description=REPLACE('$web_and_app_description', '\'', '') WHERE 
id=1";