php mysqli中的like问题
这个代码怎么了php mysqli中的like问题,php,mysql,Php,Mysql,这个代码怎么了 SELECT alias FROM Articls WHERE (title LIKE '%key%' OR sections LIKE '%key%' OR description LIKE '%key%' OR category LIKE '%key%' OR subcategory LIKE '%key%) AND MATCH (content) AGAINST ('key') GROUP BY alias ORDER BY alias DESC LIMIT
SELECT alias FROM Articls
WHERE (title LIKE '%key%'
OR sections LIKE '%key%'
OR description LIKE '%key%'
OR category LIKE '%key%'
OR subcategory LIKE '%key%)
AND MATCH (content) AGAINST ('key')
GROUP BY alias ORDER BY alias DESC LIMIT 500
但我遇到了一个错误…所以我将代码简化为:
SELECT alias FROM Articls
WHERE (title LIKE '%key%'
OR content LIKE '%key%'
OR sections LIKE '%key%'
OR description LIKE '%key%'
OR category LIKE '%key%'
OR subcategory LIKE '%key%)
ORDER BY alias DESC LIMIT 500
但我仍然得到一个错误:
警告:mysqli_fetch_all()要求参数1为mysqli_结果,
中给出的布尔值
/home/admin/domains/example.com/public_html/lib/core/p.php在线
141-您的SQL语法有错误;请查看与您的MySQL服务器版本对应的手册,以了解在“%news%”附近使用的正确语法
你在最后的LIKE语句末尾遗漏了一个撇号
SELECT alias FROM Articls
WHERE (title LIKE '%key%'
OR content LIKE '%key%'
OR sections LIKE '%key%'
OR description LIKE '%key%'
OR category LIKE '%key%'
OR subcategory LIKE '%key%')
ORDER BY alias DESC
LIMIT 500
此处的语法突出显示(颜色编码)指示缺少引号的位置。你看到了吗?哦,沃尔多游戏在哪里!它位于子类别后面,如“%key%”