我在数据库中有两个与名称和地址匹配的信息。这段代码不是返回2,而是返回21。请参见下文;(表)雇员id名字地址
$select = mysql_query("
SELECT *
FROM employees
WHERE name LIKE '%John%'
OR name LIKE '%Johanson%'
AND address='Streetford End'
");
$count = mysql_num_rows($select);
echo $count
这可能会有帮助:
SELECT * FROM employees WHERE (name LIKE '%John%' OR name LIKE '%Johanson%')
AND address='Streetford End'