Question

我想从表中选择匹配的确切单词，但我遇到了一个问题

如果行像：Angel和我确实从表中选择*，其中row =“angel”成功并且第一次

字母很小，而db是其首都，

$r = mysql_fetch_row(mysql_query("SELECT ID 
                                  FROM Login
                                  WHERE Username = 'angel'
                                    And Password = 'zxc'"));
if($r[0])
    die("success");
else
    die("failed");

In mysql Table
Username : varchar(50) : Angel
Password varchar(50) : zxc

结果应该是虚假的

因为它的天使不是天使

所以任何解决方案

Answer 1

您可以使用BINARY。

SELECT *  FROM `table` WHERE BINARY `row` = 'angel'

这将使区分大小写匹配。

Answer 2

是的，这是由于您的表字段的整理。您应该将其设置为区分大小写的排序规则，通常以cs 为latin1_swedish_cs后缀

Answer 3

试试这个，

$r = mysql_fetch_row(mysql_query("SELECT ID FROM Login WHERE  
            Password = 'zxc' AND Username collate latin1_general_cs = 'angel'"));

阅读Case Sensitive Database Query和MySQL case sensitive query

Answer 4

尝试

SELECT ID
FROM Login
WHERE Username = 'angel' COLLATE utf8_bin;

Answer 5

mysql_query("
    ALTER TABLE `Login`
    CHANGE `Username` `Username` VARCHAR(250) CHARACTER SET latin1 COLLATE latin1_swedish_ci NOT NULL
");    

$r = mysql_fetch_row(mysql_query("SELECT ID FROM Login WHERE  Username = 'angel' And Password = 'zxc'"));
if($r[0])
    die("success");
else
    die("failed");

现在它会起作用。

Answer 6

尝试

SELECT ID
FROM Login
WHERE upper(Username) = upper('angel')

php mysql查询选择匹配确切的单词

6 个答案: