PHP-如何根据26个字母分类列出用户名？

Question

有个用户表，我要把所有用户名根据26个英文字母分类列出？如何sql语句并输出？如图。

Answer 1

我觉得可以在sql中使用case when语句来实现，只是有些死板。假设表中有个字段为name，sql如下：

select
case when head BETWEEN 'A' AND 'G' THEN 'AG'
when head BETWEEN 'H' AND 'N' THEN 'HN'
when head BETWEEN 'O' AND 'T' THEN 'OT'
END AS array,name
from (
select UPPER(substr(name,1,1)) as head,`name` from edu_user) temp

取出结果后，根据array的值来处理。楼主看一下怎么样，我不知道是否合适。

Answer 2

<?php
$names = array (
"bill gates",
"tanteng",
"wangjun",
"renhong",
"zhangyong",
"liyifei",
"wangwei",
"zhoutong",
"wangcheng",
"liuyingying",
"yuanbo",
"liudehua",
"liangchaowei",
"zhangxueyou",
"zhaowei",
"jiangjieshi",
"maozedong",
"liuyifei",
"Pony",
"Xieyangjun",
"Bianxiaojun"
);

$AG = array (
"A",
"B",
"C",
"D",
"E",
"F",
"G"
);

$HN = array (
"H",
"I",
"J",
"K",
"L",
"M",
"N"
);
$OT = array (
"O",
"P",
"Q",
"R",
"S",
"T"
);
$UZ = array (
"U",
"V",
"W",
"X",
"Y",
"Z"
);
$designer = '';
//自然排序算法对数组中名称排序（不区分大小写）
natcasesort($names);
foreach ( $names as $name ) {
$firstn = strtoupper ( substr ( $name, 0, 1 ) );
if (in_array ( $firstn, $AG )) {
$designer ['AG'] [] = $name;
} elseif (in_array ( $firstn, $HN )) {
$designer ['HN'] [] = $name;
} elseif (in_array ( $firstn, $OT )) {
$designer ['OT'] [] = $name;
} elseif (in_array ( $firstn, $UZ )) {
$designer ['UZ'] [] = $name;
}
}
//按照键名排序，正确排列
ksort($designer);
print_r ( $designer );

这样就ok了，第一个数组$names假设是从数据库取出来的。