PHP动态字母列表，带标题

如何解决PHP动态字母列表，带标题

| 所以我想生产的是这样的： 除非我不想重复，否则当有多个符号时，它就会重复，并重新创建\“ C \”列表视图标题。我的SQL查询正在创建称为first_char的列，该列保存其他列之一的第一个字符。......下面是我的代码：

<?php foreach ($data as $name=>$signs): ?>
    <!-- If name is not empty,create the header -->
        <?php if (isset($name) && !empty($signs)): ?> 
            <div data-role=\'collapsible-set\' data-theme=\'d\'>

                <h2>Username:&nbsp;<strong id=\"headerName\"><?=$name?></strong></h2>

        <?php endif; ?>

    <?php endforeach; ?>
    <!-- for every sign,take the first char and create a list view -->
    <?php foreach ($signs as $char):
            $current_char = \'\';
            if ($char[\'first_char\'] != $current_char){
                $current_char = $char[\'first_char\'];
                echo \'<ul data-role=\"listview\" data-dividertheme=\"d\" style=\"margin-top: 0;\"><li data-role=\"list-divider\">\'.$current_char.\'</li></ul><br>\';
            }

            foreach ($signs as $sign):
                if($current_char == $sign[\'first_char\'] and !empty($signs)):                           
    ?>

                    <div data-role=\'collapsible\' data-collapsed=\'true\' data-icon=\'arrow-l\'>
                        <h3><?=$sign[\'sign_name\']?></h3>
                        <div class=\'ui-grid-a\'>                 
                            <div class=\"ui-block-a\">
                                <div class=\"ui-bar ui-bar-d\" style=\"height:25px;\">
                                    <strong>Last Seen:</strong><strong id=\"text\"><?=$sign[\'last_connected\']?></strong>
                                </div>
                            </div>
                            <div class=\"ui-block-a\">
                                <div class=\"ui-bar ui-bar-d\" style=\"height:25px;\">
                                    <strong>Resolution:</strong><strong id=\"text\"><?=$sign[\'resolution_x\'].\'x\'.$sign[\'resolution_y\']?></strong>
                                </div>
                            </div>
                            <div class=\"ui-block-a\">
                                <div class=\"ui-bar ui-bar-d\" style=\"height:25px;\">
                                    <strong>Group Name:</strong><strong id=\"text\"><?=$sign[\'group_name\']?></strong>
                                </div>
                            </div>
                            <div class=\"ui-block-a\">
                                <div class=\"ui-bar ui-bar-d\" style=\"height:25px;\">
                                    <strong>Sign Number:</strong><strong id=\"text\"><?=$sign[\'sign_id\']?></strong>
                                </div>
                            </div>
                       </div>
                    </div>
            <?php endif; ?>
        <?php endforeach; ?>
    <?php endforeach; ?>

提前致谢     

解决方法

这可以修复先前的错误并消除重复。 $ current_char需要在foreach循环之外，并删除了第二个foreach。

<?php $current_char = \'\';?>
        <?php foreach ($signs as $sign):?>
                <?php                       
                    if ($sign[\'first_char\'] != $current_char){
                        $current_char = $sign[\'first_char\'];
                        echo \'<ul data-role=\"listview\" data-dividertheme=\"b\" style=\"margin-top: 0;padding-top: 3px;\"><li data-role=\"list-divider\">\'.$current_char.\'</li></ul><br>\';                     
                    }
                ?>

上面用于SQL的查询是这样的：

SELECT 
    user.username,groups.group_id,groups.group_name,sign.last_connected,sign.sign_id,sign.sign_name,sign.resolution_x,sign.resolution_y,LEFT(sign.sign_name,1) AS first_char
FROM
    user
LEFT JOIN(
    groups,sign
)ON(
    user.user_id = groups.userID AND
    groups.group_id = sign.groupID
)
WHERE
    username = ? AND
    UPPER(sign.sign_name) BETWEEN \"A\" AND \"Z\"
    OR sign.sign_name BETWEEN \"0\" AND \"9\" ORDER BY sign.sign_name