MYSQL/PHP 选择不同

发布于 2024-12-05 01:01:20 字数 1570 浏览 5 评论 0原文

我在另一个页面上调用了这个函数。

function adminnav (){
    $pagewcoms = mysql_query("SELECT DISTINCT pageid FROM comments") or die(mysql_error());
    $idnavrow = mysql_fetch_row($pagewcoms);
    while ($itest = mysql_fetch_row($pagewcoms)) {
        echo "$itest[0] <br />";
    }
}

adminnav('');

该表看起来像这样

CREATE TABLE `comments` (
  `commentid` int(5) NOT NULL auto_increment,
  `pageid` int(5) NOT NULL default '0',
  `name` text NOT NULL,
  `email` text NOT NULL,
  `comment` text NOT NULL,
  `date` datetime NOT NULL default '0000-00-00 00:00:00',
  PRIMARY KEY  (`commentid`)
) ENGINE=MyISAM  DEFAULT CHARSET=utf8 AUTO_INCREMENT=481 ;

INSERT INTO `comments` VALUES(480, 1, '3', '3', '3', '2011-09-13 22:43:06');
INSERT INTO `comments` VALUES(479, 1, '2', '2', '2', '2011-09-13 22:43:01');
INSERT INTO `comments` VALUES(476, 1, '1', '1', '1', '2011-09-13 17:22:49');
INSERT INTO `comments` VALUES(477, 2, 'taylordcraig', '[email protected]', 'this is page two', '2011-09-13 17:26:09');
INSERT INTO `comments` VALUES(478, 3, 'this is page3', 'this is page3', 'this is page3', '2011-09-13 22:28:59');

我的问题是函数打印 “大批大批” 我可以处理数据，但是不应该有 3 个不同的结果吗？这是我最近的一次。这个想法是搜索表格并找到带有评论的页面，我稍后会将其列为链接。

所以我正在寻找的结果是

“1 2 3”

提前谢谢你。

编辑：

很抱歉我的问题措辞不好。问题不是“数组数组”，而是只有两个。我可以让它显示，但这并不重要。我仍然只显示“2 3”，如前所述，我只有两个数组数据。

[** 我，我现在必须重新问这个问题吗？]

原文

I have this function called on another page.

function adminnav (){
    $pagewcoms = mysql_query("SELECT DISTINCT pageid FROM comments") or die(mysql_error());
    $idnavrow = mysql_fetch_row($pagewcoms);
    while ($itest = mysql_fetch_row($pagewcoms)) {
        echo "$itest[0] <br />";
    }
}

adminnav('');

The table looks like this

CREATE TABLE `comments` (
  `commentid` int(5) NOT NULL auto_increment,
  `pageid` int(5) NOT NULL default '0',
  `name` text NOT NULL,
  `email` text NOT NULL,
  `comment` text NOT NULL,
  `date` datetime NOT NULL default '0000-00-00 00:00:00',
  PRIMARY KEY  (`commentid`)
) ENGINE=MyISAM  DEFAULT CHARSET=utf8 AUTO_INCREMENT=481 ;

INSERT INTO `comments` VALUES(480, 1, '3', '3', '3', '2011-09-13 22:43:06');
INSERT INTO `comments` VALUES(479, 1, '2', '2', '2', '2011-09-13 22:43:01');
INSERT INTO `comments` VALUES(476, 1, '1', '1', '1', '2011-09-13 17:22:49');
INSERT INTO `comments` VALUES(477, 2, 'taylordcraig', '[email protected]', 'this is page two', '2011-09-13 17:26:09');
INSERT INTO `comments` VALUES(478, 3, 'this is page3', 'this is page3', 'this is page3', '2011-09-13 22:28:59');

My problem is the function prints
"array
array"
I can deal with the data, but shouldn't there be 3 distinct results?
This is the closest I've been. The idea is to search the table and find pages with comments, which I will later list as links.

So the result I'm looking for would be

"1
2
3"

thank you in advance.

EDIT:

I'm sorry I worded my question poorly. The porlem isnt "array array" it's that there's only two. I can make it display but it doesnt matter.
I still only have "2 3" displaying, as said before, I only had two array data.

[** me, am I going to have to re-ask this now?]

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葵雨 2024-12-12 01:01:20

应该是 echo "$itest[0] >";

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吃不饱 2024-12-12 01:01:20

检查 PHP 文档： http://php.net/manual/en/ function.mysql-fetch-row.php

mysql_fetch_row — 以枚举数组的形式获取结果行

您不是在获取值，而是在获取数组。您只看到“array array”的原因是因为第一个结果在 $idnavrow = mysql_fetch_row($pagewcoms); 行上“消失”。

将您的函数更改为以下内容，我相信您会看到您期望的所有 3 个数字：

    $pagewcoms = mysql_query("SELECT DISTINCT pageid FROM comments") or die(mysql_error());
    while ($itest = mysql_fetch_row($pagewcoms)) {
        echo $itest[0] . ' <br />';
    }

Check the PHP docs: http://php.net/manual/en/function.mysql-fetch-row.php

mysql_fetch_row — Get a result row as an enumerated array

You're not fetching a value, you're fetching an array. The reason you only see "array array" is because the first result is getting "disappeared" on the $idnavrow = mysql_fetch_row($pagewcoms); line.

Change your function to the following and I'm sure you'll see all 3 numbers you expect:

    $pagewcoms = mysql_query("SELECT DISTINCT pageid FROM comments") or die(mysql_error());
    while ($itest = mysql_fetch_row($pagewcoms)) {
        echo $itest[0] . ' <br />';
    }

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