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我是新手PHP，所以如果我犯了一个愚蠢的错误，请耐心等待。我的问题是，如果 2 个用户打开了两个以上的对话，我的消息系统将显示来自第一个主题与第二个主题的对话。我不允许发布图片，所以我会尝试输入解释：现在我得到的结果看起来像这样：

对象1：你好-嘿，你今天好吗？- 我很好，谢谢！

对象2：再见——嘿，你今天好吗？- 我很好，谢谢！- 和你聊天真是太好了！- 小心！

在上面的示例中，您可以看到来自第一个对话的消息也显示在第二个对话中。

我希望每个主题的对话只显示在其适当的主题中。我试图只获取每个循环中的最后一个主题以在我的MySQL查询中使用，但是当我从 foreach 循环中获取最后一个数组值时，它仍然返回MySQL. 如果我的请求或我的代码难以理解，我很抱歉。几天来，我一直在不停地工作，但离我的目标还很遥远。请帮忙！

这是我的代码片段：

$sql = mysql_query("SELECT * FROM private_messages WHERE to_id='$my_id' AND         recipientDelete='0' GROUP BY subject ORDER BY id DESC LIMIT 200");

$resultArrays2 = array();
while($row = mysql_fetch_array($sql)){
$resultArrays2 = array( 
$subject = $row['subject']);

foreach ($resultArrays2 as $value2 => $i2){
//foreach ($i as $second){
$array2[] = $i2;
    }

$my_uname = $row['message'];
$date = strftime("%b %d, %Y",strtotime($row['time_sent']));
if($row['opened'] == "0"){ 
        $textWeight = 'msgDefault';
} else {
        $textWeight = 'msgRead';
}
$fr_id = $row['from_id'];
$stringsubfirst = (end($array2));    
// SQL - Collect username for sender inside loop
$more = mysql_query("SELECT id, username FROM mymembers WHERE id='$p' LIMIT 1");
while($rz = mysql_fetch_array($more)){ $MEid = $rz['id']; $ME = $rz['username']; }
$ret = mysql_query("SELECT id, username FROM mymembers WHERE id='$fr_id' LIMIT 1");
while($raw = mysql_fetch_array($ret)){ $Sid = $raw['id']; $Sname = $raw['username']; }

$reachingsub = mysql_query("SELECT * FROM private_messages WHERE     subject='$stringsubfirst' AND recipientDelete !='1' LIMIT 100") or die(mysql_error());
$resultArrays = array();
while($rn = mysql_fetch_array($reachingsub)){
$resultArrays = array(
$finalsub = $rn["subject"]);    
foreach ($resultArrays as $value => $i){

$array[] = $i;
    }
}
$stringsub = (end($array));
$stringcur = (current($array));
$stringnext = (next($array));
$stringprev = (prev($array));
$stringfirst = (reset($array));



$reaching = mysql_query("SELECT * FROM private_messages WHERE subject='$stringsub' AND recipientDelete !='1' ORDER BY id DESC LIMIT 200") or die(mysql_error());
    $resultArrays2 = array();
while($rw = mysql_fetch_array($reaching)){
    $resultArrays2 = array(
    $subsub = $rw["subject"]);
        foreach ($resultArrays2 as $value2 => $i2){

        $array2[] = $i2;
        $array3[] = $value2;
        }
        $stringsub2 = (end($array2));
        $stringcur2 = (current($array2));
        $stringnext2 = (next($array2));
        $stringprev2 = (prev($array2));
        $stringfirst2 = (reset($array2));
    $subsub2 = $rw["subject"];
    $usingid = $rw["id"];
    $insidemessage = $rw["message"];
    $fi = $rw["from_id"];
    $ti = $rw["to_id"];
    $minutes = strftime("%b %d, %r",strtotime($rw['time_sent']));
    if($p !== $ti){$listname = $ME;}else{$listname = $Sname;}

    if($subject !== $stringsub2 || $stringcur2 !== $stringnext2){$othermessage .= '<div  style="display:none;">'.$ti.$subsub2.$subject.$stringsub2.$stringsubfirst.'</div>';}

    if($subject == $stringsub2 || $stringcur2 == $stringnext2){$othermessage .= '<br />'.'<div style="display:none;">'.$ti.$subsub2.$subject.$stringsub2.$stringsubfirst.'</div>'.
    '<div style="font-weight:bold;">'.$listname.': '.$minutes.'</div>'.$insidemessage.'<br />';}
}
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1 回答 1

2

我不确定我是否正确理解了你的整个问题，但是......一个非常简单的解决方案是给每个对话一个唯一的标识符并使​​用它而不是可能不是唯一的主题进行选择

于 2012-09-16T15:06:12.677 回答