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表输出php和mysql未获得结果_Php_Mysql - Fatal编程技术网

表输出php和mysql未获得结果

表输出php和mysql未获得结果,php,mysql,Php,Mysql,我有以下几点:我得到的结果是附加的图像。我想用户名只列出一次,然后赢,在整个页面松散。不是每个结果对应一行 $sql\u events=mysql\u query(“按“用户名”asc从工作日订单中选择*) 或者死(mysql_error()); while($row=mysql\u fetch\u数组($sql\u事件)){ $username=$row[“username”]; $week=$row[“赢”]; $row\U color=($row\U count++%2==0?$colo

我有以下几点:我得到的结果是附加的图像。我想用户名只列出一次,然后赢,在整个页面松散。不是每个结果对应一行

$sql\u events=mysql\u query(“按“用户名”asc从工作日订单中选择*)
或者死(mysql_error());
while($row=mysql\u fetch\u数组($sql\u事件)){
$username=$row[“username”];
$week=$row[“赢”];
$row\U color=($row\U count++%2==0?$color1:$color2);
回声';
回显“.$username.”;
回音“.$week.”;
回音“.$week.”;
回音“.$week.”;
回音“.$week.”;

您正在对尚未获得数据的几周进行预回显。我们的想法是,每周打印一个单元格,只要它指向同一用户:

$sql_events = mysql_query("SELECT * FROM weekpicks ORDER BY username asc ")
        or die(mysql_error());

$current_username = null;
while ($row = mysql_fetch_array($sql_events)) {
    $username = $row["username"];    
    //User changed
    if ($current_username == null || $current_username != $username) {        
        if ($current_username != null) {
            echo '</tr>'; //Had another user before so end the row
        }
        $row_color = ($row_count++ % 2 == 0 ? $color1 : $color2);       
        echo '<tr style="background-color: ' . $row_color . ';">';
        echo '<td style="width: 100" align="center"><font size="2">' . $username . '</td>';
        $current_username = $username;
    }    
    $week = $row["win_loose"];
    echo '<td style="width: 50" align="center"><font size="2">' . $week . '</td>';
}
echo '</tr>';
$sql\u events=mysql\u query(“按用户名asc从工作日订单中选择*)
或者死(mysql_error());
$current_username=null;
while($row=mysql\u fetch\u数组($sql\u事件)){
$username=$row[“username”];
//用户更改
如果($current_username==null | |$current_username!=$username){
如果($current_username!=null){
echo“”;//之前有另一个用户,因此结束该行
}
$row\U color=($row\U count++%2==0?$color1:$color2);
回声';
回显“.$username.”;
$current_username=$username;
}    
$week=$row[“赢”];
回音“.$week.”;
}
回声';

您正在对尚未获得数据的几周进行预回显。我们的想法是,每周打印一个单元格,只要它指向同一用户:

$sql_events = mysql_query("SELECT * FROM weekpicks ORDER BY username asc ")
        or die(mysql_error());

$current_username = null;
while ($row = mysql_fetch_array($sql_events)) {
    $username = $row["username"];    
    //User changed
    if ($current_username == null || $current_username != $username) {        
        if ($current_username != null) {
            echo '</tr>'; //Had another user before so end the row
        }
        $row_color = ($row_count++ % 2 == 0 ? $color1 : $color2);       
        echo '<tr style="background-color: ' . $row_color . ';">';
        echo '<td style="width: 100" align="center"><font size="2">' . $username . '</td>';
        $current_username = $username;
    }    
    $week = $row["win_loose"];
    echo '<td style="width: 50" align="center"><font size="2">' . $week . '</td>';
}
echo '</tr>';
$sql\u events=mysql\u query(“按用户名asc从工作日订单中选择*)
或者死(mysql_error());
$current_username=null;
while($row=mysql\u fetch\u数组($sql\u事件)){
$username=$row[“username”];
//用户更改
如果($current_username==null | |$current_username!=$username){
如果($current_username!=null){
echo“”;//之前有另一个用户,因此结束该行
}
$row\U color=($row\U count++%2==0?$color1:$color2);
回声';
回显“.$username.”;
$current_username=$username;
}    
$week=$row[“赢”];
回音“.$week.”;
}
回声';

您必须更新代码。您的查询正确,但获取数据的方式错误。您必须尝试使用关联数组来解决问题。请尝试以下代码-

$sql_events = mysql_query("SELECT * FROM weekpicks ORDER BY 'username' asc ") or die (mysql_error()); 

while ($row = mysql_fetch_array($sql_events)) {
    $username[$row["username"]][] = $row["win_loose"];
}

foreach($username as $key=>$val){
    echo '<td style="width: 100" align="center"><font size="2">'.$username.'</td>';
    foreach($val as $value){
        echo '<td style="width: 50" align="center"><font size="2">'.$value.'</td>';
    }
}
$sql\u events=mysql\u query(“按'username'asc'从周picks订单中选择*)或die(mysql\u error());
while($row=mysql\u fetch\u数组($sql\u事件)){
$username[$row[“username”][]=$row[“win_lose”];
}
foreach($key=>$val的用户名){
回显“.$username.”;
foreach($val作为$value){
回显“.$value.”;
}
}
还有一件事要提的是,在每个表列上使用内联css不是一个好的实践。您可以使用一个类,并将css添加到附加到该类的css文件中


希望它能帮助…:)

您必须更新代码。您的查询正确,但获取数据的方式错误。您必须尝试关联数组来解决问题。请尝试以下代码-

$sql_events = mysql_query("SELECT * FROM weekpicks ORDER BY 'username' asc ") or die (mysql_error()); 

while ($row = mysql_fetch_array($sql_events)) {
    $username[$row["username"]][] = $row["win_loose"];
}

foreach($username as $key=>$val){
    echo '<td style="width: 100" align="center"><font size="2">'.$username.'</td>';
    foreach($val as $value){
        echo '<td style="width: 50" align="center"><font size="2">'.$value.'</td>';
    }
}
$sql\u events=mysql\u query(“按'username'asc'从周picks订单中选择*)或die(mysql\u error());
while($row=mysql\u fetch\u数组($sql\u事件)){
$username[$row[“username”][]=$row[“win_lose”];
}
foreach($key=>$val的用户名){
回显“.$username.”;
foreach($val作为$value){
回显“.$value.”;
}
}
还有一件事要提的是,在每个表列上使用内联css不是一个好的实践。您可以使用一个类,并将css添加到附加到该类的css文件中


希望它能帮助…:)

我想你也需要从引号中获取
'username'
,否则你在所有东西上都是按相同的字符串排序的。我想你也需要从引号中获取
'username'
,否则你在所有东西上都是按相同的字符串排序的。