Question

我正在通过php打印棒球队阵容。我想为丢失的玩家6（或任何缺失的位置）打印占位符

所以如果玩家1 - ＆gt;播放器5是OK打印，NO播放器＃6打印占位符，播放器7 - ＆gt;播放器9可以打印。我试图简化代码。我已经尝试过解决这个问题但我一直陷入困境。

CODE:

$rot        = array();
$pos        = array();
$jn             = array();

$x = 1;
// loads up the arrays from the db
while ( $rot[$x], $pos[$x], $jn[$x])= $r->fetch_row() ) {
$x++;
}

// counts the actual number of players in linuep
// used for validation and error display
$num_players = mysqli_num_rows($r);

// controls the lineup position
for ($i = 1; $i <= 15; $i++){
    if($rot[$i] == $i) {
    //prints player             
$lineup .= "<div data-fp='" . $pos[$i] . "'>" .$jn[$i]. "</div>";
    } else {
        // prints place holder
        $text = "This Position needs to be filled before the next game.";
        $lineup .= "<div id='pid' data-rel='".$text."' data-fp='' data-pid='' data-jn='' title=''>x</div>";
    }
}

我也试过这个迭代数组rot []找到匹配的位置并打印出来但实际上它会重复打印持有者。

// controls the lineup position
for ($x = 1; $x <= 15; $x++){

for ($i = 1; $i <= ($num_players+1); $i++) {
  if ($x == $i) {  
       //prints player              
        $lineup .= "<div data-fp='" . $pos[$i] . "'>" .$jn[$i]. "</div>";
    } else {
        // prints place holder
        $text = "This Position needs to be filled before the next game.";
        $lineup .= "<div id='pid' data-rel='".$text."' data-fp='' data-pid='' data-jn='' title=''>x</div>";
    }
  }
}

Answer 1

怎么样：

# index all players by position while taking them from the database

$players = array();
while ( $row = $r->fetch_row() ) {
    list($rot, $pos, $jn) = $row;
    $players[$pos] = compact(array('rot', $pos, $jn);
}

...

# line-up players
for ($pos = 1; $pos <= 15; $pos++)
{
    $playerExists = isset($players[$pos]);

    if ($playerExists)
    {
       # do this ...
    }
    else
    {
       # do that ...
    }
}

Answer 2

我认为你正在创建一个数组，其中所有数字元素都被填充（即你总是有1到15）并且你的错误在于 if（$ rot [$ i] == $ i）{

从数据库填充数组时，添加以下行： $playertoid = array_flip($pos); # pos is the player number array?

即。 while ( ($rot[$x], $pos[$x], $jn[$x])= $r->fetch_row() ) { $x++; } $playertoid = array_flip($pos);

现在您已经创建了一个反向查找表，其中索引是玩家编号。

替换 if($rot[$i] == $i) { 符合： if (isset($playertoid[$i])) {

如果条件存在则打印阵列

2 个答案: