Question

我对准备好的陈述很陌生，我不确定我是否正确这样做。

以下是我的尝试：

$currgame = 310791;

$sql = "SELECT fk_player_id, player_tiles, player_draws, player_turn, player_passes, swapped FROM ".$prefix."_gameplayer WHERE fk_game_id = ?";
$stmt = $mysqli->stmt_init();

$data = array();
if($stmt->prepare($sql)){
    $stmt->bind_param('i', $currgame);
    $stmt->execute();

    $fk_player_id = null; $player_tiles = null; $player_draws = null; $player_turn = null; $player_passes = null; $swapped = null;
    $stmt->bind_result($fk_player_id, $player_tiles, $player_draws, $player_turn, $player_passes, $swapped);

    $res = $stmt->get_result();

    while ($row = $res->fetch_assoc()){
        $data[] = $row;
    }
    $stmt->close(); 
}

// to display own games
foreach ($data as $row) {
    if ($row['fk_player_id'] == $playerid) {

        $udraws = $row['player_draws']+1; 
        $upass = $row['player_passes'];
        $uswaps = $row['swapped'];

        echo 'uDraws: '.$udraws.'<br>';
        echo 'uPass: '.$upass.'<br>';
        echo 'uSwaps: '.$uswaps.'<br><br>';
    }
}
// to display other games
foreach ($data as $row) {
    if ($row['fk_player_id'] != $playerid) {

        $opponent = $row['fk_player_id'];
        $oppTiles = $row['player_tiles'];

        $odraws = $row['player_draws']+1;
        $opass = $row['player_passes'];
        $oswaps = $row['swapped'];

        echo 'oID: '.$opponent.'<br>';
        echo 'oTiles: '.$oppTiles.'<br>';

        echo 'oDraws: '.$odraws.'<br>';
        echo 'oPass: '.$opass.'<br>';
        echo 'oSwaps: '.$oswaps.'<br><br>';

    }
}

尝试运行时遇到“ServerError”：它是“$ res = $ stmt-＆gt; get_result（）;”这会导致错误，但不确定原因。请帮忙。

提前致谢： - ）

####编辑

$sql = "SELECT fk_player_id, player_tiles, player_draws, player_turn, player_passes, swapped FROM ".$prefix."_gameplayer WHERE fk_game_id = ?";
$stmt = $mysqli->stmt_init();

$data = array();
if($stmt->prepare($sql)){
    $stmt->bind_param('i', $currgame);
    $stmt->execute();

    $fk_player_id = null; $player_tiles = null; $player_draws = null; $player_turn = null; $player_passes = null; $swapped = null;
    $stmt->bind_result($fk_player_id, $player_tiles, $player_draws, $player_turn, $player_passes, $swapped);

    while ($row = $stmt->fetch()){
        $data[] = $row;
    }
    $stmt->close(); 
}

echo '<pre>';
print_r($data);
echo '</pre>';

Answer 1

根据您的PHP / MySQL设置，您可能无法使用get_result（）。

解决这个问题的方法是绑定结果。

例如：

$stmt->execute();

$fk_player_id = null; $player_tiles = null; $player_draws = null; $player_turn = null; $player_passes = null; $swapped = null;

$stmt->bind_result($fk_player_id, $player_tiles, $player_draws, $player_turn, $player_passes, $swapped);

while ($stmt->fetch()) { // For each row
    /* You can then use the variables declared above, which will have the 
    new values from the query every time $stmt->execute() is ran.*/
}

For more information click here

Answer 2

由于我在您的代码中没有看到它，请确保在尝试查询之前实例化mysqli对象：

$mysqli = new mysqli("127.0.0.1", "user", "password", "mydb"); 
if($mysqli->connect_error){
    die("$mysqli->connect_errno: $mysqli->connect_error");
}

此外，ServerError肯定会显示在您的日志中并指向正确的方向。

Answer 3

while (mysqli_stmt_fetch($stmt)) {
        printf ("%s (%s)\n", $name, $code);
    }

这可能会对您有所帮助：

http://php.net/manual/en/mysqli-stmt.fetch.php

使用预准备语句获取多行

3 个答案: