php - 一个页面上的多个 MySQL/PHP 查询，不公平!

Question

我有以下 MySQL 查询:

<div class="box" id="settlement1">
        <?
            $query  = "SELECT name, pub, hospital, trade FROM settlements WHERE settlementID = 1";
            $result = mysql_query($query);

            while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
                $settlementName = $row['name'];
                $pubExists = $row['pub'];
                $hospitalExists = $row['hospital'];
                $tradeExists = $row['trade'];
            }
        ?>
        <div class="settlement-name"><? echo $settlementName; ?></div>
        <div class="settlement-contents">
            <?
                if ($hospitalExists == '1') { echo "<div class='hospital'>HOSPITAL</div>";  }
                if ($pubExists == '1') { echo "<div class='pub'>PUB</div>"; }
                if ($tradeExists == '1') { echo "<div class='trade'>TRADE</div>"; }
            ?>
        </div>
    </div>
    <div class="box" id="settlement2">
        <?
            $query  = "SELECT name, shop, hospital, trade FROM settlements WHERE settlementID = 2";
            $result = mysql_query($query);

            while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
                $settlementName = $row['name'];
                $pubExists = $row['pub'];
                $hospitalExists = $row['hospital'];
                $tradeExists = $row['trade'];
            }
        ?>
        <div class="settlement-name"><? echo $settlementName; ?></div>
        <div class="settlement-contents">
            <?
                if ($hospitalExists == '1') { echo "<div class='hospital'>HOSPITAL</div>";  }
                if ($pubExists == '1') { echo "<div class='pub'>PUB</div>"; }
                if ($tradeExists == '1') { echo "<div class='trade'>TRADE</div>"; }
            ?>
        </div>
    </div>

现在，第一个查询工作正常，但第二个查询返回与第一个相同的结果。

查询 1 到查询 2 中的所有字段都不同

^_^ <(求助!)

Answer 1

我不确定这是否是原因，但是......

第一个SQL:

$query  = "SELECT name, pub, hospital, trade FROM settlements WHERE settlementID = 1";

第二个:

$query  = "SELECT name, shop, hospital, trade FROM settlements WHERE settlementID = 2";

第一个是酒吧，第二个是商店。但是在第二种情况下，您不在 PHP 中使用 shop 值，仅使用 pub(不在此 SQL 字段中。

这里可能有 SQL 错误(数据库中没有字段)或 PHP 错误(检查结果数组中不存在的字段)。