Question

我正在尝试将用户输入从下拉框插入到我的数据库中。我使用此代码表单从数据库动态生成选项。并且它继续播放器1到播放器11，但是所有都使用相同的代码，因此我将仅为了空间的利益包含一个播放器1的示例：

<form action="teamCreate.php" onsubmit="return validateForm();" name="teamCreation" method="post">                      
<table>
    <tr>
        <td>
            Player 1:
        </td>
        <td>
            <select id="player1" name="player1">'; 
            $sql = mysql_query("SELECT playerName FROM Player");
            while ($row = mysql_fetch_array($sql))
            {
                echo "<option value=\"player\">" . $row['playerName'] . "</option>";
            }
            echo '</select>
        </td>
    </tr>
</table>

这就是我在teamCreate.php中用它做的事情：

$player=array(); //Creates an array of players, using the values posted from editTeam.php
$player[1]=$_POST['player1'];
$player[2]=$_POST['player2'];
$player[3]=$_POST['player3'];
$player[4]=$_POST['player4'];
$player[5]=$_POST['player5'];
$player[6]=$_POST['player6'];
$player[7]=$_POST['player7'];
$player[8]=$_POST['player8'];
$player[9]=$_POST['player9'];
$player[10]=$_POST['player10'];
$player[11]=$_POST['player11'];

$teamName = $_SESSION['teamName'];
$counter=1;
while ($counter < 12) //Creates a loop that goes through the player array, adding them to the database under the user's team name
        {
            $currentPlayer = $player['$counter'];
            mysql_query("INSERT INTO PlayerListing(teamName, playerName)VALUES('$teamName', '$currentPlayer')");
            $counter = $counter + 1;
        }

当我查看数据库中的结果时，为所有添加的行正确设置了teamName列，但是playerName列为空。几个小时我一直在困惑，我不明白问题是什么。

谢谢，沃尔夫。

Answer 1

看一下你的代码块：

while ($counter < 12) //Creates a loop that goes through the player array, adding them to the database under the user's team name
{
     $currentPlayer = $player['$counter'];
     mysql_query("INSERT INTO PlayerListing(teamName, playerName)VALUES('$teamName', '$currentPlayer')");
     $counter = $counter + 1;
}

当您分配$currentPlayer时，您使用单引号并在其中包含变量名称。看起来你想在那里看$counter的值，但你不能用单引号这样做，只有双引号会为你插入变量值。要解决此问题，您必须完全删除引号（这是我建议的）或用双引号括起该变量。

while ($counter < 12) //Creates a loop that goes through the player array, adding them to the database under the user's team name
{
     $currentPlayer = $player[$counter];
     mysql_query("INSERT INTO PlayerListing(teamName, playerName)VALUES('$teamName', '$currentPlayer')");
     $counter = $counter + 1;
}

问题是你的$counter索引（字面意思，不作为变量）不包含任何值，因此没有值插入到数据库中。

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