php - 如何通过php中的URL传递多个变量？

我试图在PHP的URL中传递多个变量来获取一些信息，但我认为这不起作用。

$allowedFunctions = array(
   'returnAllProducts',
   'refreshCurrentProduct'

);


$IDNUM = $_GET[ 'idNum' ];


$functionName = $_GET[ 'func' ];

if( in_array( $functionName, $allowedFunctions ) && function_exists( $functionName ) )
{
    $functionName();
}

然后我有refreshCurrentProduct函数：

function refreshCurrentProduct() {
$dbh=mysql_connect ("DATABASE","USER", "PASS") or die('I cannot connect to the database because:'. mysql_error());

mysql_select_db("TABLE");

$query = "SELECT `ID` FROM `PRODUCTS`";

$result = mysql_query($query) or die('Query failed:'.mysql_error());

$DB_STOCK = mysql_query("SELECT `STOCK` FROM `PRODUCTS`") or die('Query failed:'.mysql_error());

$DB_SHORT = mysql_query("SELECT `MYNAME` FROM `PRODUCTS`") or die('Query failed:'.mysql_error());

$DB_LONG = mysql_query("SELECT `DESCRIPTION` FROM `PRODUCTS`") or die('Query failed:'.mysql_error());

$DB_PRICE = mysql_query("SELECT `PRICE` FROM `PRODUCTS`") or die('Query failed:'.mysql_error());

$DB_SHIP = mysql_query("SELECT `SHIPPING` FROM `PRODUCTS`") or die('Query failed:'.mysql_error());


$ID = mysql_result($result,$IDNUM,"ID");
    $STOCK = mysql_result($DB_STOCK,$IDNUM,"STOCK");
    $SHORT = mysql_result($DB_SHORT,$IDNUM,"MYNAME");
    $LONG = mysql_result($DB_LONG,$IDNUM,"DESCRIPTION");
    $PRICE = mysql_result($DB_PRICE,$IDNUM,"PRICE");
    $SHIP = mysql_result($DB_SHIP,$IDNUM,"SHIPPING");

    echo '
    //echo $STOCK, $SHORT, etc....

    ';
 }

我使用的URL是products.php?func=refreshCurrentProduct&idNum=4
理论上，它应该从包含4的行中显示，但是，它只显示第一行中的信息。如果我在函数中执行$IDNUM=5操作，它将显示第5行，因此我传递信息的方式有问题。
另外，如何创建（例如）$STOCK而不必在$DB_STOCK中包含这么多代码？好像有更好的办法。。。

最佳答案

为什么不这样做（正如其他人已经提到的，$IDNUM不在函数的范围内）：

function refreshCurrentProduct() {
    $dbh=mysql_connect ("DATABASE","USER", "PASS") or die('I cannot connect to the database because:'. mysql_error());
    mysql_select_db("TABLE");

    // If $_GET['idNum'] is not a number use 0
    $rowNumber = is_numeric($_GET['idNum']) ? $_GET['idNum'] : 0;

    $query = "SELECT ID, STOCK,  MYNAME, DESCRIPTION, PRICE, SHIPPING FROM `PRODUCTS`";
    $result = mysql_query($query);

    if(mysql_data_seek($result,  $rowNumber)) {
        // The result set has indeed at least $rowNumber rows

        $row = mysql_fetch_assoc($result);

        echo $row['ID'];
        echo $row['STOCK'];
        // ... etc ....
    }
    else {
        echo "No such row!";
    }
}

不需要打六次数据库！当然，您需要添加错误处理。
顺便问一下，参数idNum是否与数据库中记录的ID相同？如果是这样，您甚至可以进一步简化：

function refreshCurrentProduct() {
    $dbh=mysql_connect ("DATABASE","USER", "PASS") or die('I cannot connect to the database because:'. mysql_error());
    mysql_select_db("TABLE");

    // If $_GET['idNum'] is not a number use 0
    $id = is_numeric($_GET['idNum']) ? $_GET['idNum'] : 0;

    $query = "SELECT ID, STOCK,  MYNAME, DESCRIPTION, PRICE, SHIPPING FROM `PRODUCTS` WHERE ID = $id";
    $result = mysql_query($query);

    if (mysql_num_rows($result) == 0) {
       echo "No rows found, nothing to print";
       return;
    }

    $row = mysql_fetch_assoc($result);

    echo $row['ID'];
    echo $row['STOCK'];
    // ... etc ....
}

idnum

php - 如何通过php中的URL传递多个变量？