Question

我试图在PHP中的URL中传递多个变量以获取一些信息，但我认为它不起作用。

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

);


$IDNUM = $_GET[ 'idNum' ];


$functionName = $_GET[ 'func' ];

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

然后我有了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....

    ';
 }

我使用的网址是products.php?func=refreshCurrentProduct&idNum=4

理论上，它应该显示在其中包含4的行中，但它只显示第一行的信息。如果我在函数中执行$IDNUM=5，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而无需在$ DB_STOCK中拥有如此多的代码？似乎必须有更好的方式...

Answer 1

看看call_user_func。

$functionName = $_GET[ 'func' ];

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

此外，如果我正确地阅读您的代码，您可以在一个查询中获取所有信息：

$query = "SELECT `ID`,`STOCK`,`MYNAME`,`DESCRIPTION`,`PRICE`,`SHIPPING` FROM `PRODUCTS`";
$result = mysql_query($query) or die('Query failed:'.mysql_error());
while ($row = mysql_fetch_assoc($result)) {
    $ID=$row['ID'];
    //etc.
}

Answer 2

为什么不这样做（正如其他人已经提到的那样，$IDNUM不在函数范围内）：

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!";
    }
}

无需打六次数据库！当然，您需要添加错误处理。

顺便说一下。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 ....
}

Answer 3

您的$IDNUM变量超出了您的功能范围。你需要将它作为一个变量传递给你的函数，或者你应该能够通过设置它来在函数中设置它。

function refreshCurrentProduct() { 
  $IDNUM = $_GET[ 'idNum' ];
  ...
}

如何通过php中的URL传递多个变量？

3 个答案: