如何将数据从一个php数据库填充下拉到另一个脚本？

Question

我创建了一个PHP下拉菜单，它是从MySql数据库中填充的，工作正常，当我想要在另一个脚本中发布所选内容时，出现了问题。问题是如何将数据发布到另一个脚本？这是实现下拉菜单的脚本的源代码。请帮帮我！

<?php
$conn = mysql_connect("localhost", "admin", "admin");

if (!$conn) {
    echo "Unable to connect to DB: " . mysql_error();
    exit;
}

if (!mysql_select_db("ekupuvac")) {
    echo "Unable to select EKupuvac: " . mysql_error();
    exit;
}

$query = "SELECT ImeK, KupuvacID FROM kupuvac ORDER BY Saldo DESC";
$result = mysql_query($query) or die(mysql_error());

if (!$result) {
    echo "Could not successfully run query ($query) from DB: " . mysql_error();
    exit;
}

if (mysql_num_rows($result) == 0) {
    echo "No rows found, nothing to print so I am exiting";
    exit;
}

$dropdown = "<select name='ImeK'>";
while($row = mysql_fetch_assoc($result)) {
$dropdown.= "\r\n<option value='{$row['KupuvacID']}'>{$row['ImeK']}</option>";
}
$dropdown .= "\r\n</select>";

echo"Izberi Kupuvac:";
echo $dropdown;

// Second Combo

$conn = mysql_connect("localhost", "admin", "admin");

if (!$conn) {
    echo "Unable to connect to DB: " . mysql_error();
    exit;
}

if (!mysql_select_db("ekupuvac")) {
    echo "Unable to select EKupuvac: " . mysql_error();
    exit;
}

$query2 = "SELECT ImeP, ProzivodID FROM proizvod ORDER BY ImeP";
$result2 = mysql_query($query2) or die(mysql_error());

if (!$result2) {
    echo "Could not successfully run query ($query2) from DB: " . mysql_error();
    exit;
}

if (mysql_num_rows($result2) == 0) {
    echo "No rows found, nothing to print so I am exiting";
    exit;
}

$dropdown2 = "<select name='ImeP'>";
while($row = mysql_fetch_assoc($result2)) {
$dropdown2.= "\r\n<option value='{$row['ProzivodID']}'>{$row['ImeP']}</option>";
}
$dropdown2.= "\r\n</select>";

echo"<br> Izberi Proizvod:";
echo $dropdown2;
echo"<br>";

mysql_free_result($result);
?>

Answer 1

一个<select>框是不够的，您需要将它封装在一个表单中

?>
<form method="post" action="somescript.php">
<?
//your controls go here
?>
</form>

然后使用$_POST创建somescript.php并访问表单变量

还要使用PDO而不是mysql_函数，因为这些函数不安全