duanpa5237 2015-04-03 20:13
浏览 49

来自动作值的php从下拉列表中获取

My Goal is get two drop down values and 1. pass them to action and 2. pass them to $_POST to next .php page. In short i want to select action php page using drop-down menu.

What i am doing wrong here, whatever i select from page drop-down menu it always default select page1.php

<html>
<body>

<select name='page'>
<option value='page1.php'>page1</option>
<option value='page2.php'>page2</option>
</select>

<form name="page" method="post" action="<?php echo $_POST['page']; ?>">
<?php
// DB connection info:
include 'db_connection.php';

$sql = "SHOW tables";
$result = mysql_query($sql);

echo "<select name='param1'>";
while ($row = mysql_fetch_array($result)) {
    echo "<option value='" . $row['Tables_in_radius'] . "'>" . $row['Tables_in_radius'] . "</option>";
}
echo "</select>";

?>

<input type="submit" value="send">
</form>
<?php
if(isset($_POST['param1'])){
}
?>

</body>
</html>
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1条回答 默认 最新

  • dougou5844 2015-04-03 22:10
    关注

    1) Place the drop down inside the form.

    2) if(isset($_POST['param1'])) <-- What is param1 ? change it to : $_POST['page']

    评论

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