我试图从表单的下拉列表中获取值并将此值打印到屏幕。从设置为"自身"的表单操作。在 MySQL 语句中使用此变量
//load drop downs
$result = mysqli_query($db, "SELECT * FROM actor_id");
//loads actor_name to drop down
print("<select name='"actor_name'">'n");
while($row = mysqli_fetch_array($result))
{
print ("<option value=" . $row[0] . ">" . $row[1] . "</option>");
$row = mysql_fetch_row($result);
}
print("</select>");
//close DB
mysqli_close($db);
?>
<form method = "post" action="actors.php" >
<br />
<input type="hidden" name="stage" value="1" />
<input type="submit" name="submit" value="Search" >
</form>
//after submit reloads this
if(isset($_POST['actor_name']))
{
$star = $_POST['actor_name'];
print "Star: " . $star;
}//want to use this variable in a mysql statement
我想你想展示POST演员在这次尝试的drop down list中被选中，
你必须add actor list form element喜欢
<form method = "post" action="actors.php" >
<?php
print("<select name='"actor_name'">'n");
while($row = mysqli_fetch_array($result))
{
$sel='';
if($row[0]==$_POST['actor_name'])
$sel='selected="selected"';
print ("<option value=" . $row[0] . " ".$sel.">" . $row[1] . "</option>");
//$row = mysql_fetch_row($result); remove this line it has no meaning
}
print("</select>");
?>
..........