[英]Populating Drop-Down in PHP using Array
I am trying to populate a drop-down menu with values from an array. 我正在尝试使用数组中的值填充下拉菜单。
I have tried to follow other answers but the syntax doesn't seem to be working. 我尝试遵循其他答案,但是语法似乎不起作用。 (I'm still relatively new to PHP).
(我对PHP还是比较陌生的)。
The following code which I am working on was produced by someone else. 我正在处理的以下代码是由其他人生成的。
$sqlite_query = "SELECT * FROM dis_kind";
$result = $db->query($sqlite_query);
$array = $result->fetchArray();
$output = "<select name=\"kind\" class=\"dis\" >\n";
$output .= "<option value=\"$this->wildcard_value\"></option>\n";
foreach ($result as $array) {
$value = $array['kind'];
$output .= "<option value=\"";
$output .= $value;
$output .= "\">";
$output .= $value;
$output .= " - ";
$output .= $array['description'];
$output .= "</option>\n";
}
$output .= "</select>\n";
I don't know why it has been done the way it has but I am stumped as to getting my drop-down to work. 我不知道为什么要按原样进行,但是我对下拉菜单的使用感到沮丧。
Currently, the box appears but is populated with no values. 当前,该框出现,但没有任何值。
Thanks. 谢谢。
Eventually managed to solve the problem myself - before I pulled my hair out! 最终我自己解决了问题-在拔出头发之前!
Instead of using the foreach
loop, I used the following: 我没有使用
foreach
循环,而是使用了以下内容:
while($array = $result->fetchArray())
{
// Output code.
}
Which populated the drop-down with values from the array. 其中使用数组中的值填充了下拉菜单。
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