需要将变量 = 设置为 php/html 中的下拉选择
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Need to set variable = to dropdown selection in php/html
提问by bneigher
I am new to php and html, and I am required to make a drop down list along with other inputs to create parameters which will be used in a perl script.
我是 php 和 html 的新手，我需要制作一个下拉列表以及其他输入来创建将在 perl 脚本中使用的参数。
I need the drop down to upon submission set a variable = to what the user selected.
我需要在提交时下拉设置一个变量 = 到用户选择的内容。
My code so far:
到目前为止我的代码：
<HTML>
<?php
$value0 = "";
$value1 = "Banana";
$value2 = "Apple";
$value3 = "Orange";
$submittedValue;
?>
<select project="FruitList" id="FruitList">
<option value = <?php echo $value0; ?>><?php echo $value0; ?></option>
<option value = <?php echo $value1; ?>><?php echo $value1; ?></option>
<option value = <?php echo $value2; ?>><?php echo $value2; ?></option>
<option value = <?php echo $value3; ?>><?php echo $value3; ?></option>
<input type="submit" name="submit" id="submit" value="Submit" />
</form>
</HTML>
I am not sure how to use the submit button so that I can set the $submittedValue variable = to the selected value in the dropdown. HELP!!
我不确定如何使用提交按钮，以便我可以将 $submittedValue 变量 = 设置为下拉列表中的选定值。帮助！！
回答by Hüseyin BABAL
You can check submitted value with your existing options like below;
您可以使用现有选项检查提交的值，如下所示；
<HTML>
<?php
$submittedValue = "";
$value0 = "";
$value1 = "Banana";
$value2 = "Apple";
$value3 = "Orange";
if (isset($_POST["FruitList"])) {
$submittedValue = $_POST["FruitList"];
}
?>
<form action="" name="fruits" method="post">
<select project="FruitList" id="FruitList" name="FruitList">
<option value = "<?php echo $value0; ?>"<?php echo ($value0 == $submittedValue)?" SELECTED":""?>><?php echo $value0; ?></option>
<option value = "<?php echo $value1; ?>"<?php echo ($value0 == $submittedValue)?" SELECTED":""?>><?php echo $value1; ?></option>
<option value = "<?php echo $value2; ?>"<?php echo ($value0 == $submittedValue)?" SELECTED":""?>><?php echo $value2; ?></option>
<option value = "<?php echo $value3; ?>"<?php echo ($value0 == $submittedValue)?" SELECTED":""?>><?php echo $value3; ?></option>
</select>
<input type="submit" name="submit" id="submit" value="Submit" />
</form>
</HTML>
EDIT: I have corrected your html and add some functionality to check submitted value
编辑：我已经更正了您的 html 并添加了一些功能来检查提交的值
回答by Mina Abadir
First of all you are missing the form action:
首先，您缺少表单操作：
<HTML>
<?php
$value0 = "";
$value1 = "Banana";
$value2 = "Apple";
$value3 = "Orange";
$submittedValue;
?>
<form action"formProcessing.php" method="POST">
<select project="FruitList" id="FruitList">
<option value = <?php echo $value0; ?>><?php echo $value0; ?></option>
<option value = <?php echo $value1; ?>><?php echo $value1; ?></option>
<option value = <?php echo $value2; ?>><?php echo $value2; ?></option>
<option value = <?php echo $value3; ?>><?php echo $value3; ?></option>
<input type="submit" name="submit" id="submit" value="Submit" />
</form>
</HTML>
Then you need to create a new script formProcessing.php, you can access the form variables using $_POST variable.
然后你需要创建一个新的脚本 formProcessing.php，你可以使用 $_POST 变量访问表单变量。
<?php
function filter($data) {
$data = trim(htmlentities(strip_tags($data)));
if (get_magic_quotes_gpc())
$data = stripslashes($data);
$data = mysql_real_escape_string($data);
return $data;
}
foreach($_POST as $key => $value) {
$mydata[$key] = filter($value);
}
echo $mydata['FruitList'];
?>
The filter method is used to escape the $_POST variable, in order to protect your Application from MySQL and XSS attacks. This should create a new copy of your $_POST variable however accessible through $mydata array.
filter 方法用于转义 $_POST 变量，以保护您的应用程序免受 MySQL 和 XSS 攻击。这应该创建 $_POST 变量的新副本，但可以通过 $mydata 数组访问。
