<分区>
我在表单中有一个隐藏的输入。当我提交表单时,我想在名为“answers”的表中创建一个新列,其标题基于隐藏输入的值。
表单代码 + 隐藏输入
<form action="frameworkplayground.php" method="POST">
<input type="hidden" name="LevelColumnAdder" value="Simplifying Fractions">
<input type="submit" id="samplesubmitbutton" value="Click Me">
</form>
获取隐藏输入值的代码(名为“LevelColumnAdder”)将“测试”一词添加到该值并将其用作新列的标题。
<?php
if(isset($_POST['LevelColumnAdder'])){
$LevelColumnAdder=$_POST['LevelColumnAdder']; //Here, I'm trying to get the value of the input named LevelColumnAdder
$db->query("ALTER TABLE answers ADD $LevelColumnAdder+"Test" VARCHAR( 255 ) NOT NULL"); //I know the +"Test" part is wrong but I don't know how to add "Test" to the value and use it as the new title
}
?>
提交此表单后,我最终希望它形成一个名为“Simplifying Fractions Test”的新列,但什么也没有发生。